Question

Let \( A \) be a \( 2 \times 2 \) real matrix such that \( AB = BA \) for all \( 2 \times 2 \) real matrices \( B \). If the trace of \( A \) equals 5, then the determinant of \( A \) (rounded off to two decimal places) equals ................

Hint:

- If \( AB = BA \) for all matrices \( B \), then \( A \) must be a scalar matrix.
- The trace of a scalar matrix is the scalar multiplied by the size of the matrix.
- The determinant of a scalar matrix is the scalar raised to the power of the matrix size.

Answer 1

1) Understanding the problem:
The condition \( AB = BA \) implies that the matrix \( A \) is a scalar multiple of the identity matrix. This is a well-known result from matrix theory: if a matrix commutes with every matrix of the same size, it must be a scalar matrix. Therefore, we can conclude that \( A = \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is a scalar.
2) Determining the scalar \( \lambda \):
Since the trace of \( A \) is given as 5, and the trace of \( A = \lambda I \) is simply \( 2\lambda \) (since the trace of the identity matrix is 2), we have: \[ 2\lambda = 5 ⇒ \lambda = \frac{5}{2} \] 3) Finding the determinant:
The determinant of a scalar matrix \( A = \lambda I \) is simply \( \lambda^2 \). Therefore: \[ \text{det}(A) = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25 \] Final Answer: The determinant of \( A \) is \( \boxed{6.25} \) (rounded to two decimal places).