Question

For the matrix $A=\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]$, the inverse of $\frac{A}{2}$ will be

• A. $2A$
• B. $\frac{A^{-1}}{4}$
• C. $\frac{A}{2}$
• D. $A^2$

Answer 1

The correct option is A $2A$

Given $A=\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]$
$A^2=A.A=\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]=I$

$\Rightarrow AA=I$

So, for $A,$ its inverse is $A$ itself

Now $\frac{A}{2}\cdot 2A=I$

So, the inverse of $\frac{A}{2}$ is $2A$
Alternate solution : we know, $(kA{)}^{-1}=\frac{1}{k}\cdot A^{-1}=2A$