Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Question:

The minimum value of $\begin{vmatrix}2&2&2\\2&2+x&2\\2&2&2+x\end{vmatrix},x∈R$ is

Options:

1

-2

0

$-\frac{1}{2}$

Correct Answer:

0

Explanation:

The correct answer is Option (3) → 0

Given determinant:

$D = \begin{vmatrix} 2 & 2 & 2 \\ 2 & 2+x & 2 \\ 2 & 2 & 2+x \end{vmatrix}$

Subtract first row from second and third rows:

Row2 → Row2 − Row1 = [0, x, 0]

Row3 → Row3 − Row1 = [0, 0, x]

Determinant becomes:

$D = \begin{vmatrix} 2 & 2 & 2 \\ 0 & x & 0 \\ 0 & 0 & x \end{vmatrix} = 2 \cdot x \cdot x = 2x^2$

Minimum value occurs at $x = 0$: $D_{\min} = 0$