Matrices & Determinants
Matrices
nta_pyq_2025_jan
Grade 12
Question:
Let $A=[a_{ij}]=\begin{pmatrix}\log_{5}128 & \log_{4}5\\ \log_{5}8 & \log_{4}25\end{pmatrix}$. If $A_{ij}$ is the cofactor of $a_{ij}$, and $C_{ij}=\displaystyle\sum_{k=1}^{2}a_{ik}A_{jk},\ 1\le i,j\le 2,\,C=[C_{ij}]$, then $8|C|$ is equal to:
Step-by-Step Solution
Key Concept: The defining identity $\sum_{k}a_{ik}A_{jk}=|A|\delta_{ij}$ gives $C=|A|I.$ Hence $|C|=|A|^{2}.$ Compute $|A|$ using change-of-base for logs.
Let $L=\log_{2}5.$ Then $\log_{5}128=\dfrac{7}{L},\,\log_{4}5=\dfrac{L}{2},\,\log_{5}8=\dfrac{3}{L},\,\log_{4}25=L.$
$|A|=\dfrac{7}{L}\cdot L-\dfrac{L}{2}\cdot\dfrac{3}{L}=7-\dfrac{3}{2}=\dfrac{11}{2}.$
By the cofactor identity $\sum_{k}a_{ik}A_{jk}=|A|\delta_{ij}$, $C=|A|\,I=\dfrac{11}{2}I.$
$|C|=\left(\dfrac{11}{2}\right)^{2}=\dfrac{121}{4}.$ So $8|C|=242.$
Correct Answer: 3