Q.Using the properties of determinants, prove that: $\begin{vmatrix} a^2 + 2a & 2a + 1 & 1 \\ 2a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{vmatrix} = (a - 1)^3$

Concept: Determinant Evaluation Using Identities (Row/Column Operations)

Step 1: Apply $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$.

Step 2: Factor $(a-1)$ from $R_1$ and $R_2$:

$(a-1)^2 \begin{vmatrix} a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{vmatrix}$ …

We want to show that

The direct expansion would be messy because of the $a^2$ term. Instead, we use row operations that do not change the determinant’s value. The key idea: if we subtract one row from another, the determinant stays the same. This can create zeros or common factors, making the determinant much simpler.

1. Subtract the third row from the first two rows The operation $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$ leaves the determinant unchanged.

Notice the third column now has two zeros and a 1. This is promising.

2. Factor common terms from the first two rows In the first row, $2a - 2 = 2(a-1)$. In the second row, $2a - 2 = 2(a-1)$ and $a-1$ is already a factor. Also, $a^2 + 2a - 3$ factors as $(a+3)(a-1)$ because $a^2 + 2a - 3 = (a+3)(a-1)$. So we can write:

Factor $(a-1)$ from the first row and $(a-1)$ from the second row. Since factoring a constant from a row multiplies the determinant by that constant, we get:

3. Expand along the third column The third column has entries $0, 0, 1$. Expanding along this column is the fastest route. The determinant becomes: …

Method: Row Subtraction to Expose Repeated Factors

When a determinant's entries are polynomials in one variable (here, expressions in $a$) and the target answer is a clean power like $(a-1)^3$, subtracting a "reference" row from the others is the standard way to surface the repeated factor.

Steps

Step 1: Pick a reference row to subtract

Look for the row with the simplest entries (often the one with constants only) and subtract it from the other rows: $R_1 \to R_1-R_3$, $R_2 \to R_2-R_3$. This does not change the determinant's value.

Step 2: Factor each modified row

After subtraction, the entries in the modified rows should factor — typically each will share a linear factor like $(a-1)$. Factor that constant out of each row separately (each factoring multiplies the determinant by that constant).

Step 3: Expand along a column with zeros …

Common Mistakes

Mistake 1: Factoring from two rows but only counting the factor once

$(a-1)$ is pulled out of both Row 1 and Row 2 separately, so it must appear as $(a-1)^2$ in front of the determinant — not $(a-1)$. Each row that contributes a common factor multiplies the determinant by that factor independently.

Mistake 2: Misfactoring the quadratic

$a^2+2a-3$ factors as $(a+3)(a-1)$. A sign slip here (e.g. writing $(a-3)(a+1)$) breaks the common-factor cancellation with the other row's $(a-1)$ even if every row operation before it was correct.

Mistake 3: Expanding along the wrong row or column …