Q.Evaluate the determinant $\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$.

Concept: Determinant Evaluation Using Identities – we can expand along a row/column with zeros to simplify.

Step 1: Notice the third column has two zeros (at positions $a_{23}$ and $a_{33}$). Expanding along column 3 is efficient.

Step 2: The determinant is

Step 3: Compute the $2\times2$ determinant:

$(-1)(1) - (3)(4) = -1 - 12 = -13$

Step 4: Multiply: $4 \times (-13) = -52$

The key insight here is not to blindly apply the full $3 \times 3$ formula. Instead, look for rows or columns with zeros — they make expansion much faster. In this determinant, the third column has two zeros (in the second and third rows). That means only one term survives when we expand along that column.

Let’s walk through it.

1. Choose the best expansion path.

   The third column is $(4, 0, 0)^T$. Expanding along this column means we multiply each entry by its cofactor and sum. Since the second and third entries are zero, only the first entry ($4$) contributes.

2. Write the expansion.

   Expanding along column 3:

where $C_{13}$ is the cofactor of the entry in row 1, column 3.

3. Find the cofactor $C_{13}$. The cofactor is $(-1)^{1+3} = (-1)^4 = 1$ times the minor $M_{13}$. The minor is the determinant of the $2 \times 2$ matrix left after deleting row 1 and column 3:

Compute this:

So $C_{13} = 1 \times (-13) = -13$.

4. Finish the calculation.

Method: Expansion Along the Row or Column with the Most Zeros

This method evaluates a $3\times3$ (or larger) determinant efficiently by choosing to expand along whichever row or column already contains the most zero entries, so most of the cofactor terms vanish automatically.

Steps

Step 1: Scan every row and column for zeros

Before expanding along the default first row, check every row and column of the determinant — the one with the most zeros needs the least computation.

Step 2: Choose that row/column for the expansion

If a column (or row) has two zero entries, only one cofactor term survives — the other two vanish because they're multiplied by $0$.

Step 3: Write the expansion for the single surviving term

where $M_{ij}$ is the $2\times2$ minor left after deleting row $i$ and column $j$.

Step 4: Get the sign right

Compute $(-1)^{i+j}$ carefully — even $i+j$ gives $+$, odd gives $-$. This is the step students most often get wrong.

Step 5: Evaluate the surviving $2\times2$ minor and multiply

Apply $ad-bc$ to the minor, then multiply by the nonzero entry and its sign.

Always scan for zeros before committing to an expansion row — it turns a $3\times3$ (or bigger) determinant into a single $2\times2$ calculation whenever the matrix has that structure.

Common Mistakes

Mistake 1: Getting the cofactor sign wrong when expanding along the third column

Why it's wrong: the sign attached to the surviving term is $(-1)^{i+j}$ for its position, not always $+1$ — picking the wrong sign flips the final answer's sign (here it happens to be $+$ since $1+3=4$ is even, but this must be checked, not assumed). Correct approach: explicitly compute $(-1)^{i+j}$ for the exact row/column of the nonzero entry being expanded, every time.

Mistake 2: Expanding along the first row out of habit instead of scanning for zeros first

Why it's wrong: expanding along row 1 here requires evaluating three separate $2\times2$ minors instead of just one, tripling the arithmetic and the chances of a slip. Correct approach: always scan every row and column for zeros before choosing where to expand — here column 3 (with two zeros) is far faster.