Question

Consider the matrix \( A = \begin{bmatrix} 4 & -1 \\ 12 & -3 \end{bmatrix} \).
The value of the determinant of \( A^5 \) is .......... (rounded off to two decimal places).

Hint:

If the determinant of a matrix is zero, the determinant of any power of that matrix will also be zero.

Answer 1

We are given the matrix \( A = \begin{bmatrix} 4 & -1 \\ 12 & -3 \end{bmatrix} \).

The determinant of matrix \( A \) is given by:
\[ \text{det}(A) = (4 \times -3) - (-1 \times 12) = -12 + 12 = 0 \]

Since the determinant of \( A \) is 0, the determinant of any power of \( A \), including \( A^5 \), will also be 0:
\[ \text{det}(A^5) = (\text{det}(A))^5 = 0^5 = 0 \]

Thus, the determinant of \( A^5 \) is 0.

Final Answer:
\[ \boxed{0.00} \]