If k is a scalar and I is a unit matrix of order 3,

Given:

$l$ is a unit matrix of order 3

As we know,

$A(\operatorname{adj} A)=|A| I$

$\Rightarrow A^{-1} A(\operatorname{adj} A)=A^{-1}|A| I$

$\Rightarrow\left(A^{-1} A\right)(\operatorname{adj} A)=A^{-1}|A| I$

$\Rightarrow I(\operatorname{adj} A)=|A|\left(A^{-1} I\right)$

$\Rightarrow \operatorname{adj} A=|A| A^{-1}$

$\Rightarrow \operatorname{adj}(k I)=|k I|(k I)^{-1}$

$\Rightarrow \operatorname{adj}(k I)=k^{3}|I|(k I)^{-1} \quad(\because$ order of $I$ is 3$)$

$\Rightarrow \operatorname{adj}(k I)=k^{3} \times 1 \times(k I)^{-1}$

$\Rightarrow \operatorname{adj}(k I)=k^{3} k^{-1}(I)^{-1}$

$\Rightarrow \operatorname{adj}(k I)=k^{2}(I)^{-1}$

$\Rightarrow \operatorname{adj}(k I)=k^{2} I \quad\left(\because I^{-1}=I\right)$

$\Rightarrow \operatorname{adj}(k I)=k^{2} I$

Hence, $\operatorname{adj}(k l)=\underline{k}^{2} \underline{I} .$