Question

The inverse of the matrix $\left[\begin{array}{ccc}7& -3& -3\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]$ is

• A. $\left[\begin{array}{ccc}1& 1& 1\\ 3& 4& 3\\ 3& 3& 4\end{array}\right]$
• B. $\left[\begin{array}{ccc}1& 3& 1\\ 4& 3& 8\\ 3& 4& 1\end{array}\right]$
• C. $\left[\begin{array}{ccc}1& 1& 1\\ 3& 3& 4\\ 3& 4& 3\end{array}\right]$
• D. $\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]$

Given $A=\left[\begin{array}{ccc}7& -3& -3\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]$
$A^{-1}=\frac{\text{Adj}A}{\text{det}A}$
We know that $\text{Adj}A=C^T$
where $C=$Co-factor matrix of $A$
$C=\left[\begin{array}{ccc}1-0& -(-1-0)& 0+1\\ -(-3)& 7-3& -(0-3)\\ 0+3& -(0-3)& 7-3\end{array}\right]$
$=\left[\begin{array}{ccc}1& 1& 1\\ 3& 4& 3\\ 3& 3& 4\end{array}\right]$
Now $C^T=\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]=\text{Adj}A$
$|A|=7\left(1\right)-(-3)(-1)+(-3)\left(1\right)=1$
$\Rightarrow A^{-1}=\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]$