CBSE Sample Paper 2022-23 · Section A · 1 mark

Q12: $x$ from an equation of two $2 \times 2$ determinants

Sample 2022-23Ch 4 · Determinants

The question

\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix},

then the possible value(s) of 'x' is/are

(a) $3$

(b) $\sqrt 3$

(d) $\sqrt 3, -\sqrt 3$

Exam answer

Expanding both determinants,

\begin{aligned} 2(1) - 4(5) &= 2x(x) - 4(6) \\ 2 - 20 &= 2x^2 - 24 \end{aligned}

Why?

A $2 \times 2$ determinant is the product of the main diagonal minus the product of the other diagonal:

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

On the right, $a = 2x$ ⁠, $b = 4$ ⁠, $c = 6$ and $d = x$ ⁠.

\begin{aligned} \Rightarrow 2x^2 &= 6 \\ \Rightarrow x^2 &= 3 \end{aligned}

Why?

Add 24 to both sides: $2x^2 = -18 + 24 = 6$ ⁠. Then divide by 2.

\begin{aligned} 2 - 20 &= 2x^2 - 24 \\ -18 &= 2x^2 - 24 \\ -18 + 24 &= 2x^2 \\ 6 &= 2x^2 \\ \frac{6}{2} &= x^2 \\ 3 &= x^2 \end{aligned}

$\therefore\ x = \pm\sqrt 3$

Why?

Watch out: $x^2 = 3$ has two roots, $\sqrt 3$ and $-\sqrt 3$ ⁠, and both satisfy the equation, because $x$ appears only as $x^2$ ⁠. So option (d), which lists both, is the answer, not (b) or (c) alone.

Answer: (d) $\sqrt 3, -\sqrt 3$

Where this comes from

Chapter 4, Determinants: Equation in 2 × 2 determinants.

Question type: Find $x$

How to spot it: click to see

Type 2. Find $x$ ⁠: two determinants set equal

Spot it: two determinants with an equals sign between them, and $x$ inside.

Here: Two $2 \times 2$ determinants are set equal: cross-multiply each, $2 - 20 = 2x^2 - 24$ ⁠, so $x^2 = 3$ and $x = \pm\sqrt3$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

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