Let A = cos α - sin α sin α cos α , a ∈ R such that A 32 = 0 - 1 1 0 . Then, a value of α is:

∵ A = cos α - sin α sin α cos α ⇒ A 2 = cos 2 α - sin 2 α sin 2 α cos 2 α ⇒ A 3 = cos 3 α - sin 3 α sin 3 α cos 3 α Similarly, A 32 = cos 32 α - sin 32 α sin 32 α cos 32 α Given, A 32 = 0 - 1 1 0 ⇒ cos 32 α - sin 32 α sin 32 α cos 32 α = 0 - 1 1 0 On comparing cos 32 α = 0 & sin 32 α = 1 , we get, α = π 64

Home

Mathematics - Matrices Question with Solution | TestHub

MathematicsMatricesProduct of MatricesMedium2 minPYQ_2019

MathematicsMediumsingle choice

Let $A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}], (a \in R)$ such that $A^{32} = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] .$ Then, a value of $α$ is:

Options:

Answer:

Solution:

$∵ A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}]$

${\Rightarrow A}^{2} = [\begin{matrix} \cos 2 α & - \sin 2 α \\ \sin 2 α & \cos 2 α \end{matrix}]$

$\Rightarrow A^{3} = [\begin{matrix} \cos 3 α & - \sin 3 α \\ \sin 3 α & \cos 3 α \end{matrix}]$

Similarly, $A^{32} = [\begin{matrix} \cos 32 α & - \sin 32 α \\ \sin 32 α & \cos 32 α \end{matrix}]$
Given,

$A^{32} = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} \cos 32 α & - \sin 32 α \\ \sin 32 α & \cos 32 α \end{matrix}] = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]$

On comparing $\cos 32 α = 0$ & $\sin 32 α = 1$ , we get,
$α = \frac{π}{64}$

Stream:JEESubject:MathematicsTopic:MatricesSubtopic:Product of Matrices

⏱ 2mℹ️ Source: PYQ_2019

Doubts & Discussion

Loading discussions...