Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]=\left[\begin{array}{cc}\log _{5} 128 & \log _{4} 5 \\ \log _{5} 8 & \log _{4} 25\end{array}\right]$. If $\mathrm{A}_{\mathrm{ij}}$ is the cofactor of $\mathrm{a}_{\mathrm{ij}}, \mathrm{C}_{\mathrm{ij}}=\sum_{\mathrm{k}=1}^{2} \mathrm{a}_{\mathrm{ik}} \mathrm{A}_{\mathrm{jk}}, 1 \leq \mathrm{i}$, $\mathrm{j} \leq 2$, and $\mathrm{C}=\left[\mathrm{C}_{\mathrm{ij}}\right]$, then $8|\mathrm{C}|$ is equal to :

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Determinants

Let

A = [a_{ij}] = [lo g_{5} 128 lo g_{5} 8 lo g_{4} 5 lo g_{4} 25]

.

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 :

Answer: (C)

For any square matrix

A

, the sum

\sum_{k} a_{ik} A_{j k}

(a row of

A

contracted against another row's cofactors) equals

∣ A ∣

when

i = j

and

0

when

i \neq = j

(this is a determinant with a repeated row). So

C = [C_{ij}] = ∣ A ∣ \cdot I

, and

∣ C ∣ = ∣ A ∣^{2}

.

Here

a_{11} = lo g_{5} 128 = 7 lo g_{5} 2, a_{22} = lo g_{4} 25 = 2 lo g_{4} 5, a_{12} = lo g_{4} 5, a_{21} = lo g_{5} 8 = 3 lo g_{5} 2.

∣ A ∣ = a_{11} a_{22} - a_{12} a_{21} = 14 lo g_{5} 2 lo g_{4} 5 - 3 lo g_{4} 5 lo g_{5} 2 = 11 lo g_{4} 5 lo g_{5} 2 = 11 lo g_{4} 2 = \frac{11}{2},

using

lo g_{4} 5 \cdot lo g_{5} 2 = lo g_{4} 2 = \frac{1}{2}

.

Therefore

8∣ C ∣ = 8∣ A ∣^{2} = 8 (\frac{11}{2})^{2} = 8 \cdot \frac{121}{4} = 242.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option