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CBSE Board Examination 2026 · Section A · 1 mark

Q4: Comparing the determinants Δ1\Delta_1 and Δ2\Delta_2

Board 2026Ch 4 · Determinants
The question

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

(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

Exam answer

Δ1\Delta_1 is a diagonal determinant, so Δ1=1×2×3=6\Delta_1 = 1 \times 2 \times 3 = 6⁠.

Why?

Every entry off the leading diagonal is 0. Expanding along R1R_1 leaves only one term:

Δ1=1∣2003∣=1((2)(3)−(0)(0))=1(6−0)=6\begin{aligned} \Delta_1 &= 1\begin{vmatrix} 2 & 0 \\ 0 & 3 \end{vmatrix} \\ &= 1\big((2)(3) - (0)(0)\big) \\ &= 1(6 - 0) = 6 \end{aligned}

So a diagonal determinant is the product of its diagonal entries.

Expanding Δ2\Delta_2 along R3R_3⁠:

Δ2=6∣0210∣=6(0−2)=−12\begin{aligned} \Delta_2 &= 6\begin{vmatrix} 0 & 2 \\ 1 & 0 \end{vmatrix} \\ &= 6(0 - 2) = -12 \end{aligned}
Why?

R3R_3 is 0,0,60, 0, 6⁠. Its two zeros give no terms, so only the 6 counts. It sits in position (3,3)(3, 3)⁠, with sign (−1)3+3=+1(-1)^{3+3} = +1⁠.

Its minor deletes R3R_3 and C3C_3⁠. A 2×22 \times 2 determinant is

∣abcd∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

so the minor is

∣0210∣=(0)(0)−(2)(1)=0−2=−2\begin{aligned} \begin{vmatrix} 0 & 2 \\ 1 & 0 \end{vmatrix} &= (0)(0) - (2)(1) \\ &= 0 - 2 = -2 \end{aligned}

Then Δ2=(+1)(6)(−2)=−12\Delta_2 = (+1)(6)(-2) = -12⁠.

∴ Δ2=−12=−2Δ1\therefore\ \Delta_2 = -12 = -2\Delta_1

Why?

−2Δ1=−2×6=−12-2\Delta_1 = -2 \times 6 = -12⁠, which is Δ2\Delta_2⁠.

Another way to see it: Δ2\Delta_2 is the diagonal determinant with entries 1, 2, 6 after swapping R1R_1 and R2R_2⁠. That diagonal determinant is 1×2×6=121 \times 2 \times 6 = 12⁠, and a row swap changes the sign, so Δ2=−12\Delta_2 = -12⁠.

Check the others: 2Δ2=−24≠Δ12\Delta_2 = -24 \ne \Delta_1⁠, Δ1≠Δ2\Delta_1 \ne \Delta_2 and −Δ1=−6≠Δ2-\Delta_1 = -6 \ne \Delta_2⁠.

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

Where this comes from

Chapter 4, Determinants: Evaluating 3 × 3 determinants.

Question type: Evaluate a determinant

How to spot it: click to see

Type 1. Evaluate a determinant: cross for 2×22\times2⁠, expand for 3×33\times3

Spot it: "evaluate" and a square block of numbers between straight bars.

Here: Two 3×33 \times 3 blocks to evaluate: Δ1\Delta_1 is diagonal, so it is 1×2×3=61 \times 2 \times 3 = 6⁠. Expand Δ2\Delta_2 along R3R_3⁠, which has two zeros: 6(0−2)=−126(0 - 2) = -12⁠.

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