First slide

Applications of determinant

Question

If $A (θ) = [\begin{array}{cc} \sin θ & icos θ \\ icos θ & \sin θ \end{array}]$ , then which of the following is not true?

Moderate

A

$A (θ)^{- 1} = A (π - θ)$
B

$A (θ) + A (π + θ)$ is a null matrix
C

$A (θ)$ is invertible for all $θ \in R$
D

$A (θ)^{- 1} = A (- θ)$

Solution

We have, $| A (θ) | = 1$
Hence, A is invertible.
$\begin{matrix} A (π + θ) = A (π + θ) = [\begin{array}{cc} \sin (π + θ) & icos (π + θ) \\ icos (π + θ) & \sin (π + θ) \end{array}] \\ = [\begin{array}{cc} - \sin θ & - icos θ \\ - icos θ & - \sin θ \end{array}] = - A (θ) \\ adj (A (θ)) = [\begin{array}{cc} \sin θ & - icos θ \\ - icos θ & \sin θ \end{array}] \\ \Rightarrow A (θ)^{- 1} = [\begin{array}{cc} \sin θ & - icos θ \\ - icos θ & \sin θ \end{array}] = A (π - θ) \end{matrix}$

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