linear algebra example problems with solutions

Linear Algebra - Questions with Solutions

Linear algebra questions with solutions and detailed explanations on matrices , spaces, subspaces and vectors , determinants , systems of linear equations and online linear algebra calculators are included.

• Matrices with Examples and Questions with Solutions .
• Transpose of a Matrix .
• Symmetric Matrix .
• Identity Matrix .
• Diagonal Matrices .
• Triangular Matrix .
• Diagonalization of Matrices .
• Inverse Matrix Questions with Solutions .
• Find Matrix Inverse Using Row Operations
• Add, Subtract and Scalar Multiply Matrices .
• Multiplication and Power of Matrices
• Eigenvalues and Eigenvectors Questions with Solutions
• Find Eigenvectors and Eigenvalues of a 3 by 3 Matrix on Video (Video)
• Find Eigenvectors and Eigenvalues of a 2 by 2 Matrix on Video (Video)
• Row Operations and Elementary Matrices .
• The Three Row Operations on Augmented Matrices .
• Write a Matrix in Reduced Row Echelon Form .
• Pivots of a Matrix in Row Echelon Form - Examples with Solutions .
• Null Space and Nullity of a Matrix .
• Column and Row Spaces and Rank of a Matrix .
• Free and Basic Variables of a Matrix - Examples with Solutions .
• Orthogonal Matrices - Examples with Solutions .
• The QR Decomposition of a Matrix .
• LU Decomposition of a Matrix .
• Properties of Matrix Operations .

Least Squares Problems

• Solve Least Squares Problems by the Normal Equations .
• Solve Least Squares Problems by the QR Decomposition .

Spaces, Subspaces and Vectors

• Vector Spaces - Examples with Solutions .
• Subspaces - Examples with Solutions .
• Vectors in ℝ n .
• Inner Product, Orthogonality and Length of Vectors .
• Orthogonal Vectors - Examples with Solutions .
• Linear Combinations of Vectors .
• Span of Vectors .
• Linearly Independent and Dependent Vectors - Examples with Solutions .
• Testing for Linearity of Vectors in a Subspace - Examples with Solutions .
• Basis, Coordinates and Dimension of Vector Spaces .
• Change of Basis - Examples with Solutions .
• Orthonormal Basis - Examples with Solutions .
• The Gram Schmidt Process for Orthonormal Basis . Examples with Solutions

determinants

• Determinant of a Square Matrix .
• Find Determinant Using Row Reduction .

Systems of Linear Equations

• The Three Elementary Operations on Systems .
• Gaussian Elimination to Solve Systems - Questions with Solutions .
• The Elimination Method in Systems - Questions with Solutions .
• Cramer's Rule with Questions and Solutions .

Videos on Linear Algebra

• Find Eigevectors and Eigenvalues of a 2 by 2 Matrix .
• Solve a 2 by 2 System of Equations by Elimination .
• Gaussian Elimination to Solve a 3 by 3 System of Equations .
• Inverse of 3 by 3 Matrix Using Gauss-Jordan .

Linear Algebra Calculators

• Find the Inverse of a Matrix Using Row Reduction .
• Multiply Matrix Calculator .
• Add Matrices Calculator .
• Row Echelon Form Calculator .
• Row Reduce Agmented Matrices - Calculator .
• Online Calculator for The Determinant of a Matrix of Any Size .
• QR Decomposition of Matrices Calculator .
• Online LU Decomposition of a Matrix Calculator .
• Systems of Equations with Complex Coefficients Solver .
• Matrix Calculator Multiplication ; step by step calculator for educational purposes.

Browse Course Material

Course info.

• Prof. Gilbert Strang

Departments

• Mathematics

As Taught In

• Linear Algebra

Learning Resource Types

Problem sets with solutions.

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Problems in Mathematics

Category: Linear Algebra

Linear Algebra Problems and Solutions.

Popular topics in Linear Algebra are

• Vector Space
• Linear Transformation
• Diagonalization

Gauss-Jordan Elimination   Inverse Matrix     Eigen Value   Caley-Hamilton Theorem   Caley-Hamilton Theorem

Linear Algebra

 by Yu · Published 04/30/2018

If the Nullity of a Linear Transformation is Zero, then Linearly Independent Vectors are Mapped to Linearly Independent Vectors

Problem 722.

Let $T: \R^n \to \R^m$ be a linear transformation. Suppose that the nullity of $T$ is zero.

If $\{\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k\}$ is a linearly independent subset of $\R^n$, then show that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.

  Read solution

 by Yu · Published 04/22/2018

Find All Values of $x$ such that the Matrix is Invertible

Problem 721.

 by Yu · Published 04/15/2018

Find All Eigenvalues and Corresponding Eigenvectors for the $3\times 3$ matrix

Problem 720, find all values of $a$ which will guarantee that $a$ has eigenvalues 0, 3, and -3., problem 719.

Let $A$ be the matrix given by \[ A= \begin{bmatrix} -2 & 0 & 1 \\ -5 & 3 & a \\ 4 & -2 & -1 \end{bmatrix} \] for some variable $a$. Find all values of $a$ which will guarantee that $A$ has eigenvalues $0$, $3$, and $-3$.

 by Yu · Published 04/03/2018

Compute the Determinant of a Magic Square

Problem 718.

Let \[ A= \begin{bmatrix} 8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2 \end{bmatrix} . \] Notice that $A$ contains every integer from $1$ to $9$ and that the sums of each row, column, and diagonal of $A$ are equal. Such a grid is sometimes called a magic square.

Compute the determinant of $A$.

 by Yu · Published 03/25/2018

Are These Linear Transformations?

Problem 717.

Define two functions $T:\R^{2}\to\R^{2}$ and $S:\R^{2}\to\R^{2}$ by \[ T\left( \begin{bmatrix} x \\ y \end{bmatrix} \right) = \begin{bmatrix} 2x+y \\ 0 \end{bmatrix} ,\; S\left( \begin{bmatrix} x \\ y \end{bmatrix} \right) = \begin{bmatrix} x+y \\ xy \end{bmatrix} . \] Determine whether $T$, $S$, and the composite $S\circ T$ are linear transformations.

Using Gram-Schmidt Orthogonalization, Find an Orthogonal Basis for the Span

Problem 716.

 by Yu · Published 03/21/2018

Normalize Lengths to Obtain an Orthonormal Basis

Problem 715.

Let \[ \mathbf{v}_{1} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} ,\; \mathbf{v}_{2} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} . \] Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?

If not, then find an orthonormal basis for $V$.

 by Yu · Published 03/19/2018

Find a Spanning Set for the Vector Space of Skew-Symmetric Matrices

Problem 714.

Let $W$ be the set of $3\times 3$ skew-symmetric matrices. Show that $W$ is a subspace of the vector space $V$ of all $3\times 3$ matrices. Then, exhibit a spanning set for $W$.

 by Yu · Published 03/07/2018

Determine Bases for Nullspaces $\calN(A)$ and $\calN(A^{T}A)$

Problem 713.

Determine bases for $\calN(A)$ and $\calN(A^{T}A)$ when \[ A= \begin{bmatrix} 1 & 2 & 1 \\ 1 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix} . \] Then, determine the ranks and nullities of the matrices $A$ and $A^{\trans}A$.

 by Yu · Published 03/05/2018

In which $\R^k$, are the Nullspace and Range Subspaces?

Problem 712.

Let $A$ be an $m \times n$ matrix. Suppose that the nullspace of $A$ is a plane in $\R^3$ and the range is spanned by a nonzero vector $\mathbf{v}$ in $\R^5$. Determine $m$ and $n$. Also, find the rank and nullity of $A$.

 by Yu · Published 02/28/2018

Prove Vector Space Properties Using Vector Space Axioms

Problem 711.

Using the axiom of a vector space, prove the following properties. Let $V$ be a vector space over $\R$. Let $u, v, w\in V$.

(a) If $u+v=u+w$, then $v=w$.

(b) If $v+u=w+u$, then $v=w$.

(c) The zero vector $\mathbf{0}$ is unique.

(d) For each $v\in V$, the additive inverse $-v$ is unique.

(e) $0v=\mathbf{0}$ for every $v\in V$, where $0\in\R$ is the zero scalar.

(f) $a\mathbf{0}=\mathbf{0}$ for every scalar $a$.

(g) If $av=\mathbf{0}$, then $a=0$ or $v=\mathbf{0}$.

(h) $(-1)v=-v$.

The first two properties are called the cancellation law .

 by Yu · Published 02/26/2018

Find a basis for $\Span(S)$, where $S$ is a Set of Four Vectors

Problem 710.

Find a basis for $\Span(S)$ where $S= \left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} , \begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix} , \begin{bmatrix} 2 \\ 6 \\ -2 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} \right\}$.

Find a Basis for the Subspace spanned by Five Vectors

Problem 709.

Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where \[ \mathbf{v}_{1}= \begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{2}= \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix} ,\;\mathbf{v}_{3}= \begin{bmatrix} 1 \\ 5 \\ -1 \\ 5 \end{bmatrix} ,\;\mathbf{v}_{4}= \begin{bmatrix} 1 \\ 1 \\ 4 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{5}= \begin{bmatrix} 2 \\ 7 \\ 0 \\ 2 \end{bmatrix} .\] Find a basis for the span $\Span(S)$.

How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix

Problem 708.

Let $A=\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 &3 & 0 & 5 \\ 1 & 1 & 6 & 3 \end{bmatrix}$.

(a) Find a basis for the nullspace of $A$.

(b) Find a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Can We Reduce the Number of Vectors in a Spanning Set?

Problem 707.

Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^3$. Is it possible that $S_2=\{\mathbf{v}_1\}$ is a spanning set for $V$?

 by Yu · Published 02/25/2018

Does an Extra Vector Change the Span?

Problem 706.

Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set \[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\] still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample.

 by Yu · Published 02/22/2018

Vector Space of Functions from a Set to a Vector Space

Problem 705.

(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?

(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.

(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.

(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?

(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.

 by Yu · Published 02/21/2018

Find a Basis for Nullspace, Row Space, and Range of a Matrix

Problem 704.

Let $A=\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 &3 & 0 & 5 \\ 1 & 1 & 6 & 3 \end{bmatrix}$. (a) Find a basis for the nullspace of $A$.

 by Yu · Published 02/15/2018

Describe the Range of the Matrix Using the Definition of the Range

Problem 703.

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Linear Algebra Problems by Topics

The list of linear algebra problems is available here.

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• Introduction to Matrices
• Elementary Row Operations
• Gaussian-Jordan Elimination
• Solutions of Systems of Linear Equations
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• Bases and Dimension of Subspaces in $\R^n$
• General Vector Spaces
• Subspaces in General Vector Spaces
• Linearly Independency of General Vectors
• Bases and Coordinate Vectors
• Dimensions of General Vector Spaces
• Linear Transformation from $\R^n$ to $\R^m$
• Linear Transformation Between Vector Spaces
• Orthogonal Bases
• Determinants of Matrices
• Computations of Determinants
• Introduction to Eigenvalues and Eigenvectors
• Eigenvectors and Eigenspaces
• Diagonalization of Matrices
• The Cayley-Hamilton Theorem
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Linear Algebra Exercises

Do you want to engage your students more when teaching linear algebra grasple offers a selection of online exercises and openly licensed material to enhance your education..

Open Exercises

Linear systems i.

• 9 Linear System Definition and Properties
• 26 Solving Linear Systems (one solution)
• 14 Solving Linear Systems (general)

Linear Systems II

• 16 Vector Definition and Arithmetic
• 5 Vector Equations
• 15 Linear Combinations and Span
• 17 Matrix Equations
• 18 Solution Set Structure
• 35 Linear Independence

Linear Transformation

• 21 Linear Transformations Definition and Properties
• 13 Standard Matrix
• 3 Linear Transformations One-to-One and Onto

Matrix Algebra

• 11 Addition, Scalar Multiplication and Transposition
• 31 Matrix Operations
• 3 Elementary Matrices
• 14 Inverse matrices (Theory)
• 20 Inverse Matrices (Computing the Inverse)
• 4 Partitioned Matrices
• 6 LU-Factorization
• 11 Subspaces Definition and Properties
• 3 Basis (Theory)
• 9 Finding a Basis
• 6 Coordinates
• 8 Dimension and Rank

Determinants

• 14 Cofactor-Expansion
• 9 Determinants Using Row and Column Operations
• 15 Determinants: Rules of Calculation
• 5 Applications of Determinants - Area and Volume
• 8 Applications of Determinants - Cramer's Rule

Eigenvectors

• 5 Markov Chains
• 13 Definition Eigenvector and Eigenvalue
• 6 Eigenspaces
• 8 The Characteristic Equation
• 4 Similarity
• 13 Diagonalization and Diagonalizability
• 13 Complex Eigenvalues
• 9 Systems of Linear Differential Equations

Orthogonality

• 10 Inner Product
• 4 Orthogonal Projections on a Line
• 11 Orthogonal and Orthonormal Sets
• 4 Orthogonal Projections
• 4 QR Factorization
• 5 The Gram-Schmidt Process
• 2 Least-Squares Method
• 5 Regression

Symmetric Matrices

• 15 Symmetric Matrices: Definitions and Properties
• 6 Orthogonal Diagonalization
• 15 Quadratic Forms
• 6 Constrained Optimization
• 8 Singular Value Decomposition

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Linear Algebra Questions

Linear algebra questions with solutions are provided here for practice and to understand what is linear algebra and its application to solving problems. Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear functions which operate on vectors and follow vector addition.

Following are the main topics under linear algebra:

• Matrices and determinants
• Vector Spaces
• System of linear equations
• Linear transformations
• Inner product spaces
• Diagonalizations and quadratic forms

We shall practice a few problems based on these topics.

Learn more about linear algebra and its applications .

Linear Algebra Questions with Solutions

Let us solve a few questions based on linear algebra.

Question 1:

Show that the matrix A is unitary matrix

\(\begin{array}{l}A=\frac{1}{5}\begin{bmatrix}-1+2i & -4-2i \\ 2-4i& -2-i \\\end{bmatrix}\end{array} \)

A matrix is said to be unitary if and only if AA* = A*A = I, where A* is the transpose of the conjugate of A.

Transpose of A

\(\begin{array}{l}A^{T}=\frac{1}{5}\begin{bmatrix}-1+2i & 2-4i \\ -4-2i& -2-i \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}A^{*}=\overline{A^{T}}=\frac{1}{5}\begin{bmatrix}-1-2i & 2+4i \\ -4+2i& -2+i \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}AA^{*}=\frac{1}{25}\begin{bmatrix}-1+2i & -4-2i \\2-4i & -2-i \\\end{bmatrix}\begin{bmatrix}-1-2i & 2+4i \\ -4+2i& -2+i \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=\frac{1}{25}\begin{bmatrix}1+4+16+4 & 0\\0 & 4+16+4+1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}\therefore AA^{*}=\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}=I\end{array} \)

Similarly, we can show that A*A = I

Hence, A is a unitary matrix.

Also refer: Types of Matrices

Question 2:

Find the rank and the nullity of the following matrix:

\(\begin{array}{l}\begin{bmatrix}1 & -2 & -1 & 4 \\2 & -4 & 3 & 5 \\-1 & 2 & 6 & -7 \\\end{bmatrix}\end{array} \)

To find the rank and nullity of the given matrix, we transform the given matrix into a row-reduced echelon form, by performing elementary transformations.

\(\begin{array}{l}A=\begin{bmatrix}1 & -2 & -1 & 4 \\2 & -4 & 3 & 5 \\-1 & 2 & 6 & -7 \\\end{bmatrix}\end{array} \)

Applying R 2 → R 2 – 2R 1 and R 3 → R 3 + R 1

\(\begin{array}{l}A~\begin{bmatrix}1 & -2 & -1 & 4 \\0 & 0 & 5 & -3 \\0 & 0 & 5 & -3 \\\end{bmatrix}\end{array} \)

Applying C 2 → C 2 + 2C 1 , C 3 → C 3 + C 1 and C 4 → C 4 – 4C 1

\(\begin{array}{l}A~\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 0 & 5 & -3 \\0 & 0 & 5 & -3 \\\end{bmatrix}\end{array} \)

Applying R 3 → R 3 – R 2

\(\begin{array}{l}A~\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 0 & 5 & -3 \\0 & 0 & 0 & 0 \\\end{bmatrix}\end{array} \)

Applying R 2 → (⅕)R 2

\(\begin{array}{l}A~\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 0 & 1 & -3/5 \\0 & 0 & 0 & 0 \\\end{bmatrix}\end{array} \)

Applying C 4 → C 4 + (⅗)C 3

\(\begin{array}{l}A~\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 \\\end{bmatrix}\end{array} \)

∴ number of non-zero rows of the row-reduced echelon form of A = rank of A = 2

number of zero rows of the row-reduced echelon form of A = nullity of A = 2

Learn more about rank and nullity .

Question 3:

Solve the following system of linear equations:

x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

The given linear equations can be written in the form of a matrix equation AX = B, where

\(\begin{array}{l}A=\begin{bmatrix}1 & 1 & 1 \\1 & 2 & 3 \\1 & 4 & 7 \\\end{bmatrix}, X = \begin{bmatrix}x \\y \\z\end{bmatrix}\:\:and\:\:B=\begin{bmatrix}6 \\14 \\30\end{bmatrix}\end{array} \)

The augmented matrix [A| B] is-

\(\begin{array}{l}[A|B]=\begin{bmatrix}1 & 1 & 1 &|6 \\1 & 2 & 3&|14 \\1 & 4 & 7 &|30 \\\end{bmatrix}\end{array} \)

We reduce the given matrix to row echelon form by applying elementary row transformations

Applying R 2 → R 2 – R 1 , R 3 → R 3 – R 1

\(\begin{array}{l}[A|B]\sim \begin{bmatrix}1 & 1 & 1 &|6 \\0 & 1& 2&|8 \\0 & 3 & 6&|24 \\\end{bmatrix}\end{array} \)

Applying R 3 → R 3 – 3R 2

\(\begin{array}{l}[A|B]\sim \begin{bmatrix}1 & 1 & 1 &|6 \\0 & 1& 2&|8 \\0 & 0 & 0&|0 \\\end{bmatrix}\end{array} \)

Since, Rank of A = Rank of [A : B] = 2 < number of unknowns

∴ the given system of linear equations has an infinite number of solutions.

Thus, we get from the row reduced echelon form matrix

x + y + z = 6 ….(i)

⇒ y = 8 – 2z putting this value of y in (i), we get

x + 8 – 2z + z = 6

⇒ x – z = –2

⇒ x = z – 2

Now taking different values of z will give different values of the given system of equations.

• Transpose of Matrix
• Determinant of a Matrix
• Matrix Multiplication
• Matrix Operations
• Special Matrices

Question 4:

Show that the set V = {(x, y) ∈ R 2 | xy ≥ 0} is not a vector space of R 2 .

For V to be a vector space, it is required that V must be closed under addition, that is for any x and y in V, x + y ∈ V

Let ( – 1, 0) and (0, 1) ∈ V

Now, ( – 1, 0) + (0, 1) = ( –1 + 0, 0 + 1) = ( –1, 1)

But, –1 × 1 = –1 < 0 ⇒ ( –1, 1) ∉ V.

∴ V is not a vector space in R 2 .

Question 5:

Find the eigenvalues of

\(\begin{array}{l}A= \begin{bmatrix}0 & 1 & 0 \\0 & 0 & 1 \\4 & -17 & 8 \\\end{bmatrix}\end{array} \)

The characteristic polynomial is given by

\(\begin{array}{l}det(A-\lambda I) = det\begin{bmatrix}-\lambda & -1&0 \\ 0& -\lambda & 1 \\4 & -17 & 8-\lambda \\\end{bmatrix}=\lambda ^{3}-8\lambda^{2}+17\lambda-4\end{array} \)

Eigenvalues of A are the roots of the above cubic equation,

𝜆 3 – 8𝜆 2 + 17𝜆 – 4 = 0

⇒ (𝜆 – 4)(𝜆 2 – 4𝜆 + 1) = 0

Solving this we get,

𝜆 = 4, 𝜆 = 2 ±√3

These are the eigenvalues of A.

Also check: Eigenvalues and Eigenvectors

Determine whether the following vector is linearly dependent or linearly independent: (1, 2, –3, 1), (3, 7, 1, –2), (1, 3, 7, –4).

The vectors could form the column vectors of matrix A. We shall find the rank of A by reducing it to row echelon form.

\(\begin{array}{l}A=\begin{bmatrix}1 & 3 & 1 \\2 & 7 & 3 \\-3 & 1 & 7 \\1 & -2 & -4 \\\end{bmatrix}\end{array} \)

Applying R 2 → R 2 – 2R 1 , R 3 → R 3 + 3R 1 , and R 4 → R 4 – R 1

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 3 & 1 \\0 & 1 & 1 \\0 & 10 & 10 \\0 & -5 & -5 \\\end{bmatrix}\end{array} \)

Applying R 3 → R 3 – 10R 2 , R 4 → R 4 + 5R 2

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 3 & 1 \\0 & 1 & 1 \\0 & 0 & 0 \\0 & 0 & 0 \\\end{bmatrix}\end{array} \)

Clearly, rank of A = 2 < number of column vectors. So, the given vectors are linearly dependent.

Question 6:

Verify whether the polynomials x 3 – 5x 2 – 2x + 3, x 3 – 1, x 3 + 2x + 4 are linearly independent.

We may construct a matrix with coefficients of x 3 , x 2 , x, and constant terms.

\(\begin{array}{l}A=\begin{bmatrix}1 & 1 & 1 \\-5 & 0 & 0 \\-2 & 0 & 2 \\3 & -1 & 4 \\\end{bmatrix}\end{array} \)

To find the rank of A let us reduce it to row echelon form by applying elementary transformations

Applying R 2 → R 2 + 5R 1 , R 3 → R 3 + 2R 1 , and R 4 → R 4 – 3R 1

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 1 & 1 \\0 & 5 & 5 \\0 & 2 & 4 \\0 & -4 & 1 \\\end{bmatrix}\end{array} \)

Applying R 2 → (⅕) R 2

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 1 & 1 \\0 & 1 & 1 \\0 & 2 & 4 \\0 & -4 & 1 \\\end{bmatrix}\end{array} \)

Applying R 3 → R 3 – 2R 2 , R 4 → R 4 + 4R 2

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 2 \\0 & 0 & 5 \\\end{bmatrix}\end{array} \)

Applying R 4 → R 4 – (5/2)R 3

\(\begin{array}{l}A\sim \begin{bmatrix}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 2 \\0 & 0 & 0 \\\end{bmatrix}\end{array} \)

∴ rank of A = 3 = number of column vectors. So the given vectors are linearly independent.

Question 7:

Show that the following matrix is diagonalizable:

\(\begin{array}{l}A=\begin{bmatrix}1 & 0 & -1 \\1 & 2 & 1 \\2 & 2 & 3 \\\end{bmatrix}\end{array} \)

First, we shall find the eigenvalues of A. The characteristic equation of A is given by:

\(\begin{array}{l}|A-\lambda I|=\begin{vmatrix}1-\lambda & 0 & -1 \\1 & 2-\lambda & 1 \\2 & 2 & 3-\lambda \\\end{vmatrix}=0\end{array} \)

⇒ (1 – 𝜆)(2 – 𝜆)(3 – 𝜆) = 0

⇒ 𝜆 = 1, 2, 3.

The eigenvector corresponding to 𝜆 1 = 1 is the non-zero solution of the following matrix equation:

(A – 1I)X = 0

\(\begin{array}{l}\Rightarrow \begin{bmatrix}0 & 0 & -1 \\1 & 1 & 1 \\2 & 2 & 2 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

Applying elementary transformation R 3 → R 3 – 2R 2 and R 2 → R 2 + R 1 , we get

\(\begin{array}{l}\Rightarrow \begin{bmatrix}0 & 0 & -1 \\1 & 1 & 0 \\0 & 0 & 0 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-z \\x+y \\0\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

⇒ z = 0, x + y = 0

If we take x = 1 ⇒ y = –1

Hence, the corresponding eigen-vector X 1 = [1 –1 0] T .

Similarly, the eigenvector corresponding to 𝜆 = 2 is given by:

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-1 & 0 & -1 \\1 & 0 & 1 \\2 & 0 & 1 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

Applying elementary transformation R 3 → R 3 – R 2 and R 2 → R 2 + R 1 , we get

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-1 & 0 & -1 \\0 & 0 & 0 \\1 & 2 & 0 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-x-z \\0 \\x+2y\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

⇒ x + z = 0, x + 2y = 0

⇒ x = –2, y = 1 and z = 2 {taking y = 1}

Hence, the corresponding eigen-vector X 2 = [ –2 1 2] T .

Finally, the eigenvector corresponding to 𝜆 = 3 is the non-zero solution of the following matrix equation:

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-2 & 0 & -1 \\1 & -1 & 1 \\2 & 2 & 0 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

Applying elementary transformation R 2 → R 2 + R 1 and R 3 → R 3 + 2R 2 , we get

\(\begin{array}{l}\Rightarrow \begin{bmatrix}-2 & 0 & -1 \\-1 & -1 & 0 \\0 & 0 & 0 \\\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\end{array} \)

⇒ 2x + z = 0, x + y = 0

On taking x = 1, we get x = 1, y = –1 and z = –2

Hence, the corresponding eigen-vector X 3 = [ 1 –1 –2] T .

Let us construct a matrix with these eigenvectors as its column vectors, we get

\(\begin{array}{l}P=\begin{bmatrix}1 & -2 & 1 \\-1 & 1 & -1 \\0 & 2 & -2 \\\end{bmatrix}\end{array} \)

Inverse of P is

\(\begin{array}{l}P^{-1}=\frac{1}{2}\begin{bmatrix}0 & -2 & 1 \\-2 & -2 & 0 \\-2 & -2 & -1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}P^{-1}AP=\frac{1}{2}\begin{bmatrix}0 & -2 & 1 \\-2 & -2 & 0 \\-2 & -2 & -1 \\\end{bmatrix}\begin{bmatrix}1 & 0 & -1 \\1 & 2 & 1 \\2 & 2 & 3 \\\end{bmatrix}\begin{bmatrix}1 & -2 & 1 \\-1 & 1 & -1 \\0 & 2 & -2 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=\begin{bmatrix}1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 3 \\\end{bmatrix}\end{array} \)

= diag(1, 2, 3)

Thus, A is diagonalizable.

Refer: Diagonalization

Question 8:

Show that the transformation T: V 2 ( R ) → V 2 ( R ) defined by T(a, b) = (a + b, a) ∀ a, b ∈ R is a linear transformation.

To show that T is a linear transformation, we need to prove that,

For any x, y ∈ V 2 ( R )

T( x + y ) = T( x ) + T( y ) and T(a x ) = aT( x ) where a is a scalar in field.

Let (x 1 , y 1 ) and (x 2 , y 2 ) are arbitrary elements of V 2 ( R )

T[(x 1 , y 1 ) + (x 2 , y 2 )] = T[(x 1 + x 2 , y 1 + y 2 )] = (x 1 + x 2 + y 1 + y 2 , x 1 + x 2 ) …..(i)

T(x 1 , y 1 ) + T(x 2 , y 2 ) = (x 1 + y 1 , x 1 ) + (x 2 + y 2 , x 2 ) = (x 1 + x 2 + y 1 + y 2 , x 1 + x 2 ) …..(ii)

From (i) and (ii), we get T[(x 1 , y 1 ) + (x 2 , y 2 )] = T(x 1 , y 1 ) + T(x 2 , y 2 )

Now, T[a(x 1 , y 1 )] = T(ax 1 , ay 2 ) = (ax 1 + ay 1 , ax 1 ) = a(x 1 + y 1 , x 1 ) = aT(x 1 , y 1 ).

∴ T is a linear transformation.

Question 9:

Show that the given subset of vectors of R 3 forms a basis for R 3 .

{(1, 2, 1), (2, 1, 0), (1, –1, 2)}

S = {(1, 2, 1), (2, 1, 0), (1, –1, 2)}

We know that any set of n linearly independent vectors forms the basis of n-dimensional vector space.

Now, dim R 3 = 3, we just need to prove that vectors in S are linearly independent.

Let \(\begin{array}{l}A=\begin{bmatrix}1 & 2 & 1 \\2 & 1 & -1 \\1 & 0 & 2 \\\end{bmatrix}\end{array} \)

We reduce this matrix to row echelon form to check the rank of A.

Applying R 2 → R 2 + (–2)R 1 and R 3 → R 3 + ( –1)R 1 , we get

\(\begin{array}{l}A\sim\begin{bmatrix}1 & 2 & 1 \\0 & -3 & -3 \\0 & -2 & 1 \\\end{bmatrix}\end{array} \)

Applying R 2 → ( –⅓)R 2 and R 3 → R 3 + 2R 2 , we get

\(\begin{array}{l}A\sim\begin{bmatrix}1 & 2 & 1 \\0 & 1 & 1 \\0 & 0 & 3 \\\end{bmatrix}\end{array} \)

Clearly, rank of A = 3 = number of vectors.

Thus, the given vectors are linearly independent.

⇒ S forms the basis of R 3 .

Question 10:

Given a linear transformation T on V 3 ( R ) defined by T(a, b, c) = (2b + c, a – 4b, 3a) corresponding to the basis B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Find the matrix representation of T.

Now, T(1, 0, 0) = (2 × 0 + 0, 1 – 4 × 0, 3 × 1) = (0, 1, 3)

= 0(1, 0, 0) + 1(0, 1, 0) + 3(0, 0, 1)

T(0, 1, 0) = (2 × 1 + 0, 0 – 4 × 1, 3 × 0) = (2, –4, 0)

= 2(1, 0, 0) –4(0, 1, 0) + 0(0, 0, 1)

And T(0, 0, 1) = (2 × 0 + 1, 0 – 4 × 0, 3 × 0) = (1, 0, 0)

= 1(1, 0, 0) + 0(0, 1, 0) + 0(0, 0, 1)

Then, the matrix representation of T with respect to the basis B is

\(\begin{array}{l}[T ; B] = \begin{bmatrix}0 & 2 & 1 \\1 & -4 & 0 \\3 & 0 & 0 \\\end{bmatrix}\end{array} \)

Practice Problems on Linear Algebra

1. Show that the following matrix is diagonalizable:

\(\begin{array}{l}A=\begin{bmatrix}8 & -8 & -2 \\4 & -3 & -2 \\3 & -4 & 1 \\\end{bmatrix}\end{array} \)

2. Show that the transformation T: V 3 ( R ) → V 2 ( R ) defined by T(a, b, c) = (b, c) ∀ a, b, c ∈ R is a linear transformation.

3. Show that the given subset of vectors of R 3 forms a basis for V 3 ( R ).

{(1, 0, –1), (1, 2, 1), (0, –3, 2)}.

4. Given a linear transformation T on V 3 ( R ) defined by T(a, b, c) = (2b + c, a – 4b, 3a) corresponding to the basis B = {(1, 1, 1), (1, 1, 0), (1, 0, 0)}. Find the matrix representation of T.

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2.2 Linear Equations in One Variable

Learning objectives.

In this section, you will:

• Solve equations in one variable algebraically.
• Solve a rational equation.
• Find a linear equation.
• Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
• Write the equation of a line parallel or perpendicular to a given line.

Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of $400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 1 .

Solving Linear Equations in One Variable

A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form a x + b = 0 a x + b = 0 and are solved using basic algebraic operations.

We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation.

The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for x x will make the equation true.

A conditional equation is true for only some values of the variable. For example, if we are to solve the equation 5 x + 2 = 3 x − 6 , 5 x + 2 = 3 x − 6 , we have the following:

The solution set consists of one number: { − 4 } . { − 4 } . It is the only solution and, therefore, we have solved a conditional equation.

An inconsistent equation results in a false statement. For example, if we are to solve 5 x − 15 = 5 ( x − 4 ) , 5 x − 15 = 5 ( x − 4 ) , we have the following:

Indeed, −15 ≠ −20. −15 ≠ −20. There is no solution because this is an inconsistent equation.

Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.

Linear Equation in One Variable

A linear equation in one variable can be written in the form

where a and b are real numbers, a ≠ 0. a ≠ 0.

Given a linear equation in one variable, use algebra to solve it.

The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x = _________, x = _________, if x is the unknown. There is no set order, as the steps used depend on what is given:

• We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.
• Apply the distributive property as needed: a ( b + c ) = a b + a c . a ( b + c ) = a b + a c .
• Isolate the variable on one side of the equation.
• When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.

Solving an Equation in One Variable

Solve the following equation: 2 x + 7 = 19. 2 x + 7 = 19.

This equation can be written in the form a x + b = 0 a x + b = 0 by subtracting 19 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.

The solution is 6.

Solve the linear equation in one variable: 2 x + 1 = −9. 2 x + 1 = −9.

Solving an Equation Algebraically When the Variable Appears on Both Sides

Solve the following equation: 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) . 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) .

Apply standard algebraic properties.

This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, x = − 5 3 . x = − 5 3 .

Solve the equation in one variable: −2 ( 3 x − 1 ) + x = 14 − x . −2 ( 3 x − 1 ) + x = 14 − x .

Solving a Rational Equation

In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation .

Recall that a rational number is the ratio of two numbers, such as 2 3 2 3 or 7 2 . 7 2 . A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).

Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

Solve the rational equation: 7 2 x − 5 3 x = 22 3 . 7 2 x − 5 3 x = 22 3 .

We have three denominators; 2 x , 3 x , 2 x , 3 x , and 3. The LCD must contain 2 x , 3 x , 2 x , 3 x , and 3. An LCD of 6 x 6 x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6 x . 6 x .

A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as ( x + 1 ) . ( x + 1 ) . Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x , x , x − 1 , x − 1 , and 3 x − 3. 3 x − 3. First, factor all denominators. We then have x , x , ( x − 1 ) , ( x − 1 ) , and 3 ( x − 1 ) 3 ( x − 1 ) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of ( x − 1 ) . ( x − 1 ) . The x x in the first denominator is separate from the x x in the ( x − 1 ) ( x − 1 ) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x , x , one factor of ( x − 1 ) , ( x − 1 ) , and the 3. Thus, the LCD is the following:

So, both sides of the equation would be multiplied by 3 x ( x − 1 ) . 3 x ( x − 1 ) . Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.

Another example is a problem with two denominators, such as x x and x 2 + 2 x . x 2 + 2 x . Once the second denominator is factored as x 2 + 2 x = x ( x + 2 ) , x 2 + 2 x = x ( x + 2 ) , there is a common factor of x in both denominators and the LCD is x ( x + 2 ) . x ( x + 2 ) .

Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.

We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.

Multiply a ( d ) a ( d ) and b ( c ) , b ( c ) , which results in a d = b c . a d = b c .

Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.

Rational Equations

A rational equation contains at least one rational expression where the variable appears in at least one of the denominators.

Given a rational equation, solve it.

• Factor all denominators in the equation.
• Find and exclude values that set each denominator equal to zero.
• Find the LCD.
• Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
• Solve the remaining equation.
• Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.

Solving a Rational Equation without Factoring

Solve the following rational equation:

We have three denominators: x , x , 2 , 2 , and 2 x . 2 x . No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2 x . 2 x . Only one value is excluded from a solution set, 0. Next, multiply the whole equation (both sides of the equal sign) by 2 x . 2 x .

The proposed solution is −1, which is not an excluded value, so the solution set contains one number, −1 , −1 , or { −1 } { −1 } written in set notation.

Solve the rational equation: 2 3 x = 1 4 − 1 6 x . 2 3 x = 1 4 − 1 6 x .

Solving a Rational Equation by Factoring the Denominator

Solve the following rational equation: 1 x = 1 10 − 3 4 x . 1 x = 1 10 − 3 4 x .

First find the common denominator. The three denominators in factored form are x , 10 = 2 ⋅ 5 , x , 10 = 2 ⋅ 5 , and 4 x = 2 ⋅ 2 ⋅ x . 4 x = 2 ⋅ 2 ⋅ x . The smallest expression that is divisible by each one of the denominators is 20 x . 20 x . Only x = 0 x = 0 is an excluded value. Multiply the whole equation by 20 x . 20 x .

The solution is 35 2 . 35 2 .

Solve the rational equation: − 5 2 x + 3 4 x = − 7 4 . − 5 2 x + 3 4 x = − 7 4 .

Solving Rational Equations with a Binomial in the Denominator

Solve the following rational equations and state the excluded values:

• ⓐ 3 x − 6 = 5 x 3 x − 6 = 5 x
• ⓑ x x − 3 = 5 x − 3 − 1 2 x x − 3 = 5 x − 3 − 1 2
• ⓒ x x − 2 = 5 x − 2 − 1 2 x x − 2 = 5 x − 2 − 1 2

The denominators x x and x − 6 x − 6 have nothing in common. Therefore, the LCD is the product x ( x − 6 ) . x ( x − 6 ) . However, for this problem, we can cross-multiply.

The solution is 15. The excluded values are 6 6 and 0. 0.

The LCD is 2 ( x − 3 ) . 2 ( x − 3 ) . Multiply both sides of the equation by 2 ( x − 3 ) . 2 ( x − 3 ) .

The solution is 13 3 . 13 3 . The excluded value is 3. 3.

The least common denominator is 2 ( x − 2 ) . 2 ( x − 2 ) . Multiply both sides of the equation by x ( x − 2 ) . x ( x − 2 ) .

The solution is 4. The excluded value is 2. 2.

Solve − 3 2 x + 1 = 4 3 x + 1 . − 3 2 x + 1 = 4 3 x + 1 . State the excluded values.

Solving a Rational Equation with Factored Denominators and Stating Excluded Values

Solve the rational equation after factoring the denominators: 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . State the excluded values.

We must factor the denominator x 2 −1. x 2 −1. We recognize this as the difference of squares, and factor it as ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Thus, the LCD that contains each denominator is ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.

The solution is −3. −3. The excluded values are 1 1 and −1. −1.

Solve the rational equation: 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 . 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 .

Finding a Linear Equation

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = m x + b , y = m x + b , where m = slope m = slope and b = y -intercept . b = y -intercept . Let us begin with the slope.

The Slope of a Line

The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2 . The lines indicate the following slopes: m = −3 , m = −3 , m = 2 , m = 2 , and m = 1 3 . m = 1 3 .

The slope of a line, m , represents the change in y over the change in x. Given two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) , ( x 2 , y 2 ) , the following formula determines the slope of a line containing these points:

Finding the Slope of a Line Given Two Points

Find the slope of a line that passes through the points ( 2 , −1 ) ( 2 , −1 ) and ( −5 , 3 ) . ( −5 , 3 ) .

We substitute the y- values and the x- values into the formula.

The slope is − 4 7 . − 4 7 .

It does not matter which point is called ( x 1 , y 1 ) ( x 1 , y 1 ) or ( x 2 , y 2 ) . ( x 2 , y 2 ) . As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.

Find the slope of the line that passes through the points ( −2 , 6 ) ( −2 , 6 ) and ( 1 , 4 ) . ( 1 , 4 ) .

Identifying the Slope and y- intercept of a Line Given an Equation

Identify the slope and y- intercept, given the equation y = − 3 4 x − 4. y = − 3 4 x − 4.

As the line is in y = m x + b y = m x + b form, the given line has a slope of m = − 3 4 . m = − 3 4 . The y- intercept is b = −4. b = −4.

The y -intercept is the point at which the line crosses the y- axis. On the y- axis, x = 0. x = 0. We can always identify the y- intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute x = 0 x = 0 and solve for y.

The Point-Slope Formula

Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.

This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.

Given one point and the slope, the point-slope formula will lead to the equation of a line:

Finding the Equation of a Line Given the Slope and One Point

Write the equation of the line with slope m = −3 m = −3 and passing through the point ( 4 , 8 ) . ( 4 , 8 ) . Write the final equation in slope-intercept form.

Using the point-slope formula, substitute −3 −3 for m and the point ( 4 , 8 ) ( 4 , 8 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .

Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.

Given m = 4 , m = 4 , find the equation of the line in slope-intercept form passing through the point ( 2 , 5 ) . ( 2 , 5 ) .

Finding the Equation of a Line Passing Through Two Given Points

Find the equation of the line passing through the points ( 3 , 4 ) ( 3 , 4 ) and ( 0 , −3 ) . ( 0 , −3 ) . Write the final equation in slope-intercept form.

First, we calculate the slope using the slope formula and two points.

Next, we use the point-slope formula with the slope of 7 3 , 7 3 , and either point. Let’s pick the point ( 3 , 4 ) ( 3 , 4 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .

In slope-intercept form, the equation is written as y = 7 3 x − 3. y = 7 3 x − 3.

To prove that either point can be used, let us use the second point ( 0 , −3 ) ( 0 , −3 ) and see if we get the same equation.

We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.

Standard Form of a Line

Another way that we can represent the equation of a line is in standard form . Standard form is given as

where A , A , B , B , and C C are integers. The x- and y- terms are on one side of the equal sign and the constant term is on the other side.

Finding the Equation of a Line and Writing It in Standard Form

Find the equation of the line with m = −6 m = −6 and passing through the point ( 1 4 , −2 ) . ( 1 4 , −2 ) . Write the equation in standard form.

We begin using the point-slope formula.

From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.

This equation is now written in standard form.

Find the equation of the line in standard form with slope m = − 1 3 m = − 1 3 and passing through the point ( 1 , 1 3 ) . ( 1 , 1 3 ) .

Vertical and Horizontal Lines

The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as

where c is a constant. The slope of a vertical line is undefined, and regardless of the y- value of any point on the line, the x- coordinate of the point will be c .

Suppose that we want to find the equation of a line containing the following points: ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , and ( −3 , 5 ) . ( −3 , 5 ) . First, we will find the slope.

Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x- coordinates are the same and we find a vertical line through x = −3. x = −3. See Figure 3 .

The equation of a horizontal line is given as

where c is a constant. The slope of a horizontal line is zero, and for any x- value of a point on the line, the y- coordinate will be c .

Suppose we want to find the equation of a line that contains the following set of points: ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , and ( 5 , −2 ) . ( 5 , −2 ) . We can use the point-slope formula. First, we find the slope using any two points on the line.

Use any point for ( x 1 , y 1 ) ( x 1 , y 1 ) in the formula, or use the y -intercept.

The graph is a horizontal line through y = −2. y = −2. Notice that all of the y- coordinates are the same. See Figure 3 .

Finding the Equation of a Line Passing Through the Given Points

Find the equation of the line passing through the given points: ( 1 , −3 ) ( 1 , −3 ) and ( 1 , 4 ) . ( 1 , 4 ) .

The x- coordinate of both points is 1. Therefore, we have a vertical line, x = 1. x = 1.

Find the equation of the line passing through ( −5 , 2 ) ( −5 , 2 ) and ( 2 , 2 ) . ( 2 , 2 ) .

Determining Whether Graphs of Lines are Parallel or Perpendicular

Parallel lines have the same slope and different y- intercepts. Lines that are parallel to each other will never intersect. For example, Figure 4 shows the graphs of various lines with the same slope, m = 2. m = 2.

All of the lines shown in the graph are parallel because they have the same slope and different y- intercepts.

Lines that are perpendicular intersect to form a 90° 90° -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is −1 : m 1 ⋅ m 2 = −1. −1 : m 1 ⋅ m 2 = −1. For example, Figure 5 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of − 1 3 . − 1 3 .

Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither

Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: 3 y = − 4 x + 3 3 y = − 4 x + 3 and 3 x − 4 y = 8. 3 x − 4 y = 8.

The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.

First equation:

Second equation:

See the graph of both lines in Figure 6

From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.

The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.

Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2 y − x = 10 2 y − x = 10 and 2 y = x + 4. 2 y = x + 4.

Writing the Equations of Lines Parallel or Perpendicular to a Given Line

As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.

Given an equation for a line, write the equation of a line parallel or perpendicular to it.

• Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
• Use the slope and the given point with the point-slope formula.
• Simplify the line to slope-intercept form and compare the equation to the given line.

Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point

Write the equation of line parallel to a 5 x + 3 y = 1 5 x + 3 y = 1 and passing through the point ( 3 , 5 ) . ( 3 , 5 ) .

First, we will write the equation in slope-intercept form to find the slope.

The slope is m = − 5 3 . m = − 5 3 . The y- intercept is 1 3 , 1 3 , but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y- intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.

The equation of the line is y = − 5 3 x + 10. y = − 5 3 x + 10. See Figure 7 .

Find the equation of the line parallel to 5 x = 7 + y 5 x = 7 + y and passing through the point ( −1 , −2 ) . ( −1 , −2 ) .

Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to 5 x − 3 y + 4 = 0 5 x − 3 y + 4 = 0 and passing through the point ( − 4 , 1 ) . ( − 4 , 1 ) .

The first step is to write the equation in slope-intercept form.

We see that the slope is m = 5 3 . m = 5 3 . This means that the slope of the line perpendicular to the given line is the negative reciprocal, or − 3 5 . − 3 5 . Next, we use the point-slope formula with this new slope and the given point.

Access these online resources for additional instruction and practice with linear equations.

• Solving rational equations
• Equation of a line given two points
• Finding the equation of a line perpendicular to another line through a given point
• Finding the equation of a line parallel to another line through a given point

2.2 Section Exercises

What does it mean when we say that two lines are parallel?

What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?

How do we recognize when an equation, for example y = 4 x + 3 , y = 4 x + 3 , will be a straight line (linear) when graphed?

What does it mean when we say that a linear equation is inconsistent?

When solving the following equation:

2 x − 5 = 4 x + 1 2 x − 5 = 4 x + 1

explain why we must exclude x = 5 x = 5 and x = −1 x = −1 as possible solutions from the solution set.

For the following exercises, solve the equation for x . x .

7 x + 2 = 3 x − 9 7 x + 2 = 3 x − 9

4 x − 3 = 5 4 x − 3 = 5

3 ( x + 2 ) − 12 = 5 ( x + 1 ) 3 ( x + 2 ) − 12 = 5 ( x + 1 )

12 − 5 ( x + 3 ) = 2 x − 5 12 − 5 ( x + 3 ) = 2 x − 5

1 2 − 1 3 x = 4 3 1 2 − 1 3 x = 4 3

x 3 − 3 4 = 2 x + 3 12 x 3 − 3 4 = 2 x + 3 12

2 3 x + 1 2 = 31 6 2 3 x + 1 2 = 31 6

3 ( 2 x − 1 ) + x = 5 x + 3 3 ( 2 x − 1 ) + x = 5 x + 3

2 x 3 − 3 4 = x 6 + 21 4 2 x 3 − 3 4 = x 6 + 21 4

x + 2 4 − x − 1 3 = 2 x + 2 4 − x − 1 3 = 2

For the following exercises, solve each rational equation for x . x . State all x -values that are excluded from the solution set.

3 x − 1 3 = 1 6 3 x − 1 3 = 1 6

2 − 3 x + 4 = x + 2 x + 4 2 − 3 x + 4 = x + 2 x + 4

3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 ) 3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 )

3 x x − 1 + 2 = 3 x − 1 3 x x − 1 + 2 = 3 x − 1

5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3 5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3

1 x = 1 5 + 3 2 x 1 x = 1 5 + 3 2 x

For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form.

( 0 , 3 ) ( 0 , 3 ) with a slope of 2 3 2 3

( 1 , 2 ) ( 1 , 2 ) with a slope of − 4 5 − 4 5

x -intercept is 1, and ( −2 , 6 ) ( −2 , 6 )

y -intercept is 2, and ( 4 , −1 ) ( 4 , −1 )

( −3 , 10 ) ( −3 , 10 ) and ( 5 , −6 ) ( 5 , −6 )

( 1 , 3 )  and   ( 5 , 5 ) ( 1 , 3 )  and   ( 5 , 5 )

parallel to y = 2 x + 5 y = 2 x + 5 and passes through the point ( 4 , 3 ) ( 4 , 3 )

perpendicular to 3 y = x − 4 3 y = x − 4 and passes through the point ( −2 , 1 ) ( −2 , 1 ) .

For the following exercises, find the equation of the line using the given information.

( − 2 , 0 ) ( − 2 , 0 ) and ( −2 , 5 ) ( −2 , 5 )

( 1 , 7 ) ( 1 , 7 ) and ( 3 , 7 ) ( 3 , 7 )

The slope is undefined and it passes through the point ( 2 , 3 ) . ( 2 , 3 ) .

The slope equals zero and it passes through the point ( 1 , −4 ) . ( 1 , −4 ) .

The slope is 3 4 3 4 and it passes through the point ( 1 , 4 ) ( 1 , 4 ) .

( –1 , 3 ) ( –1 , 3 ) and ( 4 , –5 ) ( 4 , –5 )

For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.

y = 2 x + 7 y = − 1 2 x − 4 y = 2 x + 7 y = − 1 2 x − 4

3 x − 2 y = 5 6 y − 9 x = 6 3 x − 2 y = 5 6 y − 9 x = 6

y = 3 x + 1 4 y = 3 x + 2 y = 3 x + 1 4 y = 3 x + 2

x = 4 y = −3 x = 4 y = −3

For the following exercises, find the slope of the line that passes through the given points.

( 5 , 4 ) ( 5 , 4 ) and ( 7 , 9 ) ( 7 , 9 )

( −3 , 2 ) ( −3 , 2 ) and ( 4 , −7 ) ( 4 , −7 )

( −5 , 4 ) ( −5 , 4 ) and ( 2 , 4 ) ( 2 , 4 )

( −1 , −2 ) ( −1 , −2 ) and ( 3 , 4 ) ( 3 , 4 )

( 3 , −2 ) ( 3 , −2 ) and ( 3 , −2 ) ( 3 , −2 )

For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.

( −1 , 3 )  and   ( 5 , 1 ) ( −2 , 3 )  and   ( 0 , 9 ) ( −1 , 3 )  and   ( 5 , 1 ) ( −2 , 3 )  and   ( 0 , 9 )

( 2 , 5 )  and   ( 5 , 9 ) ( −1 , −1 )  and   ( 2 , 3 ) ( 2 , 5 )  and   ( 5 , 9 ) ( −1 , −1 )  and   ( 2 , 3 )

For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y -intercept occurs. State your ymin and ymax values.

0.537 x − 2.19 y = 100 0.537 x − 2.19 y = 100

4,500 x − 200 y = 9,528 4,500 x − 200 y = 9,528

200 − 30 y x = 70 200 − 30 y x = 70

Starting with the point-slope formula y − y 1 = m ( x − x 1 ) , y − y 1 = m ( x − x 1 ) , solve this expression for x x in terms of x 1 , y , y 1 , x 1 , y , y 1 , and m m .

Starting with the standard form of an equation A x + B y = C A x + B y = C solve this expression for y y in terms of A , B , C A , B , C and x x . Then put the expression in slope-intercept form.

Use the above derived formula to put the following standard equation in slope intercept form: 7 x − 5 y = 25. 7 x − 5 y = 25.

Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.

( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) ( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) and ( 0 , 4 ) ( 0 , 4 )

Find the slopes of the diagonals in the previous exercise. Are they perpendicular?

Real-World Applications

The slope for a wheelchair ramp for a home has to be 1 12 . 1 12 . If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.

If the profit equation for a small business selling x x number of item one and y y number of item two is p = 3 x + 4 y , p = 3 x + 4 y , find the y y value when p = $ 453 and   x = 75. p = $ 453 and   x = 75.

For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be y = 45 + .25 x , y = 45 + .25 x , where x x is the number of miles traveled.

What is your cost if you travel 50 mi?

If your cost were $ 63.75 , $ 63.75 , how many miles were you charged for traveling?

Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel?

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• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra 2e
• Publication date: Dec 21, 2021
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable

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4.6: Solve Systems of Equations Using Matrices

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Learning Objectives

By the end of this section, you will be able to:

• Write the augmented matrix for a system of equations
• Use row operations on a matrix
• Solve systems of equations using matrices

Before you get started, take this readiness quiz.

• Solve: \(3(x+2)+4=4(2x−1)+9\). If you missed this problem, review [link] .
• Solve: \(0.25p+0.25(x+4)=5.20\). If you missed this problem, review [link] .
• Evaluate when \(x=−2\) and \(y=3:2x^2−xy+3y^2\). If you missed this problem, review [link] .

Write the Augmented Matrix for a System of Equations

Solving a system of equations can be a tedious operation where a simple mistake can wreak havoc on finding the solution. An alternative method which uses the basic procedures of elimination but with notation that is simpler is available. The method involves using a matrix . A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix with m rows and n columns has order \(m\times n\). The matrix on the left below has 2 rows and 3 columns and so it has order \(2\times 3\). We say it is a 2 by 3 matrix.

Each number in the matrix is called an element or entry in the matrix.

We will use a matrix to represent a system of linear equations. We write each equation in standard form and the coefficients of the variables and the constant of each equation becomes a row in the matrix. Each column then would be the coefficients of one of the variables in the system or the constants. A vertical line replaces the equal signs. We call the resulting matrix the augmented matrix for the system of equations.

Notice the first column is made up of all the coefficients of x , the second column is the all the coefficients of y , and the third column is all the constants.

Example \(\PageIndex{1}\)

ⓐ \(\left\{ \begin{array} {l} 5x−3y=−1 \\ y=2x−2 \end{array} \right. \) ⓑ \( \left\{ \begin{array} {l} 6x−5y+2z=3 \\ 2x+y−4z=5 \\ 3x−3y+z=−1 \end{array} \right. \)

ⓐ The second equation is not in standard form. We rewrite the second equation in standard form.

\[\begin{aligned} y=2x−2 \\ −2x+y=−2 \end{aligned} \nonumber\]

We replace the second equation with its standard form. In the augmented matrix, the first equation gives us the first row and the second equation gives us the second row. The vertical line replaces the equal signs.

ⓑ All three equations are in standard form. In the augmented matrix the first equation gives us the first row, the second equation gives us the second row, and the third equation gives us the third row. The vertical line replaces the equal signs.

Example \(\PageIndex{2}\)

Write each system of linear equations as an augmented matrix:

ⓐ \(\left\{ \begin{array} {l} 3x+8y=−3 \\ 2x=−5y−3 \end{array} \right. \) ⓑ \(\left\{ \begin{array} {l} 2x−5y+3z=8 \\ 3x−y+4z=7 \\ x+3y+2z=−3 \end{array} \right. \)

ⓐ \(\left[ \begin{matrix} 3 &8 &-3 \\ 2 &5 &−3 \end{matrix} \right] \)

ⓑ \(\left[ \begin{matrix} 2 &3 &1 &−5 \\ −1 &3 &3 &4 \\ 2 &8 &7 &−3 \end{matrix} \right] \)

Example \(\PageIndex{3}\)

ⓐ \(\left\{ \begin{array} {l} 11x=−9y−5 \\ 7x+5y=−1 \end{array} \right. \) ⓑ \(\left\{ \begin{array} {l} 5x−3y+2z=−5 \\ 2x−y−z=4 \\ 3x−2y+2z=−7 \end{array} \right. \)

ⓐ \(\left[ \begin{matrix} 11 &9 &−5 \\ 7 &5 &−1 \end{matrix} \right] \) ⓑ \(\left[ \begin{matrix} 5 &−3 &2 &−5 \\ 2 &−1 &−1 &4 \\ 3 &−2 &2 &−7 \end{matrix} \right] \)

It is important as we solve systems of equations using matrices to be able to go back and forth between the system and the matrix. The next example asks us to take the information in the matrix and write the system of equations.

Example \(\PageIndex{4}\)

Write the system of equations that corresponds to the augmented matrix:

\(\left[ \begin{array} {ccc|c} 4 &−3 &3 &−1 \\ 1 &2 &−1 &2 \\ −2 &−1 &3 &−4 \end{array} \right] \).

We remember that each row corresponds to an equation and that each entry is a coefficient of a variable or the constant. The vertical line replaces the equal sign. Since this matrix is a \(4\times 3\), we know it will translate into a system of three equations with three variables.

Example \(\PageIndex{5}\)

Write the system of equations that corresponds to the augmented matrix: \(\left[ \begin{matrix} 1 &−1 &2 &3 \\ 2 &1 &−2 &1 \\ 4 &−1 &2 &0 \end{matrix} \right] \).

\(\left\{ \begin{array} {l} x−y+2z=3 \\ 2x+y−2z=1 \\ 4x−y+2z=0 \end{array} \right.\)

Example \(\PageIndex{6}\)

Write the system of equations that corresponds to the augmented matrix: \(\left[ \begin{matrix} 1 &1 &1 &4 \\ 2 &3 &−1 &8 \\ 1 &1 &−1 &3 \end{matrix} \right] \).

\(\left\{ \begin{array} {l} x+y+z=4 \\ 2x+3y−z=8 \\ x+y−z=3 \end{array} \right.\)

Use Row Operations on a Matrix

Once a system of equations is in its augmented matrix form, we will perform operations on the rows that will lead us to the solution.

To solve by elimination, it doesn’t matter which order we place the equations in the system. Similarly, in the matrix we can interchange the rows.

When we solve by elimination, we often multiply one of the equations by a constant. Since each row represents an equation, and we can multiply each side of an equation by a constant, similarly we can multiply each entry in a row by any real number except 0.

In elimination, we often add a multiple of one row to another row. In the matrix we can replace a row with its sum with a multiple of another row.

These actions are called row operations and will help us use the matrix to solve a system of equations.

ROW OPERATIONS

In a matrix, the following operations can be performed on any row and the resulting matrix will be equivalent to the original matrix.

• Interchange any two rows.
• Multiply a row by any real number except 0.
• Add a nonzero multiple of one row to another row.

Performing these operations is easy to do but all the arithmetic can result in a mistake. If we use a system to record the row operation in each step, it is much easier to go back and check our work.

We use capital letters with subscripts to represent each row. We then show the operation to the left of the new matrix. To show interchanging a row:

To multiply row 2 by \(−3\):

To multiply row 2 by \(−3\) and add it to row 1:

Example \(\PageIndex{7}\)

Perform the indicated operations on the augmented matrix:

ⓐ Interchange rows 2 and 3.

ⓑ Multiply row 2 by 5.

ⓒ Multiply row 3 by −2−2 and add to row 1.

\( \left[ \begin{array} {ccc|c} 6 &−5 &2 &3 \\ 2 &1 &−4 &5 \\ 3 &−3 &1 &−1 \end{array} \right] \)

ⓐ We interchange rows 2 and 3.

ⓑ We multiply row 2 by 5.

ⓒ We multiply row 3 by \(−2\) and add to row 1.

Example \(\PageIndex{8}\)

ⓐ Interchange rows 1 and 3.

ⓑ Multiply row 3 by 3.

ⓒ Multiply row 3 by 2 and add to row 2.

\( \left[ \begin{array} {ccc|c} 5 &−2 &-2 &-2 \\ 4 &-1 &−4 &4 \\ -2 &3 &0 &−1 \end{array} \right] \)

ⓐ \( \left[ \begin{matrix} −2 &3 &0 &−2 \\ 4 &−1 &−4 &4 \\ 5 &−2 &−2 &−2 \end{matrix} \right] \)

ⓑ \( \left[ \begin{matrix} −2 &3 &0 &−2 \\ 4 &−1 &−4 &4 \\ 15 &−6 &−6 &−6 \end{matrix} \right] \)

ⓒ \( \left[ \begin{matrix} -2 &3 &0 &2 & \\ 3 &4 &-13 &-16 &-8 \\ 15 &-6 &-6 &-6 & \end{matrix} \right] \)

Example \(\PageIndex{9}\)

ⓐ Interchange rows 1 and 2,

ⓑ Multiply row 1 by 2,

ⓒ Multiply row 2 by 3 and add to row 1.

\( \left[ \begin{array} {ccc|c} 2 &−3 &−2 &−4 \\ 4 &1 &−3 &2 \\ 5 &0 &4 &−1 \end{array} \right] \)

ⓐ \( \left[ \begin{matrix} 4 &1 &−3 &2 \\ 2 &−3 &−2 &−4 \\ 5 &0 &4 &−1 \end{matrix} \right] \) ⓑ \( \left[ \begin{matrix} 8 &2 &−6 &4 \\ 2 &−3 &−2 &−4 \\ 5 &0 &4 &−1 \end{matrix} \right] \) ⓒ \( \left[ \begin{matrix} 14 &−7 &−12 &−8 \\ 2 &−3 &−2 &−4 \\ 5 &0 &4 &−1 \end{matrix} \right] \)

Now that we have practiced the row operations, we will look at an augmented matrix and figure out what operation we will use to reach a goal. This is exactly what we did when we did elimination. We decided what number to multiply a row by in order that a variable would be eliminated when we added the rows together.

Given this system, what would you do to eliminate x ?

This next example essentially does the same thing, but to the matrix.

Example \(\PageIndex{10}\)

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \( \left[ \begin{array} {cc|c} 1 &−1 &2 \\ 4 &−8 &0 \end{array} \right] \)

To make the 4 a 0, we could multiply row 1 by \(−4\) and then add it to row 2.

Example \(\PageIndex{11}\)

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \( \left[ \begin{array} {cc|c} 1 &−1 &2 \\ 3 &−6 &2 \end{array} \right] \)

\( \left[ \begin{matrix} 1 &−1 &2 \\ 0 &−3 &−4 \end{matrix} \right] \)

Example \(\PageIndex{12}\)

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \( \left[ \begin{array} {cc|c} 1 &−1 &3 \\ -2 &−3 &2 \end{array} \right] \)

\( \left[ \begin{matrix} 1 &−1 &3 \\ 0 &−5 &8 \end{matrix} \right] \)

Solve Systems of Equations Using Matrices

To solve a system of equations using matrices, we transform the augmented matrix into a matrix in row-echelon form using row operations. For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

ROW-ECHELON FORM

For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

Once we get the augmented matrix into row-echelon form, we can write the equivalent system of equations and read the value of at least one variable. We then substitute this value in another equation to continue to solve for the other variables. This process is illustrated in the next example.

How to Solve a System of Equations Using a Matrix

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 3x+4y=5 \\ x+2y=1 \end{array} \right. \)

Example \(\PageIndex{14}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 2x+y=7 \\ x−2y=6 \end{array} \right. \)

The solution is \((4,−1)\).

Example \(\PageIndex{15}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 2x+y=−4 \\ x−y=−2 \end{array} \right. \)

The solution is \((−2,0)\).

The steps are summarized here.

SOLVE A SYSTEM OF EQUATIONS USING MATRICES.

• Write the augmented matrix for the system of equations.
• Using row operations get the entry in row 1, column 1 to be 1.
• Using row operations, get zeros in column 1 below the 1.
• Using row operations, get the entry in row 2, column 2 to be 1.
• Continue the process until the matrix is in row-echelon form.
• Write the corresponding system of equations.
• Use substitution to find the remaining variables.
• Write the solution as an ordered pair or triple.
• Check that the solution makes the original equations true.

Here is a visual to show the order for getting the 1’s and 0’s in the proper position for row-echelon form.

We use the same procedure when the system of equations has three equations.

Example \(\PageIndex{16}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 3x+8y+2z=−5 \\ 2x+5y−3z=0 \\ x+2y−2z=−1 \end{array} \right. \)

Example \(\PageIndex{17}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 2x−5y+3z=8 \\ 3x−y+4z=7 \\ x+3y+2z=−3 \end{array} \right. \)

\((6,−1,−3)\)

Example \(\PageIndex{18}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} −3x+y+z=−4 \\ −x+2y−2z=1 \\ 2x−y−z=−1 \end{array} \right. \)

\((5,7,4)\)

So far our work with matrices has only been with systems that are consistent and independent, which means they have exactly one solution. Let’s now look at what happens when we use a matrix for a dependent or inconsistent system.

Example \(\PageIndex{19}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x+y+3z=0 \\ x+3y+5z=0 \\ 2x+4z=1 \end{array} \right. \)

Example \(\PageIndex{20}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x−2y+2z=1 \\ −2x+y−z=2 \\ x−y+z=5 \end{array} \right. \)

no solution

Example \(\PageIndex{21}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 3x+4y−3z=−2 \\ −2x+3y−z=−1 \\ 2x+y−2z=6 \end{array} \right. \)

The last system was inconsistent and so had no solutions. The next example is dependent and has infinitely many solutions.

Example \(\PageIndex{22}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x−2y+3z=1 \\ x+y−3z=7 \\ 3x−4y+5z=7 \end{array} \right. \)

Example \(\PageIndex{23}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x+y−z=0 \\ 2x+4y−2z=6 \\ 3x+6y−3z=9 \end{array} \right. \)

infinitely many solutions \((x,y,z)\), where \(x=z−3;\space y=3;\space z\) is any real number.

Example \(\PageIndex{24}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x−y−z=1 \\ −x+2y−3z=−4 \\ 3x−2y−7z=0 \end{array} \right. \)

infinitely many solutions \((x,y,z)\), where \(x=5z−2;\space y=4z−3;\space z\) is any real number.

Access this online resource for additional instruction and practice with Gaussian Elimination.

• Gaussian Elimination

Key Concepts

• Interchange any two rows
• Multiply a row by any real number except 0
• Add a nonzero multiple of one row to another row

Linear Algebra

Linear algebra is a branch of mathematics that deals with linear equations and their representations in the vector space using matrices. In other words, linear algebra is the study of linear functions and vectors. It is one of the most central topics of mathematics. Most modern geometrical concepts are based on linear algebra.

Linear algebra facilitates the modeling of many natural phenomena and hence, is an integral part of engineering and physics. Linear equations, matrices, and vector spaces are the most important components of this subject. In this article, we will learn more about linear algebra and the various associated topics.

What is Linear Algebra?

Linear algebra can be defined as a branch of mathematics that deals with the study of linear functions in vector spaces. When information related to linear functions is presented in an organized form then it results in a matrix. Thus, linear algebra is concerned with vector spaces, vectors, linear functions, the system of linear equations, and matrices. These concepts are a prerequisite for sister topics such as geometry and functional analysis.

Linear Algebra Definition

The branch of mathematics that deals with vectors, matrics, finite or infinite dimensions as well as a linear mapping between such spaces is defined as linear algebra. It is used in both pure and applied mathematics along with different technical forms such as physics, engineering, natural sciences, etc.

Branches of Linear Algebra

Linear algebra can be categorized into three branches depending upon the level of difficulty and the kind of topics that are encompassed within each. These are elementary, advanced, and applied linear algebra. Each branch covers different aspects of matrices, vectors, and linear functions.

Elementary Linear Algebra

Elementary linear algebra introduces students to the basics of linear algebra. This includes simple matrix operations, various computations that can be done on a system of linear equations, and certain aspects of vectors. Some important terms associated with elementary linear algebra are given below:

Scalars - A scalar is a quantity that only has magnitude and not direction. It is an element that is used to define a vector space. In linear algebra, scalars are usually real numbers.

Vectors - A vector is an element in a vector space. It is a quantity that can describe both the direction and magnitude of an element.

Vector Space - The vector space consists of vectors that may be added together and multiplied by scalars.

Matrix - A matrix is a rectangular array wherein the information is organized in the form of rows and columns. Most linear algebra properties can be expressed in terms of a matrix.

Matrix Operations - These are simple arithmetic operations such as addition , subtraction , and multiplication that can be conducted on matrices.

Advanced Linear Algebra

Once the basics of linear algebra have been introduced to students the focus shifts on more advanced concepts related to linear equations, vectors, and matrices. Certain important terms that are used in advanced linear algebra are as follows:

Linear Transformations - The transformation of a function from one vector space to another by preserving the linear structure of each vector space.

Inverse of a Matrix - When an inverse of a matrix is multiplied with the given original matrix then the resultant will be the identity matrix. Thus, A -1 A = I.

Eigenvector - An eigenvector is a non-zero vector that changes by a scalar factor (eigenvalue) when a linear transformation is applied to it.

Linear Map - It is a type of mapping that preserves vector addition and vector multiplication.

Applied Linear Algebra

Applied linear algebra is usually introduced to students at a graduate level in fields of applied mathematics, engineering, and physics. This branch of algebra is driven towards integrating the concepts of elementary and advanced linear algebra with their practical implications. Topics such as the norm of a vector, QR factorization, Schur's complement of a matrix, etc., fall under this branch of linear algebra.

Linear Algebra Topics

The topics that come under linear algebra can be classified into three broad categories. These are linear equations, matrices, and vectors. All these three categories are interlinked and need to be understood well in order to master linear algebra. The topics that fall under each category are given below.

Linear Equations

A linear equation is an equation that has the standard form \(a_{1}x_{1} + a_{2}x_{2} + ... + a_{n}x_{n}\). It is the fundamental component of linear algebra. The topics covered under linear equations are as follows:

• Linear Equations in One variable
• Linear Equations in Two Variables
• Simultaneous Linear Equations
• Solving Linear Equations
• Solutions of a Linear Equation
• Graphing Linear Equations
• Applications of Linear equations
• Straight Line

In linear algebra, there can be several operations that can be performed on vectors such as multiplication , addition, etc. Vectors can be used to describe quantities such as the velocity of moving objects. Some crucial topics encompassed under vectors are as follows:

• Types of Vectors
• Dot Product
• Cross Product
• Addition of Vectors

A matrix is used to organize data in the form of a rectangular array. It can be represented as \(A_{m\times n}\). Here, m represents the number of rows and n denotes the number of columns in the matrix. In linear algebra, a matrix can be used to express linear equations in a more compact manner. The topics that are covered under the scope of matrices are as follows:

• Matrix Operations
• Determinant
• Transpose of a Matrix
• Types of a Matrix

Linear Algebra Formula

Formulas form an important part of linear algebra as they help to simplify computations. The key to solving any problem in linear algebra is to understand the formulas and associated concepts rather than memorize them. The important linear algebra formulas can be broken down into 3 categories, namely, linear equations, vectors, and matrices.

Linear Equations: The important linear equation formulas are listed as follows:

• General form: ax + by = c
• Slope Intercept Form : y = mx + b
• a + b = b + a
• a + 0 = 0 + a = a

Vectors: If there are two vectors \(\overrightarrow{u}\) = (\(u_{1}\), \(u_{2}\), \(u_{3}\)) and \(\overrightarrow{v}\) = (\(v_{1}\), \(v_{2}\), \(v_{3}\)) then the important vector formulas associated with linear algebra are given below.

• \(\overrightarrow{u} + \overrightarrow{v} = (u_{1}+v_{1}, u_{2}+v_{2}, u_{3}+v_{3})\)
• \(\overrightarrow{u} - \overrightarrow{v} = (u_{1}-v_{1}, u_{2}-v_{2}, u_{3}-v_{3})\)
• \(\left \| u \right \| = \sqrt{u_{1}^{2} + u_{2}^{2} + u_{3}^{2}}\)
• \(\overrightarrow{u}.\overrightarrow{v} = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}\)
• \(\overrightarrow{u}\times \overrightarrow{v} = (u_{2}v_{3}-u_{3}v_{2}, u_{3}v_{1}-u_{1}v_{3}, u_{1}v_{2}-u_{2}v_{1})\)

Matrix: If there are two square matrices given by A and B where the elements are \(a_{ij}\) and \(b_{ij}\) respectively, then the following important formulas are used in linear algebra:

• C = A + B, where \(c_{ij}\) = \(a_{ij}\) + \(b_{ij}\)
• C = A - B, where \(c_{ij}\) = \(a_{ij}\) - \(b_{ij}\)
• kA = k\(a_{ij}\)
• C = AB = \(\sum_{k = 1}^{n}a_{ik}b_{kj}\)

Linear Algebra and its Applications

Linear algebra is used in almost every field. Simple algorithms also make use of linear algebra topics such as matrices. Some of the applications of linear algebra are given as follows:

• Signal Processing - Linear algebra is used in encoding and manipulating signals such as audio and video signals. Furthermore, it is required in the analysis of such signals.
• Linear Programming - It is an optimizing technique that is used to determine the best outcome of a linear function.
• Computer Science - Data scientists use several linear algebra algorithms to solve complicated problems.
• Prediction Algorithms - Prediction algorithms use linear models that are developed using concepts of linear algebra.

Related Articles:

• Introduction to Graphing
• One Variable Linear Equations and Inequalities
• Resolving a Vector into Components

Important Notes on Linear Algebra

• Linear algebra is concerned with the study of three broad subtopics - linear functions, vectors, and matrices
• Linear algebra can be classified into 3 categories. These are elementary, advanced, and applied linear algebra.
• Elementary linear algebra is concerned with the introduction to linear algebra. Advanced linear algebra builds on these concepts. Applied linear algebra applies these concepts to real-life situations.

Linear Algebra Examples

• Example 1: Using linear algebra add these two matrices. A = \(\begin{bmatrix} 5 & 6\\ 2& 1 \end{bmatrix}\) and B = \(\begin{bmatrix} 3 & 7\\ 5& 4 \end{bmatrix}\) Solution: C = A + B C = \(\begin{bmatrix} 5 & 6\\ 2& 1 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 7\\ 5& 4 \end{bmatrix}\) C = \(\begin{bmatrix} 8 & 13\\ 7& 5 \end{bmatrix}\) Answer: C = \(\begin{bmatrix} 8 & 13\\ 7& 5 \end{bmatrix}\)
• Example 2: Subtract the two vectors \(\vec{u}\) = (3, 7, 1) and \(\vec{v}\) = (6, 2, 8) using linear algebra Solution: \(\vec{u}\) - \(\vec{v}\) = (-3, 5, -7) Answer: (-3, 5, -7)
• Example 3: Solve the equations: x + 3 = 2(y - 1) and y + 1 = 5x Solution: Solving by substitution, x + 3 = 2(y - 1) x = 2y - 5 Putting this value in the second equation, y + 1 = 5 (2y - 5) y = 26 / 9 Now y + 1 = 5x (26 / 9) + 1 = 5x x = 7 / 9 Answer: x = 7 / 9, y = 26 / 9

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Practice Questions on Linear Algebra

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FAQs on Linear Algebra

What is the meaning of linear algebra.

Linear algebra is a branch of mathematics that deals with the study of linear functions , vectors, matrices, and other associated aspects.

Is Linear Algebra Difficult?

Linear algebra is a very vast branch of mathematics. However, with regular practice and instilling a strong conceptual foundation solving questions will be very easy.

What are the Prerequisites for Linear Algebra?

It is necessary to have a strong foundation regarding the properties of numbers and how to perform calculations before starting linear algebra.

What is a Subspace in Linear Algebra?

A vector space that is entirely contained in another vector space is known as a subspace in linear algebra.

How to Study Linear Algebra?

The first step is to instill a strong foundation in elementary algebra. Understanding concepts and regular revision of formulas are also crucial before moving on to advanced algebra. It is equally necessary to solve practice questions of various levels to succeed in this subject.

Is Linear Algebra Harder than Calculus?

Linear algebra serves as a prerequisite for calculus . It is important to develop deep-seated knowledge of this subject before moving on to calculus. Both subjects are easy as long as concepts are clear and sums are practiced regularly.

What is Linear Algebra Used for?

Linear algebra is used in several industries such as computer science, engineering as well as physics to create linear models using the algorithms outlined in this subject.

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Unit 3: Linear equations, functions, & graphs

About this unit.

This topic covers:

• Intercepts of linear equations/functions
• Slope of linear equations/functions
• Slope-intercept, point-slope, & standard forms
• Graphing linear equations/functions
• Writing linear equations/functions
• Interpreting linear equations/functions
• Linear equations/functions word problems

Two-variable linear equations intro

• Two-variable linear equations intro (Opens a modal)
• Solutions to 2-variable equations (Opens a modal)
• Worked example: solutions to 2-variable equations (Opens a modal)
• Completing solutions to 2-variable equations (Opens a modal)
• Solutions to 2-variable equations: substitution (old) (Opens a modal)
• Solutions to 2-variable equations: graphical (old) (Opens a modal)
• Solutions to 2-variable equations 4 questions Practice
• Complete solutions to 2-variable equations 4 questions Practice

x-intercepts and y-intercepts

• Intro to intercepts (Opens a modal)
• x-intercept of a line (Opens a modal)
• Intercepts from an equation (Opens a modal)
• Intercepts from a table (Opens a modal)
• Graphing using intercepts (old) (Opens a modal)
• Intercepts of lines review (x-intercepts and y-intercepts) (Opens a modal)
• Intercepts from a graph 4 questions Practice
• Intercepts from an equation 4 questions Practice
• Intercepts from a table 4 questions Practice
• Intro to slope (Opens a modal)
• Positive & negative slope (Opens a modal)
• Worked example: slope from graph (Opens a modal)
• Worked example: slope from two points (Opens a modal)
• Slope (more examples) (Opens a modal)
• Slope review (Opens a modal)
• Slope from graph 4 questions Practice
• Slope from two points 4 questions Practice

Horizontal & vertical lines

• Slope of a horizontal line (Opens a modal)
• Horizontal & vertical lines (Opens a modal)
• Horizontal & vertical lines 7 questions Practice

Intro to slope-intercept form

• Intro to slope-intercept form (Opens a modal)
• Worked examples: slope-intercept intro (Opens a modal)
• Slope-intercept intro 4 questions Practice

Graphing slope-intercept equations

• Graph from slope-intercept equation (Opens a modal)
• Graphing slope-intercept form (Opens a modal)
• Graphing lines from slope-intercept form review (Opens a modal)
• Graph from slope-intercept form 4 questions Practice

Writing slope-intercept equations

• Slope-intercept equation from graph (Opens a modal)
• Writing slope-intercept equations (Opens a modal)
• Slope-intercept equation from slope & point (Opens a modal)
• Slope-intercept equation from two points (Opens a modal)
• Slope-intercept form problems (Opens a modal)
• Slope-intercept equation from slope & point (old) (Opens a modal)
• Slope-intercept equation from slope & point: fractions (old) (Opens a modal)
• Finding y intercept given slope & point (old) (Opens a modal)
• Slope-intercept form review (Opens a modal)
• Slope-intercept equation from graph 4 questions Practice
• Slope-intercept from two points 4 questions Practice

Point-slope form

• Intro to point-slope form (Opens a modal)
• Point-slope & slope-intercept equations (Opens a modal)
• Point-slope form review (Opens a modal)
• Point-slope form 4 questions Practice

Standard form

• Intro to linear equation standard form (Opens a modal)
• Graphing a linear equation: 5x+2y=20 (Opens a modal)
• Converting from slope-intercept to standard form (Opens a modal)
• Standard form review (Opens a modal)
• Graph from linear standard form 4 questions Practice
• Convert linear equations to standard form 4 questions Practice

Summary: Forms of two-variable linear equations

• Slope from equation (Opens a modal)
• Writing linear equations in all forms (Opens a modal)
• Forms of linear equations review (Opens a modal)
• Slope from equation 4 questions Practice
• Linear equations in any form 4 questions Practice

Interpreting linear functions and equations

• Linear equations word problems: earnings (Opens a modal)
• Linear equations word problems: volcano (Opens a modal)
• Linear graphs word problems (Opens a modal)
• Linear graphs word problem: cats (Opens a modal)
• Linear equations word problem: marbles (Opens a modal)
• Linear equations word problem: file transfer (Opens a modal)
• Linear equations word problems: tables 4 questions Practice
• Linear equations word problems: graphs 4 questions Practice
• Linear equations word problems 4 questions Practice

Comparing linear functions

• Comparing linear functions: equation vs. graph (Opens a modal)
• Comparing linear functions: same rate of change (Opens a modal)
• Comparing linear functions: faster rate of change (Opens a modal)
• Comparing linear functions word problem: climb (Opens a modal)
• Comparing linear functions word problem: walk (Opens a modal)
• Comparing linear functions word problem: work (Opens a modal)
• Compare linear functions 4 questions Practice
• Comparing linear functions word problems 4 questions Practice

Constructing linear models for real-world relationships

• Linear functions word problem: fuel (Opens a modal)
• Linear functions word problem: pool (Opens a modal)
• Linear functions word problem: iceberg (Opens a modal)
• Linear functions word problem: paint (Opens a modal)
• Graphing linear relationships word problems 4 questions Practice
• Writing linear equations word problems 4 questions Practice

Linear models word problems

• Linear models word problem: book (Opens a modal)
• Linear models word problem: marbles (Opens a modal)
• Comparing linear rates example (Opens a modal)
• Linear models word problems 4 questions Practice
• Comparing linear rates word problems 4 questions Practice

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