Q.If $A = \begin{bmatrix} 2 & 0 & 0 \\ -1 & 2 & 3 \\ 3 & 3 & 5 \end{bmatrix}$, then find $A(\text{adj } A)$.

Concept: Eigenvalue Properties

For any square matrix $A$, the product $A(\text{adj } A) = (\det A) I$. So we only need $\det A$.

Step 1 – Compute $\det A$ by expanding along the first row (since it has two zeros):

Step 2 – Therefore, …

The key idea is not to compute the adjoint explicitly — that would be tedious and error-prone. Instead, we use a fundamental property of square matrices: the product of a matrix and its adjoint equals the determinant times the identity matrix. This is one of the most elegant shortcuts in linear algebra.

This works because each entry of $\text{adj } A$ is a cofactor, and the dot product of a row of $A$ with the corresponding column of cofactors gives $\det A$, while a dot product with a different row's cofactors gives zero (by the property of determinants with repeated rows).

So the entire problem reduces to one number: the determinant of $A$.

1. Compute $\det A$. The matrix is $3 \times 3$, and the first row has two zeros — perfect for expansion along the first row:

2. Apply the property. …

Method: Using A(adj A) = |A| I to Avoid Computing the Adjoint Directly

This method solves "find A(adj A)" questions without ever building the adjoint matrix explicitly — the standard CBSE route, needed even when a question's underlying theme brushes against advanced linear-algebra ideas like eigenvalues, which sit outside the Class 12 Determinants syllabus.

Steps

Step 1: Recall the identity that bypasses the adjoint

For any square matrix $A$ of order $n$,

This means the answer is always a scalar multiple of the identity matrix — you never need the individual cofactor entries.

Step 2: Compute only |A|, using cofactor expansion along the row/column with the most zeros …

Common Mistakes

Mistake 1: Computing the full adjoint matrix instead of using the shortcut

Why it's wrong: finding all $9$ cofactors of the $3\times3$ matrix, transposing them, and then multiplying by $A$ is slow and invites sign errors in the cofactors — when the question only needs $\det A$. Correct approach: use $A(\text{adj }A)=(\det A)I_n$ directly and compute only $\det A$.

Mistake 2: Picking a row/column to expand along that has no zeros

Why it's wrong: expanding $\det A$ along the second or third row (which have no zero entries) forces three separate $2\times2$ cofactor computations instead of one, multiplying the chance of an arithmetic slip. Correct approach: always expand along the row or column with the most zeros — here the first row, $[2,0,0]$, collapses the expansion to a single term. …