Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

For what value(s) of x is A = \(\begin{bmatrix}(\sqrt{2})x & 8 \\6 & (\sqrt{2})x \end{bmatrix}\) is not a singular matrix

Options:

\(\mathbb R- \{-2(\sqrt{6})\}\)

\(\mathbb R- \{2(\sqrt{6})\}\)

\(\mathbb R- \{2(\sqrt{6}),-2(\sqrt{6})\}\)

None of these

Correct Answer:

\(\mathbb R- \{2(\sqrt{6}),-2(\sqrt{6})\}\)

Explanation:

$\begin{bmatrix}\sqrt{2}x & 8 \\6 & \sqrt{2}x \end{bmatrix}=0⇒ 2x^2-48=0$

so $x=±2\sqrt{6}$

for \(R- \{2(\sqrt{6}),-2(\sqrt{6})\}\)

matrix is not singular