Mathematics GATE 2018 CH: 29

Question

For the matrix $A = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ if $\textit{det}$ stands for the determinant and $A^T$ is the transpose of $A$ then the value of $\textit{det}(A^TA)$ is __________ .

Answer 1

Answer: 1

$$A=\begin{bmatrix} \cos\theta&-\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$
$$A^{T}=\begin{bmatrix} \cos\theta& \sin\theta \\-\sin\theta & \cos\theta \end{bmatrix}$$
$$A^{T}\cdot A=\begin{bmatrix} \cos\theta& \sin\theta \\-\sin\theta & \cos\theta \end{bmatrix}\cdot\begin{bmatrix} \cos\theta&-\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$
$$A^{T}\cdot A=\begin{bmatrix} \cos^{2}\theta+\sin^{2}\theta&-\cos\theta \sin\theta +\sin\theta \cos\theta \\-\sin\theta \cos\theta +\cos\theta \sin\theta& \sin^{2}\theta+\cos^{2}\theta\end{bmatrix}$$
$$A^{T}\cdot A=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}$$
$$\text{det}(A^{T}\cdot A)=|A^{T}\cdot A|=1\times 1 = 1$$

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Answer 2

Answer =1