How to Evaluate a 3×3 Determinant: Step-by-Step NCERT Class 12 Guide

Evaluating a 3 × 3 determinant is one of the most commonly tested skills in the CBSE Class 12 Maths board exam. Once you understand the underlying method — cofactor expansion along a row or column — the calculations become systematic and reliable. This guide walks you through the complete process, explains every step, highlights smart shortcuts, and includes NCERT-aligned worked examples to strengthen your exam readiness.

Setting Up the 3 × 3 Determinant

A 3 × 3 determinant looks like this:

      | a₁₁  a₁₂  a₁₃ |
|A| = | a₂₁  a₂₂  a₂₃ |
      | a₃₁  a₃₂  a₃₃ |

The core idea is to reduce this 3 × 3 determinant into a combination of 2 × 2 determinants. This process is known as expansion along a row (or column).

A crucial fact from NCERT: expanding along any row or column always gives the same value. You do not need to expand along R₁ every time — in fact, choosing wisely can save significant calculation time.

Expansion Along the First Row (R₁)

The standard method taught in NCERT is expansion along the first row. Each element of R₁ is multiplied by its cofactor — a signed 2 × 2 determinant formed by deleting that element’s row and column.

The formula is:

|A| = a₁₁ × (−1)^(1+1) × M₁₁ + a₁₂ × (−1)^(1+2) × M₁₂ + a₁₃ × (−1)^(1+3) × M₁₃

which simplifies to:

|A| = a₁₁(a₂₂a₃₃ − a₃₂a₂₃) − a₁₂(a₂₁a₃₃ − a₃₁a₂₃) + a₁₃(a₂₁a₃₂ − a₃₁a₂₂)

Understanding the Sign Pattern

The signs follow a fixed checkerboard pattern:

+  −  +
−  +  −
+  −  +

The sign for the element in position (i, j) is (−1)^(i+j). So position (1,1) gives +1, position (1,2) gives −1, and position (1,3) gives +1.

Smart Strategy: Expand Along the Row With Most Zeros

Here is the most important exam shortcut: if any row or column contains zeros, expand along it. Each zero eliminates one 2 × 2 sub-determinant from your calculation, reducing the work significantly.

Worked Example 1 (NCERT Example 3)

Evaluate: |1 2 4; −1 3 0; 4 1 0|

The third column has two zeros. Expand along column C₃:

|A| = 4 × (−1)^(1+3) × |−1 3; 4 1| + 0 + 0

= 4 × (+1) × (−1 − 12) = 4 × (−13) = −52

Only one 2 × 2 determinant needed — from 9 multiplications down to 2.

Expansion Along the Second Row and First Column

Understanding expansion along other rows and columns builds flexibility for exam problems.

Expanding Along R₂

|A| = −a₂₁(a₁₂a₃₃ − a₃₂a₁₃) + a₂₂(a₁₁a₃₃ − a₃₁a₁₃) − a₂₃(a₁₁a₃₂ − a₃₁a₁₂)

Expanding Along C₁

|A| = a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₂₁(a₁₂a₃₃ − a₁₃a₃₂) + a₃₁(a₁₂a₂₃ − a₁₃a₂₂)

All three methods — R₁, R₂, and C₁ — produce the same final answer, confirming the consistency property of determinants.

Worked Example 2: Expansion Along R₁

Evaluate: |3 −1 −2; 0 0 −1; 3 −5 0|

Expanding along R₁:

= 3 × |0 −1; −5 0| − (−1) × |0 −1; 3 0| + (−2) × |0 0; 3 −5|

= 3(0 − 5) + 1(0 + 3) − 2(0 − 0)

= 3(−5) + 1(3) − 0 = −15 + 3 = −12

Key Rules to Remember for the Board Exam

Rule 1 — Sign depends on position: Always apply (−1)^(i+j) to get the cofactor. A common error is forgetting the sign flip.

Rule 2 — Zeros save time: Always scan all rows and columns before expanding. Pick the one with the most zeros.

Rule 3 — Scalar multiplication: If A = kB for n × n matrices, then |A| = kⁿ|B|. This appears regularly in Class 12 MCQs.

Rule 4 — Identical rows mean zero: If any two rows (or columns) of a determinant are identical, its value is zero. This is a quick shortcut for checking special determinants.

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