JEE Main 2025 Jan 29 Shift 1, Mathematics Q20: Algebra Of Determinants

JEE Main2025Jan 29, Shift 1Mathematics

Let $A = [a_{ij}] = [lo g_{5} 128 lo g_{4} 5 lo g_{5} 8 lo g_{4} 25] .$ If $A_{ij}$ is the cofactor of $a_{ij}$ , $C_{ij} = k = 1 \sum 2 a_{ik} A_{j k}, 1 \leq i, j \leq 2,$ and $C = [C_{ij}]$ , then $8∣ C ∣$ is equal to :

A
$262$
B
$288$
C
$242$
D
$222$

Solution

The expression $C_{ij} = \sum_{k} a_{ik} A_{j k}$ is precisely the expansion identity that yields $∣ A ∣ δ_{ij} .$ Hence $C = ∣ A ∣ I_{2} .$

Compute $∣ A ∣$ using change-of-base:

$lo g_{5} 128 = 7 lo g_{5} 2, lo g_{5} 8 = 3 lo g_{5} 2, lo g_{4} 5 = \frac{lo g _{2} 5}{2}, lo g_{4} 25 = lo g_{2} 5.$

$∣ A ∣ = 7 lo g_{5} 2 \cdot lo g_{2} 5 - 3 lo g_{5} 2 \cdot \frac{lo g _{2} 5}{2} .$

Since $lo g_{5} 2 \cdot lo g_{2} 5 = 1$ : $∣ A ∣ = 7 - \frac{3}{2} = \frac{11}{2} .$

So $C = \frac{11}{2} I_{2}$ and $∣ C ∣ = (\frac{11}{2})^{2} = \frac{121}{4} .$

$8∣ C ∣ = 8 \cdot \frac{121}{4} = 242.$

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q20: Algebra Of Determinants

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