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Calculate $$$\left[\begin{array}{cc}a & b\\c & d\end{array}\right]^{-1}$$$ using the Gauss-Jordan elimination.

Solution

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be the inverse matrix.

So, augment the matrix with the identity matrix:

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

Divide row $$$1$$$ by $$$a$$$: $$$R_{1} = \frac{R_{1}}{a}$$$.

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

Subtract row $$$1$$$ multiplied by $$$c$$$ from row $$$2$$$: $$$R_{2} = R_{2} - c R_{1}$$$.

$$$\left[\begin{array}{cc|cc}1 & \frac{b}{a} & \frac{1}{a} & 0\\0 & d - \frac{b c}{a} & - \frac{c}{a} & 1\end{array}\right]$$$

Divide row $$$2$$$ by $$$d - \frac{b c}{a}$$$: $$$R_{2} = \frac{R_{2}}{d - \frac{b c}{a}}$$$.

$$$\left[\begin{array}{cc|cc}1 & \frac{b}{a} & \frac{1}{a} & 0\\0 & 1 & - \frac{c}{a d - b c} & \frac{a}{a d - b c}\end{array}\right]$$$

Subtract row $$$2$$$ multiplied by $$$\frac{b}{a}$$$ from row $$$1$$$: $$$R_{1} = R_{1} - \frac{b}{a} R_{2}$$$.

$$$\left[\begin{array}{cc|cc}1 & 0 & \frac{d}{a d - b c} & - \frac{b}{a d - b c}\\0 & 1 & - \frac{c}{a d - b c} & \frac{a}{a d - b c}\end{array}\right]$$$

We are done. On the left is the identity matrix. On the right is the inverse matrix.

Answer

The inverse matrix is $$$\left[\begin{array}{cc}\frac{d}{a d - b c} & - \frac{b}{a d - b c}\\- \frac{c}{a d - b c} & \frac{a}{a d - b c}\end{array}\right]$$$A.


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