CUET Preparation Today
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 |
\(\mathbb R- \{-2(\sqrt{6})\}\) \(\mathbb R- \{2(\sqrt{6})\}\) \(\mathbb R- \{2(\sqrt{6}),-2(\sqrt{6})\}\) None of these |
\(\mathbb R- \{2(\sqrt{6}),-2(\sqrt{6})\}\) |
$\begin{bmatrix}\sqrt{2}x & 8 \\6 & \sqrt{2}x \end{bmatrix}=0⇒ 2x^2-48=0$ so $x=±2\sqrt{6}$ for \(R- \{2(\sqrt{6}),-2(\sqrt{6})\}\) matrix is not singular |
