Question

Let A be the $\mathbf{2}\mathbf{\times }\mathbf{2}$matrix.

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

Show that if $\mathit{a}\mathit{d}\mathbf{-}\mathit{b}\mathit{c}\mathbf{\ne }\mathbf{𝟘}$, then ${\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\left[\begin{array}{cc}\frac{d}{ad-bc}& \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right]$

Short answer

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

Step-by-step solution

Step 1:

Given:

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$and role="math" localid="1668443693934" $ad-bc\ne 0$.

To proof:

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

PROOF:

$A^{-1}$is the inverse of A if $A^{-1}$.

FIRST PART:

$A^{-1}A=\left[\begin{array}{cc}\frac{d}{ad-bc}& -\frac{b}{ad-bc}\\ -\frac{c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right]\left[\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right]$

The $\left(i,j\right)$element of the product AB is the sum of the products of each element in the $i^{th}$row of A with the corresponding element in the $jth$column of B.

$\begin{array}{r}=\left[\begin{array}{ll}\frac{ad}{ad-bc}-\frac{bc}{ad-bc}& \frac{bd}{ad-bc}-\frac{bd}{ad-bc}\\ \frac{cd}{ad-bc}-\frac{cd}{ad-bc}& -\frac{bc}{ad-bc}+\frac{ad}{ad-bc}\end{array}\right]\\ =\left[\begin{array}{ll}\frac{ad-bc}{ad-bc}& 0\\ 0& \frac{ad-bc}{ad-bc}\end{array}\right]\\ =\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ =I\end{array}$

Step 2:

SECOND PART:

$A{A}^{-1}=\left[\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right]\left[\begin{array}{cc}\frac{d}{ad-bc}& -\frac{b}{ad-bc}\\ -\frac{c}{ad-bc}& \frac{a}{ad-bc}\end{array}\right]$

The $(i,j)th$element of the product AB is the sum of the products of each element in the $i^{th}$row of A with the corresponding element in the j thcolumn of B.

$\begin{array}{r}=\left[\begin{array}{cc}\frac{ad}{ad-bc}-\frac{bc}{ad-bc}& -\frac{bd}{ad-bc}+\frac{bd}{ad-bc}\\ \frac{cd}{ad-bc}-\frac{cd}{ad-bc}& -\frac{bc}{ad-bc}+\frac{ad}{ad-bc}\end{array}\right]\\ =\left[\begin{array}{lc}\frac{ad-bc}{ad-bc}& 0\\ 0& \frac{ad-bc}{ad-bc}\end{array}\right]\\ =I\text{.}\end{array}$.

CONCLUSION:

We have shown $A^{-1}A=A{A}^{-1}=I_n$and thus $A^{-1}$is the inverse of A.