CUET UG 2025 — Mathematics Calculus

Given a square matrix $A$A of order 3 with $|A| = 2$∣A∣=2.

Let $C = [c_{ij}]$C=[cij​] where $c_{ij}$cij​ is the cofactor of $a_{ij}$aij​ in $A$A.

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For any square matrix $A$A of order $n$n, the relationship between the determinant of $A$A and its cofactor matrix $C$C is:

$$|C| = |A|^{n-1}$$∣C∣=∣A∣n−1

This follows from the property $A \cdot C^T = |A| \cdot I$A⋅CT=∣A∣⋅I, where $C^T$CT is the adjugate matrix and $I$I is the identity matrix.

Taking determinants:

$$|A \cdot C^T| = ||A| \cdot I|$$∣A⋅CT∣=∣∣A∣⋅I∣

$$|A| \times |C^T| = |A|^n$$∣A∣×∣CT∣=∣A∣n

$$|A| \times |C| = |A|^n$$∣A∣×∣C∣=∣A∣n

$$|C| = |A|^{n-1}$$∣C∣=∣A∣n−1

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For the given matrix:

$$n = 3$$n=3

$$|A| = 2$$∣A∣=2

Substituting into the formula:

$$|C| = |A|^{n-1}$$∣C∣=∣A∣n−1

$$|C| = 2^{3-1}$$∣C∣=23−1

$$|C| = 2^2$$∣C∣=22

$$|C| = 4$$∣C∣=4

Therefore, $|C| = 4$∣C∣=4.