Q.(a) The value of k such that the matrix [[1, k], [–k, 1]] is symmetric is

(a) uses the symmetric-matrix condition $A=A'$; (b) is a direct matrix multiplication using double-angle formulas; (c) uses $|kA| = k^n|A|$ for an $n\times n$ matrix.

(a) Find $k$ so that $A=\begin{pmatrix}1 & k\\ -k & 1\end{pmatrix}$ is symmetric.

$A$ is symmetric iff $A = A'$. Here $A' = \begin{pmatrix}1 & -k\\ k & 1\end{pmatrix}$. Comparing off-diagonal entries: $k = -k \Rightarrow 2k=0 \Rightarrow k=0$ — option (a).

(b) If $A=\begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix}$, prove $A^2=\begin{pmatrix}\cos2\theta & \sin2\theta\\ -\sin2\theta & \cos2\theta\end{pmatrix}$.

$A^2 = \begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix}$

Entry $(1,1) = \cos^2\theta - \sin^2\theta = \cos2\theta$

Entry $(1,2) = \cos\theta\sin\theta + \sin\theta\cos\theta = 2\sin\theta\cos\theta = \sin2\theta$

Entry $(2,1) = -\sin\theta\cos\theta - \cos\theta\sin\theta = -2\sin\theta\cos\theta = -\sin2\theta$

Entry $(2,2) = -\sin\theta\sin\theta + \cos\theta\cos\theta = \cos^2\theta - \sin^2\theta = \cos2\theta$

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