Question 41 markMCQSection A

If Δ1=∣100020003∣\Delta_1 = \begin{vmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 3\end{vmatrix} and Δ2=∣020100006∣\Delta_2 = \begin{vmatrix}0 & 2 & 0\\1 & 0 & 0\\0 & 0 & 6\end{vmatrix}, then

Determinants
Properties of Determinants

Options

(A)Δ1=2Δ2\Delta_1 = 2 \Delta_2
(B)Δ2=−2Δ1\Delta_2 = -2 \Delta_1
(C)Δ1=Δ2\Delta_1 = \Delta_2
(D)Δ2=−Δ1\Delta_2 = -\Delta_1

Official answer

Correct option: (B) Δ2=−2Δ1\Delta_2 = -2\Delta_1


Δ1=1⋅2⋅3=6\Delta_1 = 1\cdot 2\cdot 3 = 6 (diagonal determinant). Δ₂, expanded along the third column, equals 6×(0⋅0−2⋅1)=6×(−2)=−126 \times (0\cdot 0 - 2\cdot 1) = 6 \times (-2) = -12. Since −12=−2×6-12 = -2 \times 6, we get Δ2=−2Δ1\Delta_2 = -2\Delta_1.

Answer keywords:determinant expansiondiagonal matrix determinant3x3 determinantcofactor expansiondeterminant propertiesrow and column operations

Marking scheme

  1. 1.1 mark: correct option (B) selected — requires computing both determinants correctly.

Hint

Evaluate each 3×33\times 3 determinant by expansion (Δ₁ is diagonal, so multiply the diagonal entries directly), then compare the numeric values.

Quick oral answer

Δ₁ evaluates to 6 and Δ₂ evaluates to -12 by cofactor expansion, and since −12=−2×6-12 = -2 \times 6, the relation is Δ2=−2Δ1\Delta_2 = -2\Delta_1, option (B).

Common mistakes

  1. 1.Forgetting the sign from row interchange effects when expanding Δ₂, leading to +12+12 instead of −12-12.
  2. 2.Expanding along the wrong row/column and making arithmetic errors instead of choosing the row/column with the most zeros for easier expansion.

Analysis and explanation

This requires direct computation of two 3×3 determinants and comparing the results.


Computing Δ₁

  • Δ₁ is a diagonal matrix determinant: ∣100020003∣=1×2×3=6\begin{vmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 3\end{vmatrix} = 1 \times 2 \times 3 = 6.

Computing Δ₂

  • Δ2=∣020100006∣\Delta_2 = \begin{vmatrix}0 & 2 & 0\\1 & 0 & 0\\0 & 0 & 6\end{vmatrix}. Expanding along the third column (only the (3,3) entry = 6 is non-zero):
  • Δ2=6×∣0210∣=6×(0×0−2×1)=6×(−2)=−12\Delta_2 = 6 \times \begin{vmatrix}0 & 2\\1 & 0\end{vmatrix} = 6 \times (0\times 0 - 2\times 1) = 6 \times (-2) = -12.

Comparing the results

  • Δ1=6\Delta_1 = 6 and Δ2=−12\Delta_2 = -12, so Δ2=−2×6=−2Δ1\Delta_2 = -2 \times 6 = -2\Delta_1, confirming option (B).

Why other options are wrong

  • (A) Δ1=2Δ2\Delta_1 = 2\Delta_2 would need 6=2×(−12)=−246 = 2\times(-12) = -24, which is false.
  • (C) Δ1=Δ2\Delta_1 = \Delta_2 would need 6=−126 = -12, false.
  • (D) Δ2=−Δ1\Delta_2 = -\Delta_1 would need −12=−6-12 = -6, false.

Interesting facts

Interchanging any two rows (or columns) of a determinant changes its sign — this is exactly why Δ2\Delta_2, which is Δ1\Delta_1 with rows 1 and 2 swapped and column 2 scaled, comes out negative and scaled.

Determinants were originally developed by Japanese mathematician Seki Takakazu and German mathematician Gottfried Leibniz independently in the late 17th century, before matrix notation itself was formalized.

Question details

Q4 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Determinants
Topic
Properties of Determinants
Bloom's level
apply
Syllabus topic
Unit II: Algebra — Determinants: Evaluating Determinants and Row/Column Relations
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How do you quickly evaluate a diagonal matrix's determinant?

For any diagonal matrix, the determinant equals the product of the diagonal entries, since all cofactor expansions along rows/columns with zero off-diagonal entries collapse to just multiplying the diagonal terms.

Why did swapping rows change the sign of the determinant here?

One of the fundamental properties of determinants is that interchanging any two rows (or columns) multiplies the determinant by −1-1. Δ2\Delta_2 is related to Δ1\Delta_1 through such row/column manipulations combined with scaling, which is why the final value came out negative and scaled by a factor.