Question

The inverse of the matrix $\left[\begin{array}{ccc}1& 0& 0\\ 3& 3& 0\\ 5& 2& -1\end{array}\right]$ is

• A. $-\frac{1}{3}\left[\begin{array}{ccc}-3& 0& 0\\ 3& 1& 0\\ 9& 2& -3\end{array}\right]$
• B. $-\frac{1}{3}\left[\begin{array}{ccc}-3& 0& 0\\ 3& -1& 0\\ -9& -2& 3\end{array}\right]$
• C. $-\frac{1}{3}\left[\begin{array}{ccc}3& 0& 0\\ 3& -1& 0\\ -9& -2& 3\end{array}\right]$
• D. $-\frac{1}{3}\left[\begin{array}{ccc}-3& 0& 0\\ -3& -1& 0\\ -9& -2& 3\end{array}\right]$

Answer 1

The correct option is B $-\frac{1}{3}\left[\begin{array}{ccc}-3& 0& 0\\ 3& -1& 0\\ -9& -2& 3\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 0& 0\\ 3& 3& 0\\ 5& 2& -1\end{array}\right]|A|=-3\ne 0$

$\therefore A^{-1}$ exist
$A^{-1}=\frac{1}{|A|}AdjA$

$Adj\left(A\right)=$ Transpose of co-factor of $A$

$M_{11}=(-1{)}^{1+1}(-3-0)=-3$, $M_{12}=(-1{)}^{1+2}(-3-0)=3$, $M_{13}=(-1{)}^{1+3}(6-15)=-9$

$M_{21}=(-1{)}^{2+1}(0-0)=0$, $M_{22}=(-1{)}^{2+2}(-1+0)=-1$, $M_{23}=(-1{)}^{2+3}(2-0)=-2$

$M_{31}=(-1{)}^{3+1}(0-0)=0$, $M_{32}=(-1{)}^{3+2}(0-0)=0$, $M_{33}=(-1{)}^{3+3}(3-0)=3$

$Adj\left(A\right)=\left[\begin{array}{ccc}-3& 0& 0\\ 3& -1& 0\\ -9& -2& 3\end{array}\right]$
$\therefore A^{-1}=-\frac{1}{3}\left[\begin{array}{ccc}-3& 0& 0\\ 3& -1& 0\\ -9& -2& 3\end{array}\right]$