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Worked Examples · Example 15

Q.If P=[cos⁡xsin⁡x−sin⁡xcos⁡x]P = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}, then verify that P′P=IP'P = I, where II is an identity matrix.

Lakshadweep CbseNCERTSubjective· 3mImportance★★★★★est
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Taking the transpose P′P' and multiplying it by PP, every product simplifies through cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1, giving the 2×22\times 2 identity matrix.

P′=transpose of P(rows↔columns),(P′P)ij=∑k(P′)ik PkjP' = \text{transpose of } P \quad\text{(rows}\leftrightarrow\text{columns)},\qquad (P'P)_{ij} = \sum_k (P')_{ik}\,P_{kj}

Here P=[cos⁡xsin⁡x−sin⁡xcos⁡x]P = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} and I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

  1. Write the transpose by interchanging rows and columns:

P′=[cos⁡x−sin⁡xsin⁡xcos⁡x].P' = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix}.

  1. Multiply P′PP'P:

P′P=[cos⁡x−sin⁡xsin⁡xcos⁡x][cos⁡xsin⁡x−sin⁡xcos⁡x].P'P = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix}\begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}.

  1. Entry (1,1)(1,1): cos⁡x⋅cos⁡x+(−sin⁡x)(−sin⁡x)=cos⁡2x+sin⁡2x=1\cos x\cdot\cos x + (-\sin x)(-\sin x) = \cos^2 x + \sin^2 x = 1.
  2. Entry (1,2)(1,2): cos⁡x⋅sin⁡x+(−sin⁡x)cos⁡x=sin⁡xcos⁡x−sin⁡xcos⁡x=0\cos x\cdot\sin x + (-\sin x)\cos x = \sin x\cos x - \sin x\cos x = 0.
  3. Entry (2,1)(2,1): sin⁡x⋅cos⁡x+cos⁡x⋅(−sin⁡x)=sin⁡xcos⁡x−sin⁡xcos⁡x=0\sin x\cdot\cos x + \cos x\cdot(-\sin x) = \sin x\cos x - \sin x\cos x = 0.
  4. Entry (2,2)(2,2): sin⁡x⋅sin⁡x+cos⁡x⋅cos⁡x=sin⁡2x+cos⁡2x=1\sin x\cdot\sin x + \cos x\cdot\cos x = \sin^2 x + \cos^2 x = 1.
  5. Assemble the result:

P′P=[1001]=I.P'P = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I.

✓Final answer

P′P=[1001]=IP'P = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I is verified. (PP is an orthogonal matrix.)

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