Determinant of $2 \times 2$ block matrix

Question

How to find the determinant of the following block matrix?

$$\begin{bmatrix} A & A \\ A & kI \end{bmatrix}$$

Here, $A$ is any square matrix, $I$ is an identity matrix and $k$ is any constant.

Answer 1

$$\begin{bmatrix} A & A \\ A & kI \end{bmatrix}$$

is equivalent to the following block matrix

$$\begin{bmatrix} A & A \\ 0 & kI-A \end{bmatrix}$$

which is now a block triangular matrix with determinant

$$\det(A) \det(kI-A)$$

Answer 2

Or write your matrix like this $$\left(\begin{array}{cc} A & A \\ A & kI \end{array}\right)=\left(\begin{array}{cc} 0 & \frac{1}{k}A \\ A-kI & I \end{array}\right)\left(\begin{array}{cc} I & 0 \\ kI & kI \end{array}\right)$$ Then properties of determinants give $$\det\left(\left(\begin{array}{cc} A & A \\ A & kI \end{array}\right)\right)=\frac{1}{k}\det(A-kI)\det(A)k=\det(A-kI)\det(A)$$