CUET UG 2025 — Mathematics Algebra

For an invertible matrix $A$A, checking each statement:

$$A \times A^{-1} = I$$A×A−1=I

Taking determinant of both sides:

$$|A| \times |A^{-1}| = |I|$$∣A∣×∣A−1∣=∣I∣

$$|A| \times |A^{-1}| = 1$$∣A∣×∣A−1∣=1

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

Since $|A^{-1}| = \frac{1}{|A|}$∣A−1∣=∣A∣1​ and not $|A|$∣A∣, statement (A) is FALSE.

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If $A^{-1}$A−1 is the inverse of $A$A, then $A$A is the inverse of $A^{-1}$A−1.

By definition of inverse:

$$(A^{-1})^{-1} = A$$(A−1)−1=A

Statement (B) is TRUE.

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For an invertible matrix, the inverse formula is:

$$A^{-1} = \frac{adj A}{|A|}$$A−1=∣A∣adjA​

This is a standard property of invertible matrices.

Statement (C) is TRUE.

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Starting with $A \times A^{-1} = I$A×A−1=I, taking transpose of both sides:

$$(A \times A^{-1})^T = I^T$$(A×A−1)T=IT

Using $(AB)^T = B^T A^T$(AB)T=BTAT:

$$(A^{-1})^T \times A^T = I$$(A−1)T×AT=I

This means $(A^{-1})^T$(A−1)T is the inverse of $A^T$AT.

Therefore:

$$(A^T)^{-1} = (A^{-1})^T$$(AT)−1=(A−1)T

Statement (D) is TRUE.

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Statements (B), (C), and (D) are TRUE.

Statement (A) is FALSE.