Determinants: Tricky Questions Solved (7)

easy 2 min read

Question

Evaluate the determinant 123456789\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}.

Solution — Step by Step

Subtract R1R_1 from R2R_2 and R1R_1 from R3R_3 (this doesn’t change the determinant value when we replace a row with itself plus a multiple of another row):

R2R2R1R_2 \to R_2 - R_1 gives (3,3,3)(3, 3, 3). R3R3R1R_3 \to R_3 - R_1 gives (6,6,6)(6, 6, 6).

The matrix is now:

123333666\begin{vmatrix} 1 & 2 & 3 \\ 3 & 3 & 3 \\ 6 & 6 & 6 \end{vmatrix}

Rows 2 and 3 are proportional (R3=2R2R_3 = 2R_2). When two rows are proportional, the determinant is zero.

det=0\det = 0

Final answer: the determinant is 0\boxed{0}.

Why This Works

A determinant is zero when the rows (or columns) are linearly dependent. After our row operations, two rows became proportional, signalling dependence. The original matrix’s rows form an arithmetic progression along each column, which guarantees linear dependence — Row 2 = (Row 1 + Row 3)/2.

This pattern (three rows in AP) always gives determinant zero. Recognising this saves a lot of expansion work.

Alternative Method

Expand along the first row:

det=1(4548)2(3642)+3(3235)=3+129=0\det = 1 \cdot (45 - 48) - 2 \cdot (36 - 42) + 3 \cdot (32 - 35) = -3 + 12 - 9 = 0

Same answer, more arithmetic.

If rows or columns are in arithmetic progression, the determinant is zero. Check for AP patterns first — they appear in JEE Main 1-2 times a year.

Common Mistake

Computing a determinant straight from cofactor expansion when row operations would simplify dramatically. The exam clock matters — recognise patterns first, expand only as a last resort. Also, some students do row operations incorrectly: only RiRi+kRjR_i \to R_i + kR_j preserves the determinant. Multiplying a row by kk multiplies the determinant by kk.

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