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JEE Mathematics 2025 Question with Solution

Let A=[aij]=[log⁡5128log⁡45log⁡58log⁡425]A = \left[ a_{ij} \right] = \left[ \begin{array}{cc} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{array} \right]. If AijA_{ij} is the cofactor of aija_{ij}, Cij=∑k=12aikAjkC_{ij} = \sum_{k=1}^{2} a_{ik} A_{jk}, and C=[Cij]C = [C_{ij}], then 8∣C∣8|C| is equal to:

  • A

    242242

  • B

    222222

  • C

    262262

  • D

    288288

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

A=[log⁡5128log⁡45log⁡58log⁡425]A = \begin{bmatrix} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{bmatrix}

Find: 8∣C∣8|C| where Cij=∑k=12aikAjkC_{ij} = \sum_{k=1}^{2} a_{ik} A_{jk} and AijA_{ij} denotes the cofactor of aija_{ij}.

Using the identity A⋅(adj⁡A)=∣A∣IA \cdot (\operatorname{adj} A) = |A|I, we get

Cij=∑k=12aikAjkC_{ij} = \sum_{k=1}^{2} a_{ik} A_{jk}

which means C=A(adj⁡A)=∣A∣IC = A(\operatorname{adj} A) = |A|I.

So first compute ∣A∣|A|:

∣A∣=(log⁡5128)(log⁡425)−(log⁡45)(log⁡58)=(7log⁡2log⁡5)(log⁡25log⁡4)−(log⁡5log⁡4)(3log⁡2log⁡5)=(7log⁡2log⁡5)(2log⁡52log⁡2)−(log⁡52log⁡2)(3log⁡2log⁡5)=7−32=112\begin{aligned} |A| &= (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8) \\ &= \left(\frac{7\log 2}{\log 5}\right)\left(\frac{\log 25}{\log 4}\right) - \left(\frac{\log 5}{\log 4}\right)\left(\frac{3\log 2}{\log 5}\right) \\ &= \left(\frac{7\log 2}{\log 5}\right)\left(\frac{2\log 5}{2\log 2}\right) - \left(\frac{\log 5}{2\log 2}\right)\left(\frac{3\log 2}{\log 5}\right) \\ &= 7 - \frac{3}{2} \\ &= \frac{11}{2} \end{aligned}

Therefore,

C=∣A∣I=112I=[11200112]C = |A|I = \frac{11}{2}I = \begin{bmatrix} \frac{11}{2} & 0 \\ 0 & \frac{11}{2} \end{bmatrix}

Hence,

∣C∣=(112)2=1214|C| = \left(\frac{11}{2}\right)^2 = \frac{121}{4}

and

8∣C∣=8⋅1214=2428|C| = 8 \cdot \frac{121}{4} = 242

Therefore, the correct option is A, and the value of 8∣C∣8|C| is 242242.

Using matrix identity explicitly

Given: Cij=∑k=12aikAjkC_{ij} = \sum_{k=1}^{2} a_{ik} A_{jk}.

Here AjkA_{jk} is the cofactor of entry ajka_{jk}, so the matrix [Ajk][A_{jk}] is the cofactor matrix of AA. The expression

∑k=12aikAjk\sum_{k=1}^{2} a_{ik} A_{jk}

is exactly the entry in row ii and column jj of A⋅(adj⁡A)A \cdot (\operatorname{adj} A).

Now, for any square matrix,

A⋅(adj⁡A)=∣A∣IA \cdot (\operatorname{adj} A) = |A|I

Thus,

  • C11=∣A∣C_{11} = |A|
  • C12=0C_{12} = 0
  • C21=0C_{21} = 0
  • C22=∣A∣C_{22} = |A|

So it remains only to calculate ∣A∣|A|.

log⁡5128=log⁡527=7log⁡52,log⁡58=log⁡523=3log⁡52,log⁡425=log⁡452=2log⁡45.\begin{aligned} \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. \end{aligned}

Hence,

∣A∣=(log⁡5128)(log⁡425)−(log⁡45)(log⁡58)=(7log⁡52)(2log⁡45)−(log⁡45)(3log⁡52)=14(log⁡52)(log⁡45)−3(log⁡45)(log⁡52)\begin{aligned} |A| &= (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8) \\ &= (7\log_5 2)(2\log_4 5) - (\log_4 5)(3\log_5 2) \\ &= 14(\log_5 2)(\log_4 5) - 3(\log_4 5)(\log_5 2) \end{aligned}

Also,

(log⁡52)(log⁡45)=log⁡2log⁡5⋅log⁡5log⁡4=log⁡22log⁡2=12(\log_5 2)(\log_4 5) = \frac{\log 2}{\log 5} \cdot \frac{\log 5}{\log 4} = \frac{\log 2}{2\log 2} = \frac{1}{2}

Therefore,

∣A∣=14⋅12−3⋅12=112|A| = 14\cdot \frac{1}{2} - 3\cdot \frac{1}{2} = \frac{11}{2}

So,

C=[11200112],∣C∣=1214C = \begin{bmatrix} \frac{11}{2} & 0 \\ 0 & \frac{11}{2} \end{bmatrix}, \qquad |C| = \frac{121}{4}

Finally,

8∣C∣=2428|C| = 242

Therefore, the value required is 242242.

Common mistakes

  • Mistake: treating [Aij][A_{ij}] directly as the adjoint without noticing the index order in Cij=∑aikAjkC_{ij} = \sum a_{ik}A_{jk}. Why it is wrong: the formula uses cofactors with indices jkjk, which matches matrix multiplication with the adjugate. What to do instead: recognize the identity A(adj⁡A)=∣A∣IA(\operatorname{adj}A)=|A|I before expanding entries.

  • Mistake: calculating ∣A∣|A| using incorrect log conversions such as log⁡425=2log⁡45\log_4 25 = 2\log_4 5 but then simplifying inconsistently. Why it is wrong: a small logarithm error changes the determinant completely. What to do instead: convert each logarithm carefully using change of base and simplify step by step.

  • Mistake: assuming ∣C∣=∣A∣|C| = |A|. Why it is wrong: here C=∣A∣IC = |A|I, so for a 2×22 \times 2 matrix its determinant is ∣C∣=(∣A∣)2|C| = (|A|)^2. What to do instead: first write the full matrix CC, then take its determinant.

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