SATHEE: Matrices and Determinants 1 Question 3

Matrices and Determinants 1 Question 3

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

(a) $\frac{π}{32}$

(b) 0

(d) $\frac{π}{16}$

(2019 Main, 8 April I)

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Answer:

Correct Answer: 3. (c)

Solution:

Given, matrix $A = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]$

$\begin{aligned} ∴ A^{2} = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}] \begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}] \\ = [\begin{array}{cc} \cos^{2} α - \sin^{2} α & - \cos α \sin α - \sin α \cos α \\ \sin α \cos α + \cos α \sin α & - \sin^{2} α + \cos^{2} α \end{array}] \\ = [\begin{array}{cc} \cos 2 α & - \sin 2 α \\ \sin 2 α & \cos 2 α \end{array}] \end{aligned}$

Similarly,

$\begin{aligned} A^{n} & = [\begin{array}{cc} \cos (n α) & - \sin (n α) \\ \sin (n α) & \cos (n α) \end{array}], n \in N \\ \Rightarrow A^{32} & = [\begin{array}{cc} \cos (32 α) & - \sin (32 α) \\ \sin (32 α) & \cos (32 α) \end{array}] = [\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}] (given) \end{aligned}$

So, $\cos (32 α) = 0$ and $\sin (32 α) = 1$

$\Rightarrow 32 α = \frac{π}{2} \Rightarrow α = \frac{π}{64}$

Matrices and Determinants 1 Question 3

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

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Matrices and Determinants 1 Question 3

3. Let A=cos⁡α−sin⁡αsin⁡αcos⁡α,(α∈R) such that A32=0−110. Then, a value of α is

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