Question 41 markMCQSection A

The value of the determinant ∣123456789∣\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix} is :

Algebra (Determinants)
Determinants

Options

(A)5
(B)-7
(C)9
(D)0

Official answer

The correct option is D) 0.


Working (expansion along Row 1):


  • =1(5⋅9−6⋅8)−2(4⋅9−6⋅7)+3(4⋅8−5⋅7)= 1(5 \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - 5 \cdot 7)
  • =1(45−48)−2(36−42)+3(32−35)= 1(45 - 48) - 2(36 - 42) + 3(32 - 35)
  • =1(−3)−2(−6)+3(−3)= 1(-3) - 2(-6) + 3(-3)
  • =−3+12−9=0= -3 + 12 - 9 = 0.

Shortcut: The rows are in arithmetic progression (R2−R1=R3−R2R_2 - R_1 = R_3 - R_2), so R1+R3=2R2R_1 + R_3 = 2R_2, making one row a linear combination of the others — the determinant is 0.

Answer keywords:determinant3x3 determinantrow in arithmetic progressionlinear dependencecofactor expansionzero determinantsingular matrix

Marking scheme

  1. 1.1 mark: correct expansion or dependence argument giving determinant = 0 (option D).

Hint

Notice the rows are in AP (R1+R3=2R2R_1 + R_3 = 2R_2) — a dependent row forces the determinant to zero.

Quick oral answer

The rows are in AP so R1+R3=2R2R_1 + R_3 = 2R_2; a dependent row makes the determinant zero, which expansion confirms as −3+12−9=0-3 + 12 - 9 = 0.

Common mistakes

  1. 1.Sign errors in the alternating cofactor signs (+−++ - +) during expansion.
  2. 2.Arithmetic slips in the 2×22 \times 2 minors (e.g. 45−4845 - 48).
  3. 3.Not spotting the AP pattern that immediately gives 0.

Analysis and explanation

This tests determinant evaluation and the property that a linear dependence among rows gives a zero determinant.


Concept:


  • If any row (or column) is a linear combination of the others, the determinant is 0.
  • Here R1+R3=2R2R_1 + R_3 = 2R_2, a clear linear dependence.

Why D is correct:


  • Direct expansion gives −3+12−9=0-3 + 12 - 9 = 0, matching the dependence shortcut.

Why the distractors are wrong:


  • A (5), B (−7), C (9): all arise from sign or multiplication errors in the cofactor expansion; none respects the row-dependence property that guarantees 0.

Interesting facts

A zero determinant means the matrix is singular (non-invertible) — its rows/columns are linearly dependent.

Any 3×33 \times 3 matrix whose consecutive entries form an arithmetic progression along rows has determinant 0, a neat exam shortcut.

Question details

Q4 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Algebra (Determinants)
Topic
Determinants
Bloom's level
apply
Syllabus topic
Unit 2: Algebra — Determinants: evaluation and properties
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is the determinant exactly zero here?

Because the rows are linearly dependent: R1+R3=2R2R_1 + R_3 = 2R_2. Whenever one row is a combination of the others, the determinant is 0.

Does a zero determinant mean the matrix has no inverse?

Yes. A square matrix is invertible only if its determinant is non-zero; determinant 0 makes it singular and non-invertible.