If A = 0 sin α sin α 0 and det A 2 - 1 2 I = 0 , then a possible value of α is

A 2 = sin 2 α I So, A 2 - I 2 = sin 2 α - 1 2 2 = 0 ⇒ sin α = 1 2 ⇒ α = π 4

Mathematics - Matrices Question with Solution | TestHub

MathematicsMatricesProduct of MatricesEasy2 minPYQ_2021

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If $A = [\begin{matrix} 0 & \sin α \\ \sin α & 0 \end{matrix}]$ and $\det (A^{2} - \frac{1}{2} I) = 0,$ then a possible value of $α$ is

Answer:

Solution:

$A^{2} = \sin^{2} α I$

So, $|A^{2} - \frac{I}{2}| = {(\sin^{2} α - \frac{1}{2})}^{2} = 0$

$\Rightarrow \sin α = \frac{1}{\sqrt{2}}$

$\Rightarrow α = \frac{π}{4}$

Stream:JEESubject:MathematicsTopic:MatricesSubtopic:Product of Matrices

⏱ 2mℹ️ Source: PYQ_2021

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