Matrices & Determinants
Matrix determinant from parameter matrix
nta_pyq_2025_apr
Grade 12
Question:
Let $a \in \mathbb{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det(A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det((a+1)\operatorname{adj}((a-1)A)) = \frac{2^m}{3^n}$, $m, n \in \{0, 1, 2, \ldots, 20\}$, then $m + n$ is equal to:
Step-by-Step Solution
Key Concept: Apply the matrix property for matrix determinant from parameter matrix and reduce it to determinant or parameter equations.
$A = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} - I = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix}$
$|A| = -4 \Rightarrow 2 - 2a = -4 \Rightarrow a = 3$
$|(a+1)\operatorname{adj}((a-1)A)| = |4\operatorname{adj}(2A)|$
$= 4^3 |\operatorname{adj}(2A)| = 64|\operatorname{adj}(2A)|$
$= 64 \times |2A|^{3-1} = 64 \times (2^3|A|)^2$
$= 64 \times 2^6 \times 16 = 2^6 \times 2^6 \times 2^4 = 2^{16}$
Therefore, $2^m \cdot 3^n = 2^{16}$, which gives $m = 16$ and $n = 0$.
$\therefore m + n = 16$
Correct Answer: 4