CUET UG 2025 — Mathematics Algebra

For an invertible matrix $A$A of order $n$n:

$A^{-1} = \dfrac{\text{adj}(A)}{|A|}$A−1=∣A∣adj(A)​, i.e., $\text{adj}(A) = |A| \cdot A^{-1}$adj(A)=∣A∣⋅A−1

$$(AB)^{-1} = B^{-1}A^{-1}$$(AB)−1=B−1A−1

$$|A^{-1}| = |A|^{-1}$$∣A−1∣=∣A∣−1

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

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$$\text{adj}(A)$$adj(A)

From the inverse formula: $A^{-1} = \dfrac{\text{adj}(A)}{|A|}$A−1=∣A∣adj(A)​

$$\text{adj}(A) = |A| \cdot A^{-1}$$adj(A)=∣A∣⋅A−1

$$(A) \longrightarrow (IV)$$(A)⟶(IV)

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$$(AB)^{-1}$$(AB)−1

$$(AB)^{-1} = B^{-1}A^{-1}$$(AB)−1=B−1A−1

$$(B) \longrightarrow (I)$$(B)⟶(I)

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$$|A^{-1}|$$∣A−1∣

Taking determinant on both sides of $A \cdot A^{-1} = I$A⋅A−1=I:

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

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

$$= |A|^{-1}$$=∣A∣−1

$$(C) \longrightarrow (II)$$(C)⟶(II)

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$$|\text{adj}(A)|$$∣adj(A)∣

Since $\text{adj}(A) = |A| \cdot A^{-1}$adj(A)=∣A∣⋅A−1, taking determinant of both sides with $n = 3$n=3:

$$|\text{adj}(A)| = \big||A| \cdot A^{-1}\big|$$∣adj(A)∣=​∣A∣⋅A−1​

$$= |A|^3 \cdot |A^{-1}|$$=∣A∣3⋅∣A−1∣

$$= |A|^3 \cdot \dfrac{1}{|A|}$$=∣A∣3⋅∣A∣1​

$$= |A|^2$$=∣A∣2

$$(D) \longrightarrow (III)$$(D)⟶(III)

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