Chapter 11.4, Problem 89AYU

To determine

To show: The inverse of the matrix $A=\left[\begin{array}{ll}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right]$ is $A^{-1}=\frac{1}{D}\left[\begin{array}{ll}d\hfill & -b\hfill \\ -c\hfill & a\hfill \end{array}\right]$ where $D=ad-bc\ne 0$ .

Answer to Problem 89AYU

The inverse of the matrix $A=\left[\begin{array}{ll}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right]$ is $A^{-1}=\frac{1}{D}\left[\begin{array}{ll}d\hfill & -b\hfill \\ -c\hfill & a\hfill \end{array}\right]$ where $D=ad-bc\ne 0$ .

Explanation of Solution

Given information:

A $2\times 2$ matrix $A=\left[\begin{array}{ll}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right]$ and determinant is $D=ad-bc\ne 0$ .

Calculation:

Consider the $2\times 2$ matrix $A=\left[\begin{array}{ll}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right]$ and determinant is $D=ad-bc\ne 0$ .

The inverse of the matrix, $A=\left[\begin{array}{ll}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right]$ .

Let the inverse of the matrix be $A^{-1}=\left[\begin{array}{ll}p\hfill & r\hfill \\ q\hfill & s\hfill \end{array}\right]$ .

Using the relation, $A{A}^{-1}=I_2$ , we get,

  $\left[\begin{array}{ll}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right]\left[\begin{array}{ll}p\hfill & r\hfill \\ q\hfill & s\hfill \end{array}\right]=\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$

Now, multiplying the above two matrices, we get,

  $\left[\begin{array}{ll}ap+bq\hfill & ar+bs\hfill \\ cp+dq\hfill & cr+ds\hfill \end{array}\right]=\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$

Now, we equate the corresponding elements of the two matrices. First we write the system of equations when equated to the first column and when system of equations when equated to the second column.

Combing the above two matrices such that the identity matrix is on the right hand side, we get,

  $\left[\begin{array}{llll}a\hfill & b\hfill & 1\hfill & 0\hfill \\ c\hfill & d\hfill & 0\hfill & 1\hfill \end{array}\right]$

Now, we apply the Gauss-Jordan elimination to find the inverse of $\left[\begin{array}{llll}a\hfill & b\hfill & 1\hfill & 0\hfill \\ c\hfill & d\hfill & 0\hfill & 1\hfill \end{array}\right]$ ,

First we have to make the entry $a_{11}$ as 1, we perform the row transformation as $R_1\iff \frac{R_1}{a}$ .

  $\left[\begin{array}{llll}1\hfill & \frac{b}{a}\hfill & \frac{1}{a}\hfill & 0\hfill \\ c\hfill & d\hfill & 0\hfill & 1\hfill \end{array}\right]$

Next, we have to make the entry $a_{21}$ as 0, we perform the row transformation as $R_2\to R_2-c{R}_1$ .

  $\left[\begin{array}{llll}1\hfill & \frac{b}{a}\hfill & \frac{1}{a}\hfill & 0\hfill \\ 0\hfill & \frac{ad-bc}{a}\hfill & -\frac{c}{a}\hfill & 1\hfill \end{array}\right]$

Next, we have to make the entry $a_{22}$ as 1, we perform the row transformation as $R_2\to \frac{a{R}_2}{ad-bc}$ .

  $\left[\begin{array}{llll}1\hfill & \frac{b}{a}\hfill & \frac{1}{a}\hfill & 0\hfill \\ 0\hfill & 1\hfill & \frac{-c}{\left(ad-bc\right)}\hfill & \frac{a}{ad-bc}\hfill \end{array}\right]$

Next, we have to make the entry $a_{12}$ as 0, we perform the row transformation as $R_1\to R_1-\frac{b}{a}{R}_2$ .

  $\left[\begin{array}{llll}1\hfill & 0\hfill & \frac{d}{ad-bc}\hfill & \frac{-b}{ad-bc}\hfill \\ 0\hfill & 1\hfill & \frac{-c}{\left(ad-bc\right)}\hfill & \frac{a}{ad-bc}\hfill \end{array}\right]$

Therefore, the inverse of the matrix $A=\left[\begin{array}{ll}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right]$ is $A^{{}^{-1}}=\left[\begin{array}{cc}\frac{d}{ad-bc}& \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right]$ .

Rewrite it as,

  $\begin{array}{c}{A}^{{}^{-1}}=\left[\begin{array}{cc}\frac{d}{ad-bc}& \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right]\\ =\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\\ =\frac{1}{D}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\end{array}$

Here, determinant is $D=ad-bc\ne 0$ . So A is nonsingular matrix.

Thus, it is shown that the inverse of the matrix $A=\left[\begin{array}{ll}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right]$ is $A^{-1}=\frac{1}{D}\left[\begin{array}{ll}d\hfill & -b\hfill \\ -c\hfill & a\hfill \end{array}\right]$ where $D=ad-bc\ne 0$ .