CUET UG 2025 — Mathematics Algebra

Given $A$A is a $3 \times 3$3×3 matrix with $|adj A| = 9$∣adjA∣=9 and $|kA^{-1}| = 9$∣kA−1∣=9.

For any $n \times n$n×n matrix, $|adj A| = |A|^{n-1}$∣adjA∣=∣A∣n−1.

Since $A$A is a $3 \times 3$3×3 matrix:

$$|adj A| = |A|^{3-1}$$∣adjA∣=∣A∣3−1

$$|adj A| = |A|^2$$∣adjA∣=∣A∣2

Substituting the given value:

$$|A|^2 = 9$$∣A∣2=9

$$|A| = \pm 3$$∣A∣=±3

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For a scalar $k$k and $n \times n$n×n matrix $B$B, $|kB| = k^n|B|$∣kB∣=kn∣B∣.

Since $A$A is a $3 \times 3$3×3 matrix:

$$|kA^{-1}| = k^3|A^{-1}|$$∣kA−1∣=k3∣A−1∣

For inverse matrices, $|A^{-1}| = \frac{1}{|A|}$∣A−1∣=∣A∣1​.

Therefore:

$$|kA^{-1}| = k^3 \cdot \frac{1}{|A|}$$∣kA−1∣=k3⋅∣A∣1​

Substituting the given value:

$$k^3 \cdot \frac{1}{|A|} = 9$$k3⋅∣A∣1​=9

$$k^3 = 9|A|$$k3=9∣A∣

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When $|A| = 3$∣A∣=3:

$$k^3 = 9 \times 3$$k3=9×3

$$k^3 = 27$$k3=27

$$k = 3$$k=3

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When $|A| = -3$∣A∣=−3:

$$k^3 = 9 \times (-3)$$k3=9×(−3)

$$k^3 = -27$$k3=−27

$$k = -3$$k=−3

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Therefore, the value of $k = \pm 3$k=±3.