Question

The inverse of the matrix $\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$ is:

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

Answer 1

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

We know that the inverse of a matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is:

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

where $ad-bc$ is the determinant of matrix $A$ and the inverse of $A$ doesnot exist if the determinant of $A$ is zero.

Here, the given matrix is $A=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$, so, let us find the inverse of matrix $A$ as shown below:

$A^{-1}=\frac{1}{(1\times 3)-(2\times 2)}\left[\begin{array}{cc}3& -2\\ -2& 1\end{array}\right]=\frac{1}{3-4}\left[\begin{array}{cc}3& -2\\ -2& 1\end{array}\right]=-1\left[\begin{array}{cc}3& -2\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}-3& 2\\ 2& -1\end{array}\right]$

Hence, the inverse of the matrix $\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$ is $\left[\begin{array}{cc}-3& 2\\ 2& -1\end{array}\right]$.