Q.Let A be a nonsingular square matrix of order $3 \times 3$. Then $|\text{adj } A|$ is equal to (A) $|A|$ (B) $|A|^2$ (C) $|A|^3$ (D) $3|A|$

Concept: Adjoint Matrix Property

For any $n \times n$ matrix $A$, the determinant of its adjoint is $|\text{adj } A| = |A|^{n-1}$.

Step 1 — Recall the fundamental relation:

$A \cdot (\text{adj } A) = |A| I_n$.

Step 2 — Take determinants on both sides:

$|A \cdot \text{adj } A| = \big| |A| I_n \big|$.

Step 3 — Use $|AB| = |A||B|$ and $|k I_n| = k^n$:

$|A| \cdot |\text{adj } A| = |A|^n$. …

The key idea is that the adjugate matrix is built from cofactors, and its product with $A$ gives a scalar matrix. That relationship directly links their determinants.

We start from the fundamental property of the adjugate:

This holds for any square matrix $A$ of order $n$. For a nonsingular $A$, $|A| \neq 0$, so the adjugate is essentially $|A|$ times the inverse.

Now take determinants on both sides of $A \cdot (\text{adj } A) = |A| \, I_n$.

1. Determinant of a product — For any two square matrices $X$ and $Y$ of the same order, $|XY| = |X| \, |Y|$. So:

2. Determinant of a scalar multiple — The right side is $|A| \, I_n$. The determinant of $k I_n$ (where $k$ is a scalar) is $k^n$, because multiplying a single row by $k$ multiplies the determinant by $k$, and there are $n$ rows. So:

3. Equate the two:

4. Since $A$ is nonsingular, $|A| \neq 0$, we can divide both sides by $|A|$:

For $n = 3$, this becomes: …

Method: Determinant of the Adjoint, $|\operatorname{adj}A| = |A|^{n-1}$

This method derives (rather than memorises) the relationship between $|\operatorname{adj}A|$ and $|A|$ for any square matrix of order $n$, so it works no matter what order the question asks about.

Steps

Step 1: Start from the defining identity of the adjoint

This holds for every square matrix — it is the fact the whole "inverse via adjoint" method is built on.

Step 2: Take the determinant of both sides

Use $|XY|=|X||Y|$ on the left, and $|kI_n| = k^n$ on the right (scaling a single row by $k$ scales the determinant by $k$, and there are $n$ rows to scale):

$|A|\cdot|\operatorname{adj}A| = |A|^n$ …

Common Mistakes

Mistake 1: Using the wrong exponent — picking $|A|$ or $|A|^n$ instead of $|A|^{n-1}$

Why it's wrong: the identity $|\operatorname{adj}A|=|A|^{n-1}$ comes from dividing both sides of $|A|\cdot|\operatorname{adj}A|=|A|^n$ by $|A|$ — skipping that division, or misremembering the formula, leads directly to choosing option (A) $|A|$ or wrongly guessing $|A|^3$ (which is $|A|^n$, not $|A|^{n-1}$). Correct approach: re-derive the exponent from $A\cdot\operatorname{adj}(A)=|A|I_n$ each time rather than recalling "$n-1$" by rote — for $n=3$ this gives $|A|^{3-1}=|A|^2$.

Mistake 2: Confusing this property with $\det(A^{-1})=\tfrac{1}{\det A}$ …