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For a square matrix An×nA_{n \times n}

(A) ∣adj⁡A∣=∣A∣n−1|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}

(B) ∣A∣=∣adj⁡A∣n−1|\mathrm{A}|=|\operatorname{adj} \mathrm{A}|^{\mathrm{n}-1}

(C) A(adj⁡A)=∣A∣\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}|

(D) ∣A−1∣=1∣ A∣\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}

Solution

✅ Correct Option: 2

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

To verify this, we use the relationship between AA, its inverse, and its adjoint:

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

Taking determinants of both sides:

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

But we know ∣A−1∣=1∣A∣|A^{-1}| = \dfrac{1}{|A|}

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

Solving for ∣adj A∣|\text{adj}\ A|:

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

This confirms Option (A) is correct.


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

From our previous result, ∣adj A∣=∣A∣n−1|\text{adj}\ A| = |A|^{n-1}

If we raise both sides to power 1n−1\frac{1}{n-1} (assuming n≠1n \neq 1):

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

This is not the same as ∣A∣=∣adj A∣n−1|A| = |\text{adj}\ A|^{n-1}, so Option (B) is incorrect.


Option (C): A(adj A)=∣A∣IA(\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 II, hence, option (C) is wrong


Option (D): ∣A−1∣=1∣A∣|A^{-1}| = \dfrac{1}{|A|}

This is true as long as AA is invertible (meaning ∣A∣≠0|A| \neq 0).

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