Q.If $A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, find $|A|$.

Create zeros in the first column using $R_2\to R_2-2R_1$ and $R_3\to R_3-5R_1$: …

The idea

Rather than expand blindly, use the free row operation $R_i\to R_i-\lambda R_1$ to make zeros; if two rows end up equal, the determinant is immediately $0$.

Reduce the first column

Subtract multiples of $R_1$ from the other rows (this does not change the determinant):

The determinant becomes

$|A| = \begin{vmatrix} 1 & 1 & -2 \\ 0 & -1 & 1 \\ 0 & -1 & 1 \end{vmatrix}.$ …

Method: Spotting a Zero Determinant via Row Reduction to Identical Rows

A faster alternative to brute-force cofactor expansion when the entries look like they might collapse.

Steps

Step 1: Use row operations to create zeros in one column

Apply $R_i \to R_i - \lambda R_1$ for suitable $\lambda$ to each of the other rows — this never changes the determinant's value.

Step 2: Compare the resulting rows

After reduction, check whether two rows have become identical (or one has become a scalar multiple of another).

Step 3: If so, conclude the determinant is $0$ immediately …

Common Mistakes

Mistake 1: Expanding the original (un-reduced) determinant by brute-force cofactors instead of using row operations first

Why it's wrong: this skips the much faster route of spotting that two rows become identical after a simple row reduction, and increases the chance of an arithmetic mistake during full expansion. Correct approach: try $R_i \to R_i - \lambda R_1$ first to see if the structure simplifies before expanding anything.

Mistake 2: Making an arithmetic slip in one of the row operations and missing that two rows actually match …