Question

The inverse of the matrix $\begin{bmatrix}1&0&0\\ 3&3&0\\ 5&2&-1\end{bmatrix}$ is

• A. $ - \frac{1}{3} \begin{bmatrix} - 3 &0&0\\ 3&1&0\\ 9 &2& - 3 \end{bmatrix}$
• B. $ - \frac{1}{3} \begin{bmatrix} - 3 &0&0\\ 3& - 1&0\\ - 9 & - 2& 3 \end{bmatrix}$
• C. $ - \frac{1}{3} \begin{bmatrix} 3 &0&0\\ 3& - 1&0\\ - 9 & - 2& 3 \end{bmatrix}$
• D. $ - \frac{1}{3} \begin{bmatrix} - 3 &0&0\\ - 3& 1&0\\ - 9 & - 2& 3 \end{bmatrix}$

Answer 1

Answer: (B)

$A=\begin{bmatrix}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{bmatrix}$
$| A |=-3 \neq0$
$\therefore A ^{-1}$ exist
$A ^{-1}=\frac{1}{| A |}$ Adj $A$
$\text{Adj}( A )=$ Transpose of co-factor of $A$
$M _{11}=(-1)^{1+1}(-3-0)=-3, M _{12}=(-1)^{1+2}(-3-0)=3, M _{13}=(-1)^{1+3}(6-15)=-9$
$M _{21}=(-1)^{2+1}(0-0)=0, M _{22}=(-1)^{2+2}(-1+0)=-1, M _{23}=(-1)^{2+3}(2-0)=-2$
$M _{31}=(-1)^{3+1}(0-0)=0, M _{32}=(-1)^{3+2}(0-0)=0, M _{33}=(-1)^{3+3}(3-0)=3$
$\text{Adj}(A)=\begin{bmatrix}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{bmatrix}$
$\therefore A ^{-1}=-\frac{1}{3}\begin{bmatrix}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{bmatrix}$