Question:
If \(A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}\), then the value of \(\det(A^4) + \det\left(A^{10} - (\mathrm{Adj}(2A))^{10}\right)\) is equal to _____________.
If \(A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}\), then the value of \(\det(A^4) + \det\left(A^{10} - (\mathrm{Adj}(2A))^{10}\right)\) is equal to _____________.
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All matrices involved are upper triangular, so powers and differences are upper triangular and determinants are products of diagonal entries.
Updated On: Nov 4, 2026
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Correct Answer: 16
Solution and Explanation
\(\det A = -2\), so \(\det(A^4) = (-2)^4 = 16\).
\(2A = \begin{bmatrix} 4 & 6 \\ 0 & -2 \end{bmatrix}\), \(\mathrm{Adj}(2A) = \begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix}\).
Both \(A\) and \(\mathrm{Adj}(2A)\) are upper triangular, so the diagonal of the 10th power is the 10th power of the diagonal.
\(A^{10}\) has diagonal \((2^{10}, (-1)^{10}) = (1024, 1)\); \((\mathrm{Adj}(2A))^{10}\) has diagonal \(((-2)^{10}, 4^{10}) = (1024, 1048576)\).
The difference is upper triangular with diagonal \((0,\ 1 - 1048576)\), so its determinant is \(0\).
Total \(= 16 + 0 = 16\).
Answer: 16
\(2A = \begin{bmatrix} 4 & 6 \\ 0 & -2 \end{bmatrix}\), \(\mathrm{Adj}(2A) = \begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix}\).
Both \(A\) and \(\mathrm{Adj}(2A)\) are upper triangular, so the diagonal of the 10th power is the 10th power of the diagonal.
\(A^{10}\) has diagonal \((2^{10}, (-1)^{10}) = (1024, 1)\); \((\mathrm{Adj}(2A))^{10}\) has diagonal \(((-2)^{10}, 4^{10}) = (1024, 1048576)\).
The difference is upper triangular with diagonal \((0,\ 1 - 1048576)\), so its determinant is \(0\).
Total \(= 16 + 0 = 16\).
Answer: 16
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