Let A = \beginbmatrix a_ij \endbmatrix = \beginbmatrix \log_5 128 & \log_4 5 \log_5 8 & \log_4 25 \endbmatrix . If A_ij is the cofactor of a_ij , C_ij = \sum\limits_k=1^2 a_ik A_jk , 1 \leq i, j \leq 2 , and C=[C_ij] , then 8|C| is equal to :

Matrices and Determinants PYQ — JEE Main 2025 · 29 Jan · Shift 1 · Q28 with Solution

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Correct answer: D

Write the matrix clearly

Given

A=\begin{bmatrix} \log_5 128 & \log_4 5\\ \log_5 8 & \log_4 25 \end{bmatrix}

Let us simplify each entry:

$\log_5 128=\log_5(2^7)=7\log_5 2$
$\log_5 8=\log_5(2^3)=3\log_5 2$
$\log_4 25=\log_4(5^2)=2\log_4 5$

So if we put $x=\log_5 2,\qquad y=\log_4 5,$ then

A=\begin{bmatrix} 7x & y\\ 3x & 2y \end{bmatrix}.

Find cofactors of $A$

For a $2\times 2$ matrix

\begin{bmatrix} a&b\\ c&d\end{bmatrix},

the cofactor matrix entries are $A_{11}=d,\; A_{12}=-c,\; A_{21}=-b,\; A_{22}=a.$

Hence for

A=\begin{bmatrix}7x&y\\3x&2y\end{bmatrix},

we get

A_{11}=2y,\quad A_{12}=-3x,\quad A_{21}=-y,\quad A_{22}=7x.

Interpret $C_{ij}$

Given

C_{ij}=\sum_{k=1}^{2} a_{ik}A_{jk}.

This is exactly the $(i,j)$ -entry of the product

C=A\,(\text{cofactor matrix of }A)^T.

But transpose of the cofactor matrix is the adjugate matrix. Therefore,

C=A\,\operatorname{adj}(A).

Now we use the standard identity:

A\,\operatorname{adj}(A)=|A|I.

Hence

C=|A|I.

So for a $2\times 2$ matrix,

|C|=|A|^2.

Compute $|A|$

|A|=(7x)(2y)-(y)(3x)=14xy-3xy=11xy.

Thus

|A|=11xy.

Now evaluate $xy$ :

x=\log_5 2=\frac{\ln 2}{\ln 5},

y=\log_4 5=\frac{\ln 5}{\ln 4}=\frac{\ln 5}{2\ln 2}.

Therefore,

xy=\frac{\ln 2}{\ln 5}\cdot \frac{\ln 5}{2\ln 2}=\frac12.

|A|=11\cdot \frac12=\frac{11}{2}.

Compute $|C|$

Since

C=|A|I=\frac{11}{2}I,

we have

|C|=\left(\frac{11}{2}\right)^2=\frac{121}{4}.

Hence

8|C|=8\cdot \frac{121}{4}=2\cdot 121=242.

Match with options

8|C|=242

So the correct option is D.

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View written solutionFree

Correct answer: D

Write the matrix clearly

Given

A=\begin{bmatrix} \log_5 128 & \log_4 5\\ \log_5 8 & \log_4 25 \end{bmatrix}

Let us simplify each entry:

$\log_5 128=\log_5(2^7)=7\log_5 2$
$\log_5 8=\log_5(2^3)=3\log_5 2$
$\log_4 25=\log_4(5^2)=2\log_4 5$

So if we put $x=\log_5 2,\qquad y=\log_4 5,$ then

A=\begin{bmatrix} 7x & y\\ 3x & 2y \end{bmatrix}.

Find cofactors of $A$

For a $2\times 2$ matrix

\begin{bmatrix} a&b\\ c&d\end{bmatrix},

the cofactor matrix entries are $A_{11}=d,\; A_{12}=-c,\; A_{21}=-b,\; A_{22}=a.$

Hence for

A=\begin{bmatrix}7x&y\\3x&2y\end{bmatrix},

we get

A_{11}=2y,\quad A_{12}=-3x,\quad A_{21}=-y,\quad A_{22}=7x.

Interpret $C_{ij}$

Given

C_{ij}=\sum_{k=1}^{2} a_{ik}A_{jk}.

This is exactly the $(i,j)$ -entry of the product

C=A\,(\text{cofactor matrix of }A)^T.

But transpose of the cofactor matrix is the adjugate matrix. Therefore,

C=A\,\operatorname{adj}(A).

Now we use the standard identity:

A\,\operatorname{adj}(A)=|A|I.

Hence

C=|A|I.

So for a $2\times 2$ matrix,

|C|=|A|^2.

Compute $|A|$

|A|=(7x)(2y)-(y)(3x)=14xy-3xy=11xy.

Thus

|A|=11xy.

Now evaluate $xy$ :

x=\log_5 2=\frac{\ln 2}{\ln 5},

y=\log_4 5=\frac{\ln 5}{\ln 4}=\frac{\ln 5}{2\ln 2}.

Therefore,

xy=\frac{\ln 2}{\ln 5}\cdot \frac{\ln 5}{2\ln 2}=\frac12.

|A|=11\cdot \frac12=\frac{11}{2}.

Compute $|C|$

Since

C=|A|I=\frac{11}{2}I,

we have

|C|=\left(\frac{11}{2}\right)^2=\frac{121}{4}.

Hence

8|C|=8\cdot \frac{121}{4}=2\cdot 121=242.

Match with options

8|C|=242

So the correct option is D.

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