Question

Using elementary transformation, find the inverse of the matrix $A=\left[\begin{array}{cc}a& b\\ c& \left(\frac{1+bc}{a}\right)\end{array}\right]$.

• A. $\Rightarrow A^{-1}=\left[\begin{array}{cc}\frac{1+bc}{a}& b\\ -c& a\end{array}\right]$
• B. $\Rightarrow A^{-1}=\left[\begin{array}{cc}\frac{1+bc}{a}& -b\\ c& a\end{array}\right]$
• C. $\Rightarrow A^{-1}=\left[\begin{array}{cc}\frac{1+bc}{a}& b\\ c& a\end{array}\right]$
• D. None of these.

Answer 1

The correct option is D None of these.

$A=\left[\begin{array}{cc}a& b\\ c& \left(\frac{1+bc}{a}\right)\end{array}\right]$
We
write, $A=I.A$
$\Rightarrow \left[\begin{array}{cc}a& b\\ c& \left(\frac{1+bc}{a}\right)\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}1& \frac{b}{a}\\ c& \left(\frac{1+bc}{a}\right)\end{array}\right]=\left[\begin{array}{cc}\frac{1}{a}& 0\\ 0& 1\end{array}\right]A$
$(R_1\to \frac{R_1}{a})$
or $\left[\begin{array}{cc}1& \frac{b}{a}\\ 0& \frac{1}{a}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{a}& 0\\ \frac{-c}{a}& 1\end{array}\right]A$ $(R_2\to R_2-c{R}_1)$
or
$\left[\begin{array}{cc}1& \frac{b}{a}\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{a}& 0\\ -c& a\end{array}\right]A$
$(R_2\to a{R}_2)$
or $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1+bc}{a}& -b\\ -c& a\end{array}\right]A$ $(R_1\to R_1-\frac{b}{a}{R}_2)$
$\Rightarrow A^{-1}=\left[\begin{array}{cc}\frac{1+bc}{a}& -b\\ -c& a\end{array}\right]$