Question

If $A=\left[\begin{array}{ccc}5& 0& 2\\ 0& 3& 0\\ 2& 0& 1\end{array}\right],then{A}^{-1}=$

• A. $\left[\begin{array}{ccc}1& 0& -2\\ 0& \frac{1}{3}& 0\\ -2& 0& 5\end{array}\right]$
• B. $\left[\begin{array}{ccc}5& 0& 2\\ 0& -\frac{1}{3}& 0\\ 2& 0& 1\end{array}\right]$
• C. $\left[\begin{array}{ccc}\frac{1}{5}& 0& \frac{1}{2}\\ 0& \frac{1}{3}& 0\\ \frac{1}{2}& 0& 1\end{array}\right]$
• D. $\left[\begin{array}{ccc}\frac{1}{5}& 0& -\frac{1}{2}\\ 0& \frac{1}{3}& 0\\ -\frac{1}{2}& 0& 1\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{ccc}1& 0& -2\\ 0& \frac{1}{3}& 0\\ -2& 0& 5\end{array}\right]$

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

$\text{Determinant of matrix [A] is}\left|A\right|=(5\times 3)+(2\times -6)=3\ne 0$

$\Rightarrow A^{-1}\text{exists}.$

$\text{Co-factor matrix of}A=\left[\begin{array}{ccc}3& 0& -6\\ 0& 1& 0\\ -6& 0& 15\end{array}\right]$

Adjoint of the matrix [A] is the transposed matrix of co-factors of [A].
$\therefore i.e.,Adj\left[A\right]=\left[Cof\right(A){]}^T={\left[\begin{array}{ccc}3& 0& -6\\ 0& 1& 0\\ -6& 0& 15\end{array}\right]}^T$
=$\left[\begin{array}{ccc}3& 0& -6\\ 0& 1& 0\\ -6& 0& 15\end{array}\right]$
$\Rightarrow [A{]}^{-1}=\frac{adj\left[A\right]}{\left|A\right|}$=$\left[\begin{array}{ccc}1& 0& -2\\ 0& \frac{1}{3}& 0\\ -2& 0& 5\end{array}\right]$