Matrices & Determinants
Cofactor Matrix
nta_pyq_2025_apr
Grade 12
Question:
Let $A = [a_{ij}] = \begin{bmatrix}\log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25\end{bmatrix}$. If $A_{ij}$ is the cofactor of $a_{ij}$, $C_{ij} = \sum_{k=1}^{2} a_{ik} A_{jk}$, $1 \leq i, j \leq 2$, and $C = [C_{ij}]$, then $8|C|$ is equal to:
Step-by-Step Solution
Key Concept: $C_{ij} = \sum_k a_{ik}A_{jk}$ equals $|A|$ when $i = j$ and 0 when $i \neq j$ (this is the property of the product $A \cdot \text{adj}(A)^T$). So $C = |A| \cdot I$.
$|A| = \log_5 128 \cdot \log_4 25 - \log_4 5 \cdot \log_5 8 = \frac{7\log 2}{\log 5} \cdot \frac{2\log 5}{2\log 2} - \frac{\log 5}{2\log 2} \cdot \frac{3\log 2}{\log 5} = 7 - \frac{3}{2} = \frac{11}{2}$. $C = \begin{bmatrix}11/2 & 0\\ 0 & 11/2\end{bmatrix}$, $|C| = \frac{121}{4}$. $8|C| = 242$.
Correct Answer: 242