Q.$|\operatorname{adj} A| = |A|^2$, where $A$ is a square matrix of order two.

Use the order fact for the adjoint: for an $n\times n$ matrix, $|\operatorname{adj} A| = |A|^{\,n-1}$.

Here $A$ has order $n=2$, so $|\operatorname{adj} A| = |A|^{\,2-1} = |A|$, not $|A|^2$. (The exponent $2$ would be the …

The general rule

For any $n\times n$ matrix, $|\operatorname{adj} A| = |A|^{\,n-1}$. It comes from taking determinants of the central identity $A\,(\operatorname{adj} A) = |A|\,I_n$:

For $n=2$ this gives $|\operatorname{adj} A| = |A|^{1} = |A|$; the exponent $2$ belongs to $n=3$.

Direct check for order two

Take $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, so $|A| = ad-bc$. Its adjoint is …

Method: Using the Order-Dependent Rule $|\operatorname{adj} A| = |A|^{n-1}$

This method applies whenever a statement relates $|\operatorname{adj} A|$ to $|A|$ — don't guess the exponent, derive it from the order every time.

Steps

Step 1: Take determinants of the central adjoint identity

Every square matrix satisfies $A\cdot\operatorname{adj}(A)=|A|\,I_n$. Take the determinant of both sides:

Step 2: Solve for $|\operatorname{adj} A|$

Provided $|A|\neq 0$, divide both sides by $|A|$:

This is a general formula — the exponent is always $n-1$, one less than the matrix's own order, never $n$.

Step 3: Substitute the actual order $n$ of the matrix in the question …

Common Mistakes

Mistake 1: Assuming $|\operatorname{adj} A| = |A|^2$ for every matrix, regardless of order

Why it's wrong: the exponent $2$ is the special case for $n=3$ (where $n-1=2$), not a universal rule. Correct approach: always derive the exponent from the actual order — $|\operatorname{adj} A| = |A|^{\,n-1}$ — and substitute $n$ before comparing.

Mistake 2: Mixing up $|A|^{n}$ and $|A|^{n-1}$ when taking determinants of $A\cdot\operatorname{adj}(A) = |A|I_n$ …