Question

Given \( A = \begin{pmatrix} 2 & 5 \\ 0 & 3 \end{pmatrix} \), the value of the determinant \( |A^4 - 5A^3 + 6A^2 + 2I| \) is \(\underline{\hspace{2cm}}\). 

Hint:

When calculating the determinant of matrix expressions, simplify the matrix algebra first and then compute the determinant of the result.

Answer 1

Correct Answer: 4

We first calculate powers of the matrix \( A \): \[ A^2 = \begin{pmatrix} 2 & 5 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 2 & 5 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 15 \\ 0 & 9 \end{pmatrix} \] \[ A^3 = A \times A^2 = \begin{pmatrix} 2 & 5 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 4 & 15 \\ 0 & 9 \end{pmatrix} = \begin{pmatrix} 8 & 45 \\ 0 & 27 \end{pmatrix} \] \[ A^4 = A \times A^3 = \begin{pmatrix} 2 & 5 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 8 & 45 \\ 0 & 27 \end{pmatrix} = \begin{pmatrix} 16 & 135 \\ 0 & 81 \end{pmatrix} \] Now substitute into the given expression: \[ A^4 - 5A^3 + 6A^2 + 2I = \begin{pmatrix} 16 & 135 \\ 0 & 81 \end{pmatrix} - 5 \begin{pmatrix} 8 & 45 \\ 0 & 27 \end{pmatrix} + 6 \begin{pmatrix} 4 & 15 \\ 0 & 9 \end{pmatrix} + 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \] Simplifying: \[ = \begin{pmatrix} 16 & 135 \\ 0 & 81 \end{pmatrix} - \begin{pmatrix} 40 & 225 \\ 0 & 135 \end{pmatrix} + \begin{pmatrix} 24 & 90 \\ 0 & 54 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \] \[ = \begin{pmatrix} 16 - 40 + 24 + 2 & 135 - 225 + 90 + 0 \\ 0 & 81 - 135 + 54 + 2 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}. \] The determinant of the resulting matrix is: \[ \text{det} = (2)(2) - (0)(0) = 4. \] Thus, the value of the determinant is \( 4 \).