Let (A=[a_ i j ]=[ array cc _5 128 & _4 5 _5 8 & _4 25 array ]). If (A_ i j ) is the cofactor of (a_ i j , C_ i j =∑_ k=1 ^ 2 a_ i k A_ j k , 1 ≤ i, j ≤ 2), and (C= [C_ i j ] ), then (8| C| ) is equal to :

Correct answer: C. 242 aligned & |A|=(11)/(2) & C_ 11 =∑_ k=1 ^2 a_ 1 k · A_ 1 k =a_ 11 ~A_ 11 +a_ 12 ~A_ 12 =|A|=(11)/(2) & C_ 12 =∑_ k=1 ^2 a_ 1 k · A_ 2 k =0 & C_ 21 =∑_ k=1 ^2 a_ 2 k · A_ 1 k =0 & C_ 22 =∑_ k=1 ^2 a_ 2 k · A_ 2 k =|A|=(11)/(2) & C=[ array cc 11 / 2 & 0 0 & 11 / 2 array ] & |C|=(121)/(4) & 8|C|=242 aligned

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Matrices

Let (A=[a_ i j ]=[ array cc _5 128 & _4 5 _5 8 & _4 25 array ]). If...

Let (\mathrm{A}=\left[\mathrm{a}_{i j}\right]=\left[

\begin{array}{cc}\log _5 128 & \log _4 5 \\ \log _5 8 & \log _4 25\end{array}

\right]). If (\mathrm{A}_{i j}

) is the cofactor of (\mathrm{a}_{i j}, \mathrm{C}_{i j}=\sum_{\mathrm{k}=1}^

\mathrm{a}_{i \mathrm{k}} \mathrm{A}_{j \mathrm{k}},

1 \leq i, j \leq 2), and (\mathrm{C}= \left[\mathrm{C}_{i j}\right]

), then (8|

\mathrm{C}|$) is equal to :

Options

A288
B222
C242
D262

Correct answer

C. 242

Step-by-step solution

\begin{aligned} & |\mathrm{A}|=\frac{11}{2} \\ & \mathrm{C}_{11}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{1 \mathrm{k}} \cdot \mathrm{A}_{1 \mathrm{k}}=\mathrm{a}_{11} \mathrm{~A}_{11}+\mathrm{a}_{12} \mathrm{~A}_{12}=|\mathrm{A}|=\frac{11}{2} \\ & \mathrm{C}_{12}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{1 \mathrm{k}} \cdot \mathrm{A}_{2 \mathrm{k}}=0 \\ & \mathrm{C}_{21}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{2 \mathrm{k}} \cdot \mathrm{A}_{1 \mathrm{k}}=0 \\ & \mathrm{C}_{22}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{2 \mathrm{k}} \cdot \mathrm{A}_{2 \mathrm{k}}=|\mathrm{A}|=\frac{11}{2} \\ & \mathrm{C}=\left[\begin{array}{cc}11 / 2 & 0 \\ 0 & 11 / 2\end{array}\right] \\ & |\mathrm{C}|=\frac{121}{4}\\ & 8|\mathrm{C}|=242\end{aligned}

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