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Q.Simplify : cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\cos\theta\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix} OR Show that the matrix A=[1−15−121513]A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} is a symmetric matrix.

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2026Subjective· 2mImportance★★★★★
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Multiply the scalars into each matrix, add corresponding entries, and simplify using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1; the result is the identity matrix.

cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\cos\theta\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix}

Step 1: Scale each matrix.

=[cos⁡2θsin⁡θcos⁡θ−sin⁡θcos⁡θcos⁡2θ]+[sin⁡2θ−sin⁡θcos⁡θsin⁡θcos⁡θsin⁡2θ]= \begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix}

Step 2: Add corresponding entries.

Entry (1,1)(1,1): cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta = 1

Entry (1,2)(1,2): sin⁡θcos⁡θ−sin⁡θcos⁡θ=0\sin\theta\cos\theta - \sin\theta\cos\theta = 0

Entry (2,1)(2,1): −sin⁡θcos⁡θ+sin⁡θcos⁡θ=0-\sin\theta\cos\theta+\sin\theta\cos\theta = 0

Entry (2,2)(2,2): cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta = 1

=[1001]=I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

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