CUET UG 2025 — Mathematics Arithmetic

For an invertible matrix of order $n$n:

$$|\text{adj } A| = |A|^{n-1}$$∣adj A∣=∣A∣n−1

$$A(\text{adj } A) = |A| \cdot I$$A(adj A)=∣A∣⋅I

$$|kA| = k^n|A|$$∣kA∣=kn∣A∣

$$|A^{-1}| = \dfrac{1}{|A|}$$∣A−1∣=∣A∣1​

Since $A$A is of order 3, $n = 3$n=3.

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(A) $|\text{adj } A|$∣adj A∣

$$|\text{adj } A| = |A|^{n-1}$$∣adj A∣=∣A∣n−1

$$= |A|^{3-1}$$=∣A∣3−1

$$= |A|^2$$=∣A∣2

$$(A) \rightarrow (II)$$(A)→(II)

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(B) $|A(\text{adj } A)|$∣A(adj A)∣

Using the identity $A(\text{adj } A) = |A| \cdot I$A(adj A)=∣A∣⋅I, taking determinant on both sides:

$$|A(\text{adj } A)| = \big||A| \cdot I\big|$$∣A(adj A)∣=​∣A∣⋅I​

Since $|A|$∣A∣ is a scalar and $|kI| = k^n$∣kI∣=kn:

$$= |A|^3 \cdot |I|$$=∣A∣3⋅∣I∣

$$= |A|^3 \cdot 1$$=∣A∣3⋅1

$$= |A|^3$$=∣A∣3

$$(B) \rightarrow (IV)$$(B)→(IV)

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(C) $|2A|$∣2A∣

$$|kA| = k^n |A|$$∣kA∣=kn∣A∣

$$|2A| = 2^3 \cdot |A|$$∣2A∣=23⋅∣A∣

$$= 8|A|$$=8∣A∣

$$(C) \rightarrow (I)$$(C)→(I)

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(D) $|A^{-1}|$∣A−1∣

From $A \cdot A^{-1} = I$A⋅A−1=I, taking determinant on both sides:

$$|A| \cdot |A^{-1}| = |I| = 1$$∣A∣⋅∣A−1∣=∣I∣=1

$$|A^{-1}| = \dfrac{1}{|A|}$$∣A−1∣=∣A∣1​

$$(D) \rightarrow (III)$$(D)→(III)

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$$(A) \rightarrow (II)$$(A)→(II)
$$(B) \rightarrow (IV)$$(B)→(IV)
$$(C) \rightarrow (I)$$(C)→(I)
$$(D) \rightarrow (III)$$(D)→(III)