Question

The inverse of the matrix $\left[\begin{array}{ccc}2& 0& 0\\ 0& 3& 0\\ 0& 0& 4\end{array}\right]$ is

• A. $\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& \frac{1}{4}\end{array}\right]$
• B. $\frac{1}{24}\left[\begin{array}{ccc}2& 0& 0\\ 0& 3& 0\\ 0& 0& 4\end{array}\right]$
• C. $\frac{1}{24}\left[\begin{array}{ccc}-2& 0& 0\\ 0& -3& 0\\ 0& 0& -4\end{array}\right]$
• D. $\left[\begin{array}{ccc}-\frac{1}{2}& 0& 0\\ 0& -\frac{1}{3}& 0\\ 0& 0& -\frac{1}{4}\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& \frac{1}{4}\end{array}\right]$

We know that $A^{-1}=\frac{1}{|A|}adj\left(A\right)$

So, the determinant of matrix A= $2det\left(\begin{array}{cc}3& 0\\ 0& 4\end{array}\right)-0det\left(\begin{array}{cc}0& 0\\ 0& 4\end{array}\right)-0det\left(\begin{array}{cc}0& 3\\ 0& 0\end{array}\right)=2\times 12-0-0=24$

Cofactor of matrix $A$ =$\left(\begin{array}{ccc}12& 0& 0\\ 0& 8& 0\\ 0& 0& 6\end{array}\right)$

$adj\left(A\right)$=${\left(\begin{array}{ccc}12& 0& 0\\ 0& 8& 0\\ 0& 0& 6\end{array}\right)}^T=\left(\begin{array}{ccc}12& 0& 0\\ 0& 8& 0\\ 0& 0& 6\end{array}\right)$

Therefore,

$A^{-1}=\frac{1}{24}\left(\begin{array}{ccc}12& 0& 0\\ 0& 8& 0\\ 0& 0& 6\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& \frac{1}{4}\end{array}\right)$