Question

If A is a square matrix of order $n\times n$ and k is a scalar, then $adj\left(kA\right)$ is equal to

• A. $kadjA$
• B. $k^{n}adjA$
• C. $k^{n-1}adjA$
• D. $k^{n+1}adjA$

Answer 1

Answer: (C)

$k^{n-1}adjA$

First we have to write a formula from a general formula

$(kA{)}^{-1}=\frac{adj\left(kA\right)}{det\left(kA\right)}$ $Since$ $A^{-1}=\frac{adj\left(A\right)}{det\left(A\right)}$

$\Rightarrow$ $adj\left(kA\right)=det\left(kA\right)(kA{)}^{-1}$

$=$ $k^{n}det\left(A\right)k^{-1}.A^{-1}$

$=\frac{k^n}{k}det\left(A\right).A^{-1}$

$=k^{n-1}.det\left(A\right).\frac{adj\left(A\right)}{det\left(A\right)}$

$=k^{n-1}.adj\left(A\right)$