Question

Let $M$ be a $3 \times 3$ non-singular matrix with $det(M)=\alpha$. If $\left[M^{-1} adj(adj(M)]=K I\right.$, then the value of $K$ is

• A. 1
• B. $\alpha$
• C. $\alpha^2$
• D. $\alpha^3$

Answer 1

The Correct Option is B

Formulae:- $A ^{-1}=\frac{1}{| A |} adj( A )$
$A \cdot adj( A )=| A | I =adj( A ) \cdot A$
$adj(adj A)=| A |^{( n -2)} A$
Now, ${\left[ M ^{-1} adj(adj( M )]\right.}$
$=\frac{1}{| M |} adj( M ) \cdot| M |^{(3-2)} M$
$=adj( M ) \cdot M$
$=| M | I$
$=\alpha I$
$\therefore k =\alpha$