Question

If $A=$ $\left[\begin{array}{ccc}1& 2& 3\\ 3& -2& 1\\ 4& 2& 1\end{array}\right]$ then $A^{-1}$

• A. $A^2-23I$
• B. $\frac{A^2-23I}{40}$
• C. $23I-A^2$
• D. $\frac{23I-A^2}{40}$

Answer 1

Answer: (B)

$\frac{A^2-23I}{40}$

The inverse of matrix A is =

$A^{-1}=\frac{adj\left(A\right)}{Det\left(A\right)}$

The determinant of the matrix is

$=1(-2-2)-2(3-4)+3(6+8)$

$=-4+2+42=40$

And the cofactor of the matrix is

$=\left[\begin{array}{ccc}M11& -M12& M13\\ -M21& M22& -M23\\ M31& -M32& M33\end{array}\right]$

$=\left[\begin{array}{ccc}-4& 1& 14\\ 4& -11& 6\\ 8& 8& -8\end{array}\right]$

$\therefore$ Adjoint of the matrix is transpose of cofactor matrix

i.e. adjoint of the matrix is

$\left[\begin{array}{ccc}-4& 4& 8\\ 1& -11& 8\\ 14& 6& -8\end{array}\right]$

$\therefore A^{-1}=\frac{\left[\begin{array}{ccc}-4& 4& 8\\ 1& -11& 8\\ 14& 6& -8\end{array}\right]}{40}$

Now, $A^2-23I=\left[\begin{array}{ccc}19& 4& 8\\ 1& 12& 8\\ 14& 6& 15\end{array}\right]-\left[\begin{array}{ccc}23& 0& 0\\ 0& 23& 0\\ 0& 0& 23\end{array}\right]$

$A^2-23I=\left[\begin{array}{ccc}-4& 4& 8\\ 1& -11& 8\\ 14& 6& -8\end{array}\right]$

$\therefore A^{-1}=\frac{A^2-23I}{40}$

Hence the answer is option B