Q. 89 - Precalculus Enhanced with Graphing Utilities

Question

Consider the $2$ by $2$ squre matrix

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

If $D=ad-bc\ne 0$, show that A is nonsingular and that

$A^{-1}=\frac{1}{D}\left[\begin{array}{r}d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-b\\ -c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\end{array}\right]$

Step-by-Step Solution

Answer

The matrix A is nonsingular.

Step 1. Reduce the augmented matrix into its echelon form

We have the given matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Let us reduce the augmented matrix $\left[A\mid I_2\right]$ into its echelon form.

$A=\left[\begin{array}{cccc}a& b& 1& 0\\ c& d& 0& 1\end{array}\right]$

Now perform the operation,$R_1=\frac{R_1}{2}$

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

$R_2=R_2-C{R}_1$

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

$R_2=\frac{a{R}_2}{ad-bc}$

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

$R_1=R_1-\frac{b{R}_2}{a}$

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

Step 2. Inverse of the matrix A

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

Here, let $D=ad-bc\ne 0$

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

Hence, we have proved that matrix A is a non-singular matrix.