Q.Evaluate the determinants

Concept: Determinant Evaluation Using Identities

The key idea is to use properties like row/column operations or direct expansion to simplify. For (iii), note it's a skew-symmetric matrix of odd order, so its determinant is zero.

(i) Expand along the second row (which has two zeros):

(ii) Expand directly:

(iii) The matrix is skew-symmetric ($A^T = -A$) and of odd order $3 \times 3$. For such matrices, $\det(A) = 0$.

$\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix} = 0$ …

A $3\times 3$ determinant is fastest to evaluate by expanding along a row or column that has zeros, so the zero entries drop terms.

(i) $\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}$ — expand along row 2 (only $a_{23}=-1$ is nonzero):

(ii) $\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$ — expand along row 1:

(iii) $\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$ — expand along row 1: …

Method: Evaluating a $3\times3$ Determinant Efficiently

The general strategy for any "evaluate the determinant" question with numeric entries.

Steps

Step 1: Scan for a row or column with the most zeros

Expanding along a row/column with zeros eliminates those terms automatically, cutting the work roughly in half or more.

Step 2: Check for special structure before expanding at all

If the matrix is skew-symmetric ($A^T=-A$, i.e. $a_{ij}=-a_{ji}$ and every diagonal entry is $0$) and of odd order, its determinant is always $0$ — no computation needed. Other shortcuts: triangular (product of diagonal), two equal/proportional rows (determinant $0$).

Step 3: Expand along the chosen row/column using cofactors

$\Delta = a_{i1}C_{i1}+a_{i2}C_{i2}+a_{i3}C_{i3}, \qquad C_{ij}=(-1)^{i+j}M_{ij}.$ …

Common Mistakes

Mistake 1: Forgetting the alternating sign $(-1)^{i+j}$ when expanding along a row or column other than the first

Why it's wrong: the sign pattern is a checkerboard starting with $+$ at position $(1,1)$ — expanding along row $2$ or column $3$, for instance, starts with a different sign than row $1$ does, and dropping this flips the final answer. Correct approach: always explicitly compute $(-1)^{i+j}$ for the actual row/column index used, not just assume $+,-,+$.

Mistake 2: Not recognising a skew-symmetric matrix of odd order and doing unnecessary full expansion …