Solving Systems with Inverses: Learn It 4

Giselle Miller

Solving Systems with Inverses: Learn It 4

Solving a System of Linear Equations Using the Inverse of a Matrix

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: [latex]X[/latex] is the matrix representing the variables of the system, and [latex]B[/latex] 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 [latex]AX=B[/latex]

system of equation in matrix form

To solve a system of linear equations using an inverse matrix, let [latex]A[/latex] be the coefficient matrix, let [latex]X[/latex] be the variable matrix, and let [latex]B[/latex] be the constant matrix. Thus, we want to solve a system [latex]AX=B[/latex].

For example, look at the following system of equations.

[latex]\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}[/latex]

From this system, the coefficient matrix is

[latex]A=\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right][/latex]

The variable matrix is

[latex]X=\left[\begin{array}{c}x\\ y\end{array}\right][/latex]

And the constant matrix is

[latex]B=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right][/latex]

Then [latex]AX=B[/latex] looks like

[latex]\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right][/latex]

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, [latex]\left({2}^{-1}\right)2=\left(\frac{1}{2}\right)2=1[/latex].
[latex]\\[/latex]
To solve a single linear equation [latex]ax=b[/latex] for [latex]x[/latex], we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of [latex]a[/latex]. Thus,

[latex]\begin{array}{c}\text{ }ax=b\\ \text{ }\left(\frac{1}{a}\right)ax=\left(\frac{1}{a}\right)b\\ \left({a}^{-1}\text{ }\right)ax=\left({a}^{-1}\right)b\\ \left[\left({a}^{-1}\right)a\right]x=\left({a}^{-1}\right)b\\ \text{ }1x=\left({a}^{-1}\right)b\\ \text{ }x=\left({a}^{-1}\right)b\end{array}[/latex]

The only difference between 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.

Given a system of equations, write the coefficient matrix [latex]A[/latex], the variable matrix [latex]X[/latex], and the constant matrix [latex]B[/latex]. Then [latex]AX=B[/latex]. Multiply both sides by the inverse of [latex]A[/latex] to obtain the solution.

[latex]\begin{array}{r}\hfill \left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\\ \hfill \left[\left({A}^{-1}\right)A\right]X=\left({A}^{-1}\right)B\\ \hfill IX=\left({A}^{-1}\right)B\\ \hfill X=\left({A}^{-1}\right)B\end{array}[/latex]

Solve the given system of equations using the inverse of a matrix.

[latex]\begin{array}{r}\hfill 3x+8y=5\\ \hfill 4x+11y=7\end{array}[/latex]

Can we solve for [latex]X[/latex] by finding the product [latex]B{A}^{-1}?[/latex]

No, recall that matrix multiplication is not commutative, so [latex]{A}^{-1}B\ne B{A}^{-1}[/latex]. Consider our steps for solving the matrix equation.

[latex]\begin{array}{r}\hfill \left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\\ \hfill \left[\left({A}^{-1}\right)A\right]X=\left({A}^{-1}\right)B\\ \hfill IX=\left({A}^{-1}\right)B\\ \hfill X=\left({A}^{-1}\right)B\end{array}[/latex]

Notice in the first step we multiplied both sides of the equation by [latex]{A}^{-1}[/latex], but the [latex]{A}^{-1}[/latex] was to the left of [latex]A[/latex] on the left side and to the left of [latex]B[/latex] on the right side. Because matrix multiplication is not commutative, order matters.

Solve the following system using the inverse of a matrix.

[latex]\begin{array}{r}\hfill 5x+15y+56z=35\\ \hfill -4x - 11y - 41z=-26\\ \hfill -x - 3y - 11z=-7\end{array}[/latex]

Write the equation [latex]AX=B[/latex].

[latex]\left[\begin{array}{ccc}5& 15& 56\\ -4& -11& -41\\ -1& -3& -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{r}\hfill 35\\ \hfill -26\\ \hfill -7\end{array}\right][/latex]

First, we will find the inverse of [latex]A[/latex] by augmenting with the identity.

[latex]\left[\begin{array}{ccc|ccc}\hfill 5& \hfill 15& \hfill 56& \hfill 1& \hfill 0& \hfill0\\ \hfill -4& \hfill -11& \hfill -41& \hfill 0& \hfill 1& \hfill 0\\ \hfill-1& \hfill -3& \hfill -11& \hfill 0& \hfill 0& \hfill 1\end{array}\right][/latex]

Multiply row 1 by [latex]\frac{1}{5}[/latex].

[latex]\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 3& \hfill \frac{56}{5}& \hfill \frac{1}{5}& \hfill 0& \hfill0\\ \hfill -4& \hfill -11& \hfill -41& \hfill 0& \hfill 1& \hfill 0\\ \hfill-1& \hfill -3& \hfill -11& \hfill 0& \hfill 0& \hfill 1\end{array}\right][/latex]

Multiply row 1 by 4 and add to row 2.

[latex]\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 3& \hfill \frac{56}{5}& \hfill \frac{1}{5}& \hfill 0& \hfill0\\ \hfill 0& \hfill 1& \hfill \frac{19}{5}& \hfill \frac{4}{5}& \hfill 1& \hfill 0\\ \hfill-1& \hfill -3& \hfill -11& \hfill 0& \hfill 0& \hfill 1\end{array}\right][/latex]

Add row 1 to row 3.

[latex]\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 3& \hfill \frac{56}{5}& \hfill \frac{1}{5}& \hfill 0& \hfill0\\ \hfill 0& \hfill 1& \hfill \frac{19}{5}& \hfill \frac{4}{5}& \hfill 1& \hfill 0\\ \hfill0& \hfill 0& \hfill \frac{1}{5}& \hfill \frac{1}{5}& \hfill 0& \hfill 1\end{array}\right][/latex]

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

[latex]\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill -\frac{1}{5}& \hfill -\frac{11}{5}& \hfill -3& \hfill0\\ \hfill 0& \hfill 1& \hfill \frac{19}{5}& \hfill \frac{4}{5}& \hfill 1& \hfill 0\\ \hfill0& \hfill 0& \hfill \frac{1}{5}& \hfill \frac{1}{5}& \hfill 0& \hfill 1\end{array}\right][/latex]

Multiply row 3 by 5.

[latex]\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill -\frac{1}{5}& \hfill -\frac{11}{5}& \hfill -3& \hfill0\\ \hfill 0& \hfill 1& \hfill \frac{19}{5}& \hfill \frac{4}{5}& \hfill 1& \hfill 0\\ \hfill0& \hfill 0& \hfill 1& \hfill 1& \hfill 0& \hfill 5\end{array}\right][/latex]

Multiply row 3 by [latex]\frac{1}{5}[/latex] and add to row 1.

[latex]\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -2& \hfill -3& \hfill 1\\ \hfill 0& \hfill 1& \hfill \frac{19}{5}& \hfill \frac{4}{5}& \hfill 1& \hfill 0\\ \hfill0& \hfill 0& \hfill 1& \hfill 1& \hfill 0& \hfill 5\end{array}\right][/latex]

Multiply row 3 by [latex]-\frac{19}{5}[/latex] and add to row 2.

[latex]\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -2& \hfill -3& \hfill 1\\ \hfill 0& \hfill 1& \hfill 0& \hfill -3& \hfill 1& \hfill -19\\ \hfill0& \hfill 0& \hfill 1& \hfill 1& \hfill 0& \hfill 5\end{array}\right][/latex]

So,

[latex]{A}^{-1}=\left[\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right][/latex]

Multiply both sides of the equation by [latex]{A}^{-1}[/latex]. We want

[latex]{A}^{-1}AX={A}^{-1}B:[/latex]

[latex]\left[\begin{array}{rrr}\hfill -2& \hfill -3& \hfill 1\\ \hfill -3& \hfill 1& \hfill -19\\ \hfill 1& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill 5& \hfill 15& \hfill 56\\ \hfill -4& \hfill -11& \hfill -41\\ \hfill -1& \hfill -3& \hfill -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{rrr}\hfill -2& \hfill -3& \hfill 1\\ \hfill -3& \hfill 1& \hfill -19\\ \hfill 1& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{r}\hfill 35\\ \hfill -26\\ \hfill -7\end{array}\right][/latex]

Thus,

[latex]{A}^{-1}B=\left[\begin{array}{r}\hfill -70+78 - 7\\ \hfill -105 - 26+133\\ \hfill 35+0 - 35\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right][/latex]

The solution is [latex]\left(1,2,0\right)[/latex].

How To: Given a system of equations, solve with matrix inverses using a calculator

Save the coefficient matrix and the constant matrix as matrix variables [latex]\left[A\right][/latex] and [latex]\left[B\right][/latex].
Enter the multiplication into the calculator, calling up each matrix variable as needed.
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.

Solve the system of equations with matrix inverses using a calculator

[latex]\begin{array}{l}2x+3y+z=32\hfill \\ 3x+3y+z=-27\hfill \\ 2x+4y+z=-2\hfill \end{array}[/latex]