Matrices & Determinants
Determinant of parameter matrix
nta_pyq_2025_apr
Grade 12
Question:
Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A = \begin{bmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{bmatrix}$, $|A| = -1$. Let $B$ be the inverse of the matrix $\text{adj}(A \cdot \text{adj}(A))$. Then $|\lambda B + I|$ is equal to ______
Step-by-Step Solution
Key Concept: Apply the matrix property for determinant of parameter matrix and reduce it to determinant or parameter equations.
$\begin{vmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{vmatrix} = -1$
$|A| = \lambda(16) - 2(-34) + 3(-39) = -1$
$16\lambda - 48 = -1$
$\lambda = 3$
Let $C = A \cdot \text{adj}(A)$
$AC = A \cdot \text{adj}(A) = |A| \cdot I = -I$
$C = -A^{-1}$
Now $B = [\text{adj}(A \cdot \text{adj}(A))]^{-1} = B = \text{adj}(A)$
Now $\lambda B + I = 3B + I$
Let $P = 3B + I$
$P = 3\text{adj}(A) + I$
$AP = 3A\text{adj}(A) + A$
$AP = 3|A| \cdot I + A$
$AP = -3I + A$
$|AP| = |A - 3I|$
$|A| \cdot |P| = \begin{vmatrix} 0 & 2 & 3 \\ 4 & 2 & 6 \\ 7 & -1 & -1 \end{vmatrix} = 38$
$(-1) \cdot |P| = 38$
$|P| = -38$
Correct Answer: 38