Q.If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$, then $x$ is equal to (A) 6 (B) $\pm 6$ (C) $-6$ (D) 0

The key idea is that a determinant of a $2\times 2$ matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ equals $ad - bc$.

Step 1: Compute the left-hand side:

$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = x \cdot x - 2 \cdot 18 = x^2 - 36$.

Step 2: Compute the right-hand side: …

The problem gives you two determinants set equal to each other. The first has an unknown $x$ in two places; the second is a constant matrix. Your job is to find which $x$ makes the equality true.

The key idea is straightforward: compute each determinant using the rule for a $2\times2$ matrix, then solve the resulting equation.

For a matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, the determinant is $ad - bc$. That’s the only formula you need here.

1. Compute the left-hand determinant:

2. Compute the right-hand determinant:

3. Set them equal:

4. Solve: $x^2 = 36 \quad \Rightarrow \quad x = \pm 6.$ …

Method: Solving "Find $x$" Equations Between Two Determinants (MCQ form)

Same underlying technique as any equate-two-determinants problem, applied here to select the correct multiple-choice option.

Steps

Step 1: Evaluate the left-hand determinant in terms of $x$

Use $ad-bc$ on the matrix containing the unknown.

Step 2: Evaluate the right-hand (numeric) determinant

Reduce it to a single number using the same formula.

Step 3: Set the two equal and solve for $x$ …

Common Mistakes

Mistake 1: Treating a zero right-hand determinant as "no solution" or skipping the problem

Why it's wrong: a right-hand determinant that evaluates to $0$ is a perfectly valid target value — it simply means the left-hand determinant must also equal $0$, giving $x^2=36$, not an undefined or unsolvable case. Correct approach: evaluate the right-hand side numerically like any other determinant, then equate and solve normally.

Mistake 2: Selecting only the positive root ($x=6$, option A) instead of both signs …