CUET UG 2025 — Mathematics Algebra

For an $n \times n$n×n matrix $A$A, the determinant of a scalar multiple is given by $|kA| = k^n \times |A|$∣kA∣=kn×∣A∣.

For an $n \times n$n×n matrix $A$A, the determinant of the adjoint matrix is given by $|adj\ A| = |A|^{n-1}$∣adj A∣=∣A∣n−1.

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Given: Order of matrix $A$A is $3$3 and $|A| = 2$∣A∣=2

To find $|adj\ 2A|$∣adj 2A∣, first find $|2A|$∣2A∣.

$$|2A| = 2^3 \times |A|$$∣2A∣=23×∣A∣

$$|2A| = 8 \times 2$$∣2A∣=8×2

$$|2A| = 16$$∣2A∣=16

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Now treat $2A$2A as a single matrix. Since $2A$2A is also a $3 \times 3$3×3 matrix:

$$|adj\ 2A| = |2A|^{3-1}$$∣adj 2A∣=∣2A∣3−1

$$|adj\ 2A| = |2A|^2$$∣adj 2A∣=∣2A∣2

$$|adj\ 2A| = (16)^2$$∣adj 2A∣=(16)2

$$|adj\ 2A| = 256$$∣adj 2A∣=256

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Therefore, $|adj\ 2A| = 256$∣adj 2A∣=256.