Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: X is the matrix representing the variables of the system, and B is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as AX = B To solve a system of linear equations using an inverse matrix, let A be the coefficient matrix, let X be the variable matrix, and let B be the constant matrix. Thus, we want to solve a system AX = B. For example, look at the following system of equations. [latex-display]\displaystyle{a}_{{1}}{x}+{b}_{{1}}{y}={c}_{{1}}[/latex-display] [latex-display]\displaystyle{a}_{{2}}{x}+{b}_{{2}}{y}={c}_{{2}}[/latex-display] From this system, the coefficient matrix is [latex-display]\displaystyle{A}={\left[\matrix{{a}_{{1}}&{b}_{{1}}\{a}_{{2}}&{b}_{{2}}}]}[/latex-display] The variable matrix is [latex-display]\displaystyle{X}={\left[\matrix{{x}\{y}}]}[/latex-display] And the constant matrix is [latex-display]\displaystyle{B}={\left[\matrix{{c}_{{1}}\{c}_{{2}}}]}[/latex-display] Then looks like [latex-display]\displaystyle{\left[\matrix{{a}_{{1}}&{b}_{{1}}\{a}_{{2}}&{b}_{{2}}}]}{\left[\matrix{{x}\{y}}]}={\left[\matrix{{c}_{{1}}\{c}_{{2}}}]}[/latex-display] Recall the discussion earlier in this section regarding multiplying a real number by its inverse, [latex-display]\displaystyle{({2}^{{-{1}}})}{2}={(\frac{{1}}{{2}})}{2}={1}[/latex-display] To solve a single linear equation ax = b for x we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of a. Thus, [latex-display]\displaystyle{a}{x}={b}[/latex-display] [latex-display]\displaystyle{(\frac{{1}}{{a}})}{a}{x}={(\frac{{1}}{{a}})}{b}[/latex-display] [latex-display]\displaystyle{({a}^{{-{1}}})}{a}{x}={({a}^{{-{1}}})}{b}[/latex-display] [latex-display]\displaystyle{[{({a}^{{-{1}}})}{a}]}{x}={({a}^{{-{1}}})}{b}[/latex-display] [latex-display]\displaystyle{1}{x}={({a}^{{-{1}}})}{b}[/latex-display] [latex-display]\displaystyle{x}={({a}^{{-{1}}})}{b}[/latex-display] The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable. We will investigate this idea in detail, but it is helpful to begin with a 2 × 2 system and then move on to a 3 × 3 system.

Solving a System of Equations Using the Inverse of a Matrix

Given a system of equations, write the coefficient matrix A, the variable matrix X, and the constant matrix B. Then AX = B Multiply both sides by the inverse of A to obtain the solution. [latex-display]\displaystyle{({A}^{{-{1}}})}{A}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{[{({A}^{{-{1}}})}{A}]}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{I}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{X}={({A}^{{-{1}}})}{B}[/latex-display]

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Q&A

If the coefficient matrix does not have an inverse, does that mean the system has no solution? No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.

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Example 1

Solve the given system of equations using the inverse of a matrix. 3 x + 8y = 5 4 x + 11y = 7

Solution

Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix. [latex-display]\displaystyle{A}={\left[\matrix{{3}&{8}\{4}&{11}}]},{X}={\left[\matrix{{x}\{y}}]},{B}={\left[\matrix{{5}\{7}}]}[/latex-display] Then [latex-display]\displaystyle{\left[\matrix{{3}&{8}\{4}&{11}}]}{\left[\matrix{{x}\{y}}]}={\left[\matrix{{5}\{7}}]}[/latex-display] First, we need to calculate A^–1 Using the formula to calculate the inverse of a 2 × 2 matrix, we have: [latex-display]\displaystyle{A}^{{-{1}}}=\frac{{1}}{{{a}{d}-{b}{c}}}{\left[\matrix{{d}&-{b}\-{c}&{a}}]}[/latex-display] [latex-display]\displaystyle=\frac{{1}}{{{3}{({11})}-{8}{({4})}}}{\left[\matrix{{11}&-{8}\-{4}&{3}}]}=\frac{{1}}{{1}}{\left[\matrix{{11}&-{8}\-{4}&{3}}]}[/latex-display] So, [latex-display]\displaystyle{A}^{{-{1}}}={\left[\matrix{{11}&-{8}\-{4}&{3}}]}[/latex-display] Now we are ready to solve. Multiply both sides of the equation by A ^–1. [latex-display]\displaystyle{({A}^{{-{{1}}}})}{A}{X}={({A}^{{-{1}}})}{B}[/latex-display] [latex-display]\displaystyle{\left[\matrix{{11}&-{8}\-{4}&{3}}]}{\left[\matrix{{3}&{8}\{4}&{11}}]}{\left[\matrix{{x}\{y}}]}={\left[\matrix{{11}&-{8}\-{4}&{3}}]}{\left[\matrix{{5}\{7}}]}[/latex-display] [latex-display]\displaystyle{\left[\matrix{{1}&{0}\{0}&{1}}]}{\left[\matrix{{x}\{y}}]}={\left[\matrix{{11}{({5})}+-{8}{({7})}\-{4}{({5})}+{3}{({7})}}]}[/latex-display] [latex-display]\displaystyle{\left[\matrix{{x}\{y}}]}={\left[\matrix{-{1}\{1}}]}[/latex-display] The solution is (–1,1).

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Q&A

Can we solve for X by finding the product BA^–1? No, recall that matrix multiplication is not commutative, so A^–1B ≠ BA^–1 Consider our steps for solving the matrix equation. [latex-display]\displaystyle{({A}^{{-{1}}})}{A}{X}={({A}^{{-{1}}})}{B}[/latex-display]   [latex-display]\displaystyle{[{({A}^{{-{1}}})}{A}]}{X}={({A}^{{-{1}}})}{B}[/latex-display]   [latex-display]\displaystyle{I}{X}={({A}^{{-{1}}})}{B}[/latex-display]   [latex-display]\displaystyle{X}={({A}^{{-{1}}})}{B}[/latex-display] Notice in the first step we multiplied both sides of the equation by A^–1, but the A^–1, was to the left of A on the left side and to the left of B on the right side. Because matrix multiplication is not commutative, order matters.

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Example 2

1. Solve the following system using the inverse of a matrix. 5x + 15y + 56z = 35 −4x − 11y − 41z = −26 −x −3y − 11z = −7
2. Solve the system using the inverse of the coefficient matrix. 2x − 17y + 11z = 0 −x + 11y − 7z = 8 3y − 2z = −2

Solutions

1. Write the equation AX = B [latex]\displaystyle{\left[\matrix{{5}&{15}&{56}\-{4}&-{11}&-{41}\-{1}&-{3}&-{11}}]}{\left[\matrix{{x}\{y}\{z}}]}={\left[\matrix{{35}\-{26}\-{7}}]}[/latex] First we will find the inverse of A by augmenting with the identity. [latex-display]\displaystyle{\left[\matrix{{5}&{15}&{56}&{\mid}&{1}&{0}&{0}\-{4}&-{11}&-{41}&{\mid}&{0}&{1}&{0}\-{1}&-{3}&-{11}&{\mid}&{0}&{0}&{1}}]}[/latex-display] Multiply row 1 by [latex-display]\displaystyle\frac{{1}}{{5}}[/latex-display] Multiply row 1 by 4 and add to row 2. [latex-display]\displaystyle{\left[\matrix{{1}&{3}&frac{{56}}{{5}}&{\mid}&frac{{1}}{{5}}&{0}&{0}\{0}&{1}&frac{{19}}{{5}}&{\mid}&frac{{4}}{{5}}&{1}&{0}\-{1}&-{3}&-{11}&{\mid}&{0}&{0}&{1}}]}[/latex-display] Add row 1 to row 3. [latex-display]\displaystyle{\left[\matrix{{1}&{3}&frac{{56}}{{5}}&{\mid}&frac{{1}}{{5}}&{0}&{0}\{0}&{1}&frac{{19}}{{5}}&{\mid}&frac{{4}}{{5}}&{1}&{0}\{0}&{0}&frac{{1}}{{5}}&{\mid}&frac{{1}}{{5}}&{0}&{1}}]}[/latex-display] Multiply row 2 by –3 and add to row 1. [latex-display]\displaystyle{\left[\matrix{{1}&{0}&-\frac{{1}}{{5}}&{\mid}&-\frac{{11}}{{5}}&-{3}&{0}\{0}&{1}&frac{{19}}{{5}}&{\mid}&frac{{4}}{{5}}&{1}&{0}\{0}&{0}&frac{{1}}{{5}}&{\mid}&frac{{1}}{{5}}&{0}&{1}}]}[/latex-display] Multiply row 3 by 5. [latex-display]\displaystyle{\left[\matrix{{1}&{0}&-\frac{{1}}{{5}}&{\mid}&-\frac{{11}}{{5}}&-{3}&{0}\{0}&{1}&frac{{19}}{{5}}&{\mid}&frac{{4}}{{5}}&{1}&{0}\{0}&{0}&{1}&{\mid}&{1}&{0}&{5}}]}[/latex-display] Multiply row 3 by [latex]\displaystyle\frac{{1}}{{5}}[/latex] Multiply row 3 by [latex]\displaystyle-\frac{{19}}{{5}}[/latex] So, [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{-{2}&-{3}&{1}\-{3}&{1}&-{19}\{1}&{0}&{5}}]}[/latex-display] Multiply both sides of the equation by A^–1. We want A^–1AX = A^–1B: Thus, [latex-display]\displaystyle{A}^{{-{{1}}}}{B}={\left[\matrix{-{70}+{78}-{7}\-{105}-{26}+{133}\{35}+{0}-{35}}]}={\left[\matrix{{1}\{2}\{0}}]}[/latex-display] The solution is (1,2,0).
2. [latex]\displaystyle{X}={\left[\matrix{{4}\{38}\{58}}]}[/latex]

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How To

Given a system of equations, solve with matrix inverses using a calculator.

1. Save the coefficient matrix and the constant matrix as matrix variables [A] and [B].
2. Enter the multiplication into the calculator, calling up each matrix variable as needed.
3. If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.

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Example 3

Solve the system of equations with matrix inverses using a calculator [latex-display]\displaystyle{\left[\matrix{{2}{x}+{3}{y}+{z}={32}\{3}{x}+{3}{y}+{z}=-{27}\{2}{x}+{4}{y}+{z}=-{2}}.}[/latex-display]

Solution

On the matrix page of the calculator, enter the coefficient matrix as the matrix variable [A] and enter the constant matrix as the matrix variable [B]. [latex-display]\displaystyle{[{A}]}={\left[\matrix{{2}&{3}&{1}\{3}&{3}&{1}\{2}&{4}&{1}}]},{[{B}]}={\left[\matrix{{32}\-{27}\-{2}}]}[/latex-display] On the home screen of the calculator, type in the multiplication to solve for X, calling up each matrix variable as needed. [A]^–1 × [B] Evaluate the expression. [latex-display]\displaystyle{\left[\matrix{-{59}\-{34}\{252}}]}[/latex-display]

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Media

Access these online resources for additional instruction and practice with solving systems with inverses.  

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Key Equations

Identity matrix for a matrix	[latex]\displaystyle{I}_{{2}}={\left[\matrix{{1}&{0}\{0}&{1}}]}[/latex]
Identity matrix for a matrix	[latex]\displaystyle{I}_{{3}}={\left[\matrix{{1}&{0}&{0}\{0}&{1}&{0}\{0}&{0}&{1}}]}[/latex]
Multiplicative inverse of a matrix

Key Concepts

• An identity matrix has the property [latex]\displaystyle{A}{I}={I}{A}={A}[/latex].
• An invertible matrix has the property [latex]\displaystyle{A}{A}^{{-{{1}}}}={A}^{{-{{1}}}}{A}={I}[/latex].
• Use matrix multiplication and the identity to find the inverse of a 2 × 2 matrix.
• The multiplicative inverse can be found using a formula.
• Another method of finding the inverse is by augmenting with the identity.
• We can augment a 3 × 3 matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse.
• Write the system of equations as [latex]\displaystyle{A}{X}={B}[/latex].
• We can also use a calculator to solve a system of equations with matrix inverses.

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OpenStax, Precalculus, "Solving a System of Linear Equations Using the Inverse of a Matrix," licensed under a CC BY 3.0 license. Mathispower4u, "The Identity Matrix," licensed under a Standard YouTube license. Mathispower4u, " Determining Inverse Matrices Using Augmented Matrices," licensed under a Standard YouTube license. Mathispower4u, " Using a Matrix Equation to Solve a System of Equations," licensed under a Standard YouTube license.