First slide

Transpose of matrix

Question

If $A = [\begin{array}{rr} \cos θ - \sin θ \\ \sin θ \cos θ \end{array}]$ ,then $A^{T} + A = I_{2}$ if

Easy

A

$θ = nπ, n \in Z$
B

$θ = (2 n + 1) \frac{π}{2}, n \in Z$
C

$θ = 2 nπ + \frac{π}{3}, n \in Z$
D

None of these

Solution

We have, $A^{T} + A = I_{2}$
$\begin{array}{l} \Rightarrow [\begin{array}{rr} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}] + [\begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{cl} 2 \cos θ & 0 \\ 0 & 2 \cos θ \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] \\ \Rightarrow \cos θ = \frac{1}{2} = \cos \frac{π}{3} \Rightarrow θ = 2 nπ \pm \frac{π}{3}, n \in Z \end{array}$

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