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CBSE Board Examination 2024 · Section A · 1 mark

Q12: Value of a 2×22 \times 2 determinant in xx

Board 2024Ch 4 · Determinants
The question ∣x+1x−1x2+x+1x2−x+1∣\begin{vmatrix} x + 1 & x - 1 \\ x^2 + x + 1 & x^2 - x + 1 \end{vmatrix}

is equal to :

(A) 2x32x^3

(B) 22

(C) 00

(D) 2x3−22x^3 - 2

Exam answer

Expanding, and using the identities for a3+b3a^3 + b^3 and a3−b3a^3 - b^3⁠:

Why?

A 2×22 \times 2 determinant is the product of the leading diagonal minus the product of the other diagonal:

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

Here ad=(x+1)(x2−x+1)ad = (x + 1)(x^2 - x + 1) and bc=(x−1)(x2+x+1)bc = (x - 1)(x^2 + x + 1)⁠.

(x+1)(x2−x+1)−(x−1)(x2+x+1)=(x3+1)−(x3−1)=x3+1−x3+1=2\begin{aligned} &(x + 1)(x^2 - x + 1) \\ &\quad - (x - 1)(x^2 + x + 1) \\ &= (x^3 + 1) - (x^3 - 1) \\ &= x^3 + 1 - x^3 + 1 \\ &= 2 \end{aligned}
Why?

Each product is a standard identity, with a=xa = x and b=1b = 1⁠:

(a+b)(a2−ab+b2)=a3+b3(a−b)(a2+ab+b2)=a3−b3\begin{aligned} &(a + b)(a^2 - ab + b^2) \\ &= a^3 + b^3 \\ &(a - b)(a^2 + ab + b^2) \\ &= a^3 - b^3 \end{aligned}

If you do not remember the identities, multiply out term by term:

(x+1)(x2−x+1)=x3−x2+x+x2−x+1=x3+1(x−1)(x2+x+1)=x3+x2+x−x2−x−1=x3−1\begin{aligned} &(x + 1)(x^2 - x + 1) \\ &= x^3 - x^2 + x \\ &\quad + x^2 - x + 1 \\ &= x^3 + 1 \\[4pt] &(x - 1)(x^2 + x + 1) \\ &= x^3 + x^2 + x \\ &\quad - x^2 - x - 1 \\ &= x^3 - 1 \end{aligned}

(The x2x^2 terms cancel and the xx terms cancel in each.)

Watch out for the sign when the bracket opens: −(x3−1)=−x3+1-(x^3 - 1) = -x^3 + 1⁠. The x3x^3 terms cancel, so the value is 22 for every xx⁠.

Answer: (B) 22

Where this comes from

Chapter 4, Determinants: Evaluating a 2 × 2 determinant.

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: The bars hold a 2×22 \times 2⁠: cross-multiply to get (x+1)(x2−x+1)−(x−1)(x2+x+1)(x + 1)(x^2 - x + 1) - (x - 1)(x^2 + x + 1)⁠. The cube identities make it (x3+1)−(x3−1)=2(x^3 + 1) - (x^3 - 1) = 2⁠.

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