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.

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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..

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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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Problems in Mathematics

Linear Algebra

Linear algebra problems and solutions.

The topics in Linear Algebra are listed below. Each page contains definitions and summary of the topic followed by exercise problems.

This is version 0 (11/15/2017), that is, still work in progress.

  • Introduction to Matrices
  • Elementary Row Operations
  • Gaussian-Jordan Elimination
  • Solutions of Systems of Linear Equations
  • Linear Combination and Linear Independence
  • Nonsingular Matrices
  • Inverse Matrices
  • Subspaces in $\R^n$
  • 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
  • Dot Products and Length of Vectors
  • Eigenvalues and Eigenvectors of Linear Transformations
  • Jordan Canonical Form

Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Elimination Method of solving linear equations

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Graphical Method of Solving Linear Equations

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Cross Multiplication Method of solving linear equations

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

  • Linear Equations
  • Application of Linear Equations
  • Two-Variable Linear Equations
  • Linear Equations and Half Planes
  • One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Solving Linear Equations Example

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

  • Substitution method
  • Elimination method
  • Cross multiplication method
  • Graphical method

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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.

Figure shows two matrices. The one on the left has the numbers minus 3, minus 2 and 2 in the first row and the numbers minus 1, 4 and 5 in the second row. The rows and columns are enclosed within brackets. Thus, it has 2 rows and 3 columns. It is labeled 2 cross 3 or 2 by 3 matrix. The matrix on the right is similar but with 3 rows and 4 columns. It is labeled 3 by 4 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.

The equations are 3x plus y equals minus 3 and 2x plus 3y equals 6. A 2 by 3 matrix is shown. The first row is 3, 1, minus 3. The second row is 2, 3, 6. The first column is labeled coefficients of x. The second column is labeled coefficients of y and the third is labeled constants.

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.

The equations are 3x plus y equals minus 3 and 2x plus 3y equals 6. A 2 by 3 matrix is shown. The first row is 3, 1, minus 3. The second row is 2, 3, 6. The first column is labeled coefficients of x. The second column is labeled coefficients of y and the third is labeled constants.

ⓑ 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.

The equations are 6x minus 5y plus 2z equals 3, 2x plus y minus 4z equals 5 and 3x minus 3y plus z equals minus 1. A 4 by 3 matrix is shown whose first row is 6, minus 5, 2, 3. Its second row is 2, 1, minus 4, 5. Its third row is 3, minus 3, 1 and minus 1. Its first three columns are labeled x, y and z respectively.

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.

A 3 by 4 matrix is shown. Its first row is 4, minus 3, 3, minus 1. Its second row is 1, 2, minus 1, 2. Its third row is minus 2, minus 1, 3, minus 4. The three equations are 4x minus 3y plus 3z equals minus 1, x plus 2y minus z equals 2 and minus 2x minus y plus 3z equals minus 4.

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:

A 2 by 3 matrix is shown. Its first row, labeled R2 is 2, minus 1, 2. Its second row, labeled R1 is 5, minus 3, minus 1.

To multiply row 2 by \(−3\):

A 2 by 3 matrix is shown. Its first row is 5, minus 3, minus 1. Its second row is 2, minus 1, 2. An arrow point from this matrix to another one on the right. The first row of the new matrix is the same. The second row is preceded by minus 3 R2. It is minus 6, 3, minus 6.

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

A 2 by 3 matrix is shown. Its first row is 5, minus 3, minus 1. Its second row is 2, minus 1, 2. An arrow point from this matrix to another one on the right. The first row of the new matrix is preceded by minus 3 R2 plus R1. It is minus 1, 0, minus 7. The second row is 2, minus 1, 2.

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.

Two 3 by 4 matrices are shown. In the one on the left, the first row is 6, minus 5, 2, 3. The second row is 2, 1, minus 4, 5. The third row is 3, minus 3, 1, minus 1. The second matrix is similar except that rows 2 and 3 are interchanged.

ⓑ We multiply row 2 by 5.

Two 3 by 4 matrices are shown. In the one on the left, the first row is 6, minus 5, 2, 3. The second row is 2, 1, minus 4, 5. The third row is 3, minus 3, 1, minus 1. The second matrix is similar to the first except that row 2, preceded by 5 R2, is 10, 5, minus 20, 25.

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

In the 3 by 4 matrix, the first row is 6, minus 5, 2, 3. The second row is 2, 1, minus 4, 5. The third row is 3, minus 3, 1, minus 1. Performing the operation minus 2 R3 plus R1 on the first row, the first row becomes 6 plus minus 2 times 3, minus 5 plus minus 2 times minus 3, 2 plus minus 2 times 1 and 3 plus minus 2 times minus 1. This becomes 0, 1, 0, 5. The remaining 2 rows of the new matrix are the same.

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 ?

The two equations are x minus y equals 2 and 4x minus 8y equals 0. Multiplying the first by minus 4, we get minus 4x plus 4y equals minus 8. Adding this to the second equation we get minus 4y equals minus 8.

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.

The 2 by 3 matrix is 1, minus 1, 2 and 4, minus 8, 0. Performing the operation minus 4R1 plus R2 on row 2, the second row of the new matrix becomes 0, minus 4, minus 8. The first row remains the same.

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.

A 2 by 3 matrix is shown on the left. Its first row is 1, a, b. Its second row is 0, 1, c. An arrow points diagonally down and right, overlapping both the 1s in the matrix. A 3 by 4 matrix is shown on the right. Its first row is 1, a, b, d. Its second row is 0, 1, c, e. Its third row is 0, 0, 1, f. An arrow points diagonally down and right, overlapping all the 1s in the matrix. a, b, c, d, e, f are real numbers.

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: \(\left\{ \begin{array} {l} 3x+4y=5 \\ x+2y=1 \end{array} \right. \)

The equations are 3x plus 4y equals 5 and x plus 2y equals 1. Step 1. Write the augmented matrix for the system of equations. We get a 2 by 3 matrix with first row 3, 4, 5 and second row 1, 2, 1.

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.

The figure shows 3 steps for a 2 by 3 matrix and 6 steps for a 3 by 4 matrix. For the former, step 1 is to get a 1 in row 1 column 1. Step to is to get a 0 is row 2 column 1. Step 3 is to get a 1 in row 2 column 2. For a 3 by 4 matrix, step 1 is to get a 1 in row 1 column 1. Step 2 is to get a 0 in row 2 column 1. Step 3 is to get a 0 in row 3 column 1. Step 4 is to get a 1 in row 2 column 2. Step 5 is to get a 0 in row 3 column 2. Step 6 is to get a 1 in row 3 column 3.

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.

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  • What is a linear equation?
  • A linear equation represents a straight line on a coordinate plane. It can be written in the form: y = mx + b where m is the slope of the line and b is the y-intercept.
  • How do you find the linear equation?
  • To find the linear equation you need to know the slope and the y-intercept of the line. To find the slope use the formula m = (y2 - y1) / (x2 - x1) where (x1, y1) and (x2, y2) are two points on the line. The y-intercept is the point at which x=0.
  • What are the 4 methods of solving linear equations?
  • There are four common methods to solve a system of linear equations: Graphing, Substitution, Elimination and Matrix.
  • How do you identify a linear equation?
  • Here are a few ways to identify a linear equation: Look at the degree of the equation, a linear equation is a first-degree equation. Check if the equation has two variables. Graph the equation.
  • What is the most basic linear equation?
  • The most basic linear equation is a first-degree equation with one variable, usually written in the form of y = mx + b, where m is the slope of the line and b is the y-intercept.

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

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How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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    Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables.

  13. Algebra

    Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.

  14. Linear Algebra Problems and Solution

    Linear Algebra Problems and Solutions. The topics in Linear Algebra are listed below. Each page contains definitions and summary of the topic followed by exercise problems. This is version 0 (11/15/2017), that is, still work in progress. Introduction to Matrices.

  15. Solving Linear Equations

    Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 --------- (1) x = 6 - y. Step 2: Substitute the value of the variable found in step 1 in the second linear equation.

  16. Linear equations word problems

    Math > Algebra 1 > Forms of linear equations > Intro to slope-intercept form Linear equations word problems Google Classroom Ever since Renata moved to her new home, she's been keeping track of the height of the tree outside her window. H represents the height of the tree (in centimeters), t years since Renata moved in. H = 210 + 33 t

  17. Linear Algebra Calculator

    Free linear algebra calculator - solve matrix and vector operations step-by-step

  18. Linear Equations

    Type a math problem Solve Examples 5 = 2x + 3 5b = −2b + 3 4r − 3 = 2r 3(a − 5) = 2(6 + a) n − 43n +6 = 2 Quiz 5 = 2x+3 4r−3 = 2r n−43n+6 = 2 Learn about linear equations using our free math solver with step-by-step solutions.

  19. 4.6: Solve Systems of Equations Using Matrices

    If you missed this problem, review . Solve: \(0.25p+0.25(x+4)=5.20\). ... 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. ... How to solve a system of equations using matrices. Write the ...

  20. 3,000 Solved Problems in Linear Algebra

    Master linear algebra with Schaum's--the high-performance solved-problem guide. It will help you cut study time, hone problem-solving skills, and achieve your personal best on exams! Students love Schaum's Solved Problem Guides because they produce results. Each year, thousands of students improve their test scores and final grades with these indispensable guides. Get the edge on your ...

  21. Algebra

    A linear equation is any equation that can be written in the form. ax +b = 0 a x + b = 0. where a a and b b are real numbers and x x is a variable. This form is sometimes called the standard form of a linear equation. Note that most linear equations will not start off in this form. Also, the variable may or may not be an x x so don't get too ...

  22. How To Solve Linear Equations In Algebra

    This algebra video explains how to solve linear equations. It contains plenty of examples and practice problems.Full 1 Hour Video on YouTube: ...

  23. Linear Equation Calculator

    How do you find the linear equation? To find the linear equation you need to know the slope and the y-intercept of the line. To find the slope use the formula m = (y2 - y1) / (x2 - x1) where (x1, y1) and (x2, y2) are two points on the line. The y-intercept is the point at which x=0.

  24. Solving Equations

    Example 1: solve equations involving like terms. Solve for x. x. 5q-4q=9 5q −4q = 9. Combine like terms. Combine the q q terms on the left side of the equation. To do this, subtract 4q 4q from both sides. (5 q-4 q)=9-4 q (5q −4q) = 9− 4q. The goal is to simplify the equation by combining like terms.