Question

If matrix $A=\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]$ such that $Ax=I$, then $x=$................

• A. $\frac{1}{5}\left[\begin{array}{cc}1& 3\\ 2& -1\end{array}\right]$
• B. $\frac{1}{5}\left[\begin{array}{cc}4& 2\\ 4& -1\end{array}\right]$
• C. $\frac{1}{5}\left[\begin{array}{cc}-3& 2\\ 4& -1\end{array}\right]$
• D. $\frac{1}{5}\left[\begin{array}{cc}-1& 2\\ -1& 4\end{array}\right]$

Answer 1

The correct option is C $\frac{1}{5}\left[\begin{array}{cc}-3& 2\\ 4& -1\end{array}\right]$

Given : $A=\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]$ such that $Ax=I$ ... $\left(i\right)$

Since, $A$ and $I$ are of order $2\times 2$. So, $x$ will be a matrix of order $2\times 2$

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

From $\left(i\right)$, we get

$Ax=I$

$⟹\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$⟹\left[\begin{array}{cc}a+2c& b+2d\\ 4a+3c& 4b+3d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$⟹a+2c=1$ ...... $\left(ii\right),$ $4a+3c=0$ ........ $\left(iii\right)$

$b+2d=0$ ....... $\left(iv\right)$ and $4b+3d=1$ ....... $\left(v\right)$

Solving $\left(ii\right)$ and $\left(iii\right)$ simultaneously we get

$a=-\frac{3}{5}$ and $c=\frac{4}{5}$

Then solving $\left(iv\right)$ and $\left(v\right)$ simultaneously we get

$b=\frac{2}{5}$ and $d=-\frac{1}{5}$

Substituting all these values in $x$, we get

$x=\left[\begin{array}{cc}-\frac{3}{5}& \frac{2}{5}\\ \frac{4}{5}& -\frac{1}{5}\end{array}\right]$

$\therefore x=\frac{1}{5}\left[\begin{array}{cc}-3& 2\\ 4& -1\end{array}\right]$