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Matrices and Determinants question

2024 · 27 Jan · Shift 1 · Q41
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  5. /2024 · 27 Jan · Shift 1 · Q41

Matrices and Determinants question

2024 · 27 Jan · Shift 1 · Q41

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Consider the matrix f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]f(x)=​cosxsinx0​−sinxcosx0​001​​. Given below are two statements : Statement I : f(−x)f(-x)f(−x) is the inverse of the matrix f(x)f(x)f(x). Statement II : f(x)f(y)=f(x+y)f(x) f(y)=f(x+y)f(x)f(y)=f(x+y). In the light of the above statements, choose the correct answer from the options given below :
  1. A
    Statement I is false but Statement II is true
  2. B
    Both Statement I and Statement II are false
  3. C
    Both Statement I and Statement II are true
  4. D
    Statement I is true but Statement II is false
View written solutionFree

Correct answer: C

  1. Write the matrix clearly

Given

f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x)=\begin{bmatrix} \cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0&0&1 \end{bmatrix}f(x)=​cosxsinx0​−sinxcosx0​001​​

This is the standard rotation matrix in the xyxyxy-plane, with zzz-coordinate unchanged.

  1. Check Statement I: f(−x)f(-x)f(−x) is the inverse of f(x)f(x)f(x)

First compute f(−x)f(-x)f(−x):

f(−x)=[cos⁡(−x)−sin⁡(−x)0sin⁡(−x)cos⁡(−x)0001]f(-x)=\begin{bmatrix} \cos(-x) & -\sin(-x) & 0\\ \sin(-x) & \cos(-x) & 0\\ 0&0&1 \end{bmatrix}f(−x)=​cos(−x)sin(−x)0​−sin(−x)cos(−x)0​001​​

Using

cos⁡(−x)=cos⁡x,sin⁡(−x)=−sin⁡x\cos(-x)=\cos x, \qquad \sin(-x)=-\sin xcos(−x)=cosx,sin(−x)=−sinx

we get

f(−x)=[cos⁡xsin⁡x0−sin⁡xcos⁡x0001]f(-x)=\begin{bmatrix} \cos x & \sin x & 0\\ -\sin x & \cos x & 0\\ 0&0&1 \end{bmatrix}f(−x)=​cosx−sinx0​sinxcosx0​001​​

Now multiply f(x)f(x)f(x) and f(−x)f(-x)f(−x):

f(x)f(−x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001][cos⁡xsin⁡x0−sin⁡xcos⁡x0001]f(x)f(-x)= \begin{bmatrix} \cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0\\ -\sin x & \cos x & 0\\ 0&0&1 \end{bmatrix}f(x)f(−x)=​cosxsinx0​−sinxcosx0​001​​​cosx−sinx0​sinxcosx0​001​​

Compute entries of the top-left 2×22\times 22×2 block:

(1,1)=cos⁡2x+sin⁡2x=1,(1,2)=cos⁡xsin⁡x−sin⁡xcos⁡x=0,(2,1)=sin⁡xcos⁡x−cos⁡xsin⁡x=0,(2,2)=sin⁡2x+cos⁡2x=1.\begin{aligned} (1,1)&=\cos^2 x+\sin^2 x=1,\\ (1,2)&=\cos x\sin x-\sin x\cos x=0,\\ (2,1)&=\sin x\cos x-\cos x\sin x=0,\\ (2,2)&=\sin^2 x+\cos^2 x=1. \end{aligned}(1,1)(1,2)(2,1)(2,2)​=cos2x+sin2x=1,=cosxsinx−sinxcosx=0,=sinxcosx−cosxsinx=0,=sin2x+cos2x=1.​

And the third row/column remains unchanged, so

f(x)f(−x)=I3f(x)f(-x)=I_3f(x)f(−x)=I3​

Similarly, f(−x)f(x)=I3f(-x)f(x)=I_3f(−x)f(x)=I3​. Hence,

f(−x)=f(x)−1f(-x)=f(x)^{-1}f(−x)=f(x)−1

So Statement I is true.

  1. Check Statement II: f(x)f(y)=f(x+y)f(x)f(y)=f(x+y)f(x)f(y)=f(x+y)

Write both matrices:

f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001],f(y)=[cos⁡y−sin⁡y0sin⁡ycos⁡y0001]f(x)=\begin{bmatrix} \cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0&0&1 \end{bmatrix}, \qquad f(y)=\begin{bmatrix} \cos y & -\sin y & 0\\ \sin y & \cos y & 0\\ 0&0&1 \end{bmatrix}f(x)=​cosxsinx0​−sinxcosx0​001​​,f(y)=​cosysiny0​−sinycosy0​001​​

Now multiply:

f(x)f(y)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001][cos⁡y−sin⁡y0sin⁡ycos⁡y0001]f(x)f(y)= \begin{bmatrix} \cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0\\ \sin y & \cos y & 0\\ 0&0&1 \end{bmatrix}f(x)f(y)=​cosxsinx0​−sinxcosx0​001​​​cosysiny0​−sinycosy0​001​​

Compute the top-left 2×22\times 22×2 block:

(1,1)=cos⁡xcos⁡y−sin⁡xsin⁡y=cos⁡(x+y),(1,2)=−cos⁡xsin⁡y−sin⁡xcos⁡y=−sin⁡(x+y),(2,1)=sin⁡xcos⁡y+cos⁡xsin⁡y=sin⁡(x+y),(2,2)=−sin⁡xsin⁡y+cos⁡xcos⁡y=cos⁡(x+y).\begin{aligned} (1,1)&=\cos x\cos y-\sin x\sin y=\cos(x+y),\\ (1,2)&=-\cos x\sin y-\sin x\cos y=-\sin(x+y),\\ (2,1)&=\sin x\cos y+\cos x\sin y=\sin(x+y),\\ (2,2)&=-\sin x\sin y+\cos x\cos y=\cos(x+y). \end{aligned}(1,1)(1,2)(2,1)(2,2)​=cosxcosy−sinxsiny=cos(x+y),=−cosxsiny−sinxcosy=−sin(x+y),=sinxcosy+cosxsiny=sin(x+y),=−sinxsiny+cosxcosy=cos(x+y).​

Thus,

f(x)f(y)=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos⁡(x+y)0001]=f(x+y)f(x)f(y)=\begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0\\ \sin(x+y) & \cos(x+y) & 0\\ 0&0&1 \end{bmatrix}=f(x+y)f(x)f(y)=​cos(x+y)sin(x+y)0​−sin(x+y)cos(x+y)0​001​​=f(x+y)

So Statement II is true.

  1. Conclusion
  • Statement I: True
  • Statement II: True

Therefore, the correct option is:

C\boxed{\text{C}}C​
  1. Comparison with stored correct answer

Stored correct answer: C\text{C}C

My derived answer: C\text{C}C

So, the answer matches the stored correct answer.

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