Question

If A is a square matrix of order n x n and k is a scalar, then adj(kA) is equal to

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

Answer 1

Answer: (C)

$k^{n-1} \text{ adj A}$

Answer 2

Let A be a square matrix of order n x n and k be a scalar. We want to find adj(kA).

We know that for any square matrix M, $M (\text{adj } M) = (\text{adj } M) M = \text{det}(M) I_n$, where $I_n$ is the identity matrix of order n.

Let M = kA. Then $(kA) \text{adj}(kA) = \text{det}(kA) I_n$.

We also know the property that for a scalar k and an n x n matrix A, $\text{det}(kA) = k^n \text{det}(A)$.

Substituting this into the equation, we get:

$(kA) \text{adj}(kA) = k^n \text{det}(A) I_n$.

We also know that $A (\text{adj } A) = \text{det}(A) I_n$.

If A is invertible, then $\text{det}(A) \neq 0$, and $A^{-1} = \frac{1}{\text{det(A)}} \text{adj A}$.

Also, if $k \neq 0$, then kA is invertible if and only if A is invertible.

If $k \neq 0$ and A is invertible, then $(kA)^{-1} = k^{-1} A^{-1}$.

Multiply the equation $(kA) \text{adj}(kA) = k^n \text{det}(A) I_n$ by $(kA)^{-1}$ from the left:

$(kA)^{-1} (kA) \text{adj(kA)} = (kA)^{-1} k^n \text{det(A)} I_n$

$I_n \text{adj(kA)} = (k^{-1} A^{-1}) k^n \text{det(A)}$

$\text{adj(kA)} = k^{n-1} A^{-1} \text{det(A)}$

Since $A^{-1} \text{det(A)} = \text{adj A}$, we have:

$\text{adj(kA)} = k^{n-1} \text{adj A}$.