CUET Mathematics 2024 - For a square matrix A_n x n (A) |adj A|=|A|^n-1 (B) |A|=|adj A|^n-1 (C) A(adj A)=|A| (D) |A^-1|=1|~A| | PYQs + Solutions

CUET Mathematics 2024

Algebra

Matrices & Determinants

Easy

For a square matrix $A_{n \times n}$
(A) $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}$
(B) $|\mathrm{A}|=|\operatorname{adj} \mathrm{A}|^{\mathrm{n}-1}$
(C) $\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}|$
(D) $\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}$

(B) and (D) only

(A) and (D) only

(A), (C) and (D) only

(B), (C) and (D) only

✅ Correct Option: 2

Option (A):

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

To verify this, we use the relationship between

A

, its inverse, and its adjoint:

A^{-1} = \dfrac{\text{adj}\ A}{|A|}

Taking determinants of both sides:

|A^{-1}| = \dfrac{|\text{adj}\ A|}{|A|^n}

But we know

|A^{-1}| = \dfrac{1}{|A|}

So:

\dfrac{|\text{adj}\ A|}{|A|^n} = \dfrac{1}{|A|}

Solving for

|\text{adj}\ A|

|\text{adj}\ A| = |A|^n \cdot \dfrac{1}{|A|} = |A|^{n-1}

This confirms Option (A) is correct.

Option (B):

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

From our previous result,

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

If we raise both sides to power

\frac{1}{n-1}

(assuming

n \neq 1

(|\text{adj}\ A|)^{\frac{1}{n-1}} = |A|

This is not the same as

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

, so Option (B) is incorrect.

\newline

Option (C):

A(\text{adj}\ A) = |A|I

This is a fundamental property of the adjoint matrix. When we multiply a matrix by its adjoint, we get the determinant of the matrix times the identity matrix.

Note: The original option is missing the identity matrix

I

, hence, option (C) is wrong

Option (D):

|A^{-1}| = \dfrac{1}{|A|}

This is true as long as

A

is invertible (meaning

|A| \neq 0

CUET Mathematics 2024 PYQs

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