CUET UG 2025 — Mathematics Algebra

For any $n \times n$n×n matrix $A$A:

$$|adj A| = |A|^{n-1}$$∣adjA∣=∣A∣n−1

For any $n \times n$n×n matrix $A$A and scalar $k$k:

$$|kA| = k^n \times |A|$$∣kA∣=kn×∣A∣

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Given that $A$A is a $3 \times 3$3×3 matrix and $|adj A| = 64$∣adjA∣=64.

Using the formula for determinant of adjoint:

$$|adj A| = |A|^{n-1}$$∣adjA∣=∣A∣n−1

$$|adj A| = |A|^{3-1}$$∣adjA∣=∣A∣3−1

$$|adj A| = |A|^2$$∣adjA∣=∣A∣2

$$64 = |A|^2$$64=∣A∣2

$$|A| = \pm 8$$∣A∣=±8

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Using the formula for determinant of scalar multiple with $k = 5$k=5 and $n = 3$n=3:

$$|5A| = 5^n \times |A|$$∣5A∣=5n×∣A∣

$$|5A| = 5^3 \times |A|$$∣5A∣=53×∣A∣

$$|5A| = 125 \times (\pm 8)$$∣5A∣=125×(±8)

$$|5A| = \pm 1000$$∣5A∣=±1000

Therefore, the value of $|5A|$∣5A∣ is $\pm 1000$±1000.