9.7 Solving systems with inverses (Page 2/8)

Precalculus

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Show that the following two matrices are inverses of each other.

A = [\begin{array}{r} 1 & 4 \\ −1 & −3 \end{array}], B = [\begin{array}{r} −3 & −4 \\ 1 & 1 \end{array}]

\begin{array}{l} A B = [\begin{array}{r} 1 & 4 \\ −1 & −3 \end{array}] \begin{array}{r}  \end{array} [\begin{array}{r} −3 & −4 \\ 1 & 1 \end{array}] = [\begin{array}{r} 1 (−3) + 4 (1) & 1 (−4) + 4 (1) \\ −1 (−3) + −3 (1) & −1 (−4) + −3 (1) \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] \\ B A = [\begin{array}{r} −3 & −4 \\ 1 & 1 \end{array}] \begin{array}{r}  \end{array} [\begin{array}{r} 1 & 4 \\ −1 & −3 \end{array}] = [\begin{array}{r} −3 (1) + −4 (−1) & −3 (4) + −4 (−3) \\ 1 (1) + 1 (−1) & 1 (4) + 1 (−3) \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] \end{array}

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Finding the multiplicative inverse using matrix multiplication

We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication .

Finding the multiplicative inverse using matrix multiplication

Use matrix multiplication to find the inverse of the given matrix.

A = [\begin{array}{r} 1 & −2 \\ 2 & −3 \end{array}]

For this method, we multiply $A$ by a matrix containing unknown constants and set it equal to the identity.

[\begin{array}{r} 1 & −2 \\ 2 & −3 \end{array}] [\begin{array}{r} a & b \\ c & d \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}]

Find the product of the two matrices on the left side of the equal sign.

[\begin{array}{r} 1 & −2 \\ 2 & −3 \end{array}] [\begin{array}{r} a & b \\ c & d \end{array}] = [\begin{array}{r} 1 a −2 c & 1 b −2 d \\ 2 a −3 c & 2 b −3 d \end{array}]

Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.

\begin{matrix} 1 a −2 c = 1 R_{1} \\ 2 a −3 c = 0 R_{2} \end{matrix}

Using row operations, multiply and add as follows: $(−2) R_{1} + R_{2} \to R_{2} .$ Add the equations, and solve for $c .$

\begin{array}{r} 1 a - 2 c = 1 \\ 0 + 1 c = - 2 \\ c = - 2 \end{array}

Back-substitute to solve for $a .$

\begin{array}{r} a −2 (−2) = 1 \\ a + 4 = 1 \\ a = −3 \end{array}

Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.

\begin{array}{r} 1 b −2 d = 0 & R_{1} \\ 2 b −3 d = 1 & R_{2} \end{array}

Using row operations, multiply and add as follows: $(−2) R_{1} + R_{2} = R_{2} .$ Add the two equations and solve for $d .$

\begin{array}{r} 1 b −2 d = 0 \\ \frac{0 + 1 d = 1}{d = 1} \end{array}

Once more, back-substitute and solve for $b .$

\begin{array}{r} b −2 (1) = 0 \\ b −2 = 0 \\ b = 2 \end{array}

A^{−1} = [\begin{array}{r} −3 & 2 \\ −2 & 1 \end{array}]

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Finding the multiplicative inverse by augmenting with the identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix $A$ is transformed into $I,$ the augmented matrix $I$ transforms into $A^{−1} .$

For example, given

A = [\begin{array}{r} 2 & 1 \\ 5 & 3 \end{array}]

augment $A$ with the identity

[\begin{array}{r} 2 & 1 \\ 5 & 3 \end{array} | \begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}]

Perform row operations with the goal of turning $A$ into the identity.

Switch row 1 and row 2.
$[\begin{array}{r} 5 & 3 \\ 2 & 1 \end{array} | \begin{array}{r} 0 & 1 \\ 1 & 0 \end{array}]$
Multiply row 2 by $−2$ and add to row 1.
$[\begin{array}{r} 1 & 1 \\ 2 & 1 \end{array} | \begin{array}{r} −2 & 1 \\ 1 & 0 \end{array}]$
Multiply row 1 by $−2$ and add to row 2.
$[\begin{array}{r} 1 & 1 \\ 0 & −1 \end{array} | \begin{array}{r} −2 & 1 \\ 5 & −2 \end{array}]$
Add row 2 to row 1.
$[\begin{array}{r} 1 & 0 \\ 0 & −1 \end{array} | \begin{array}{r} 3 & −1 \\ 5 & −2 \end{array}]$
Multiply row 2 by $−1.$
$[\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array} | \begin{array}{r} 3 & −1 \\ −5 & 2 \end{array}]$

The matrix we have found is $A^{−1} .$

A^{−1} = [\begin{array}{r} 3 & −1 \\ −5 & 2 \end{array}]

Finding the multiplicative inverse of 2×2 matrices using a formula

When we need to find the multiplicative inverse of a $2 \times 2$ matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.

If $A$ is a $2 \times 2$ matrix, such as

A = [\begin{array}{r} a & b \\ c & d \end{array}]

the multiplicative inverse of $A$ is given by the formula

A^{−1} = \frac{1}{a d - b c} [\begin{array}{r} d & - b \\ - c & a \end{array}]

where $a d - b c \neq 0.$ If $a d - b c = 0,$ then $A$ has no inverse.

Using the formula to find the multiplicative inverse of matrix A

Use the formula to find the multiplicative inverse of

A = [\begin{matrix} 1 & −2 \\ 2 & −3 \end{matrix}]

Using the formula, we have

\begin{array}{l} A^{−1} = \frac{1}{(1) (−3) - (−2) (2)} [\begin{matrix} −3 & 2 \\ −2 & 1 \end{matrix}] \\ = \frac{1}{−3 + 4} [\begin{matrix} −3 & 2 \\ −2 & 1 \end{matrix}] \\ = [\begin{matrix} −3 & 2 \\ −2 & 1 \end{matrix}] \end{array}

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Source: OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6

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