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Official PYQ · 1996Q2.6Linear Algebra: Matrices, Determinants, System of Linear EquationsMCQ1 mark

GATE CS 1996 Q2.6 — Linear Algebra: Matrices, Determinants, System of Linear Equations

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The matrices [cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix} \cos\theta && -\sin\theta \\ \sin\theta && \cos\theta \end{bmatrix} and [a00b]\begin{bmatrix} a && 0\\ 0&& b \end{bmatrix} commute under multiplication

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Answer: A

Let the two matrices be defined as:

A=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]andB=[a00b]A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}

For the matrices to commute under multiplication, they must satisfy the condition AB=BAAB = BA.

Let us compute the product ABAB:

AB=[cos⁡θ−sin⁡θsin⁡θcos⁡θ][a00b]=[acos⁡θ−bsin⁡θasin⁡θbcos⁡θ]AB = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} = \begin{bmatrix} a\cos\theta & -b\sin\theta \\ a\sin\theta & b\cos\theta \end{bmatrix}

Now, let us compute the product BABA:

BA=[a00b][cos⁡θ−sin⁡θsin⁡θcos⁡θ]=[acos⁡θ−asin⁡θbsin⁡θbcos⁡θ]BA = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} = \begin{bmatrix} a\cos\theta & -a\sin\theta \\ b\sin\theta & b\cos\theta \end{bmatrix}

Equating the corresponding elements of ABAB and BABA:

  1. The diagonal elements are already equal: acos⁡θ=acos⁡θa\cos\theta = a\cos\theta and bcos⁡θ=bcos⁡θb\cos\theta = b\cos\theta.
  2. For the off-diagonal elements, we require: −bsin⁡θ=−asin⁡θ  ⟹  (a−b)sin⁡θ=0-b\sin\theta = -a\sin\theta \implies (a - b)\sin\theta = 0 and asin⁡θ=bsin⁡θ  ⟹  (a−b)sin⁡θ=0a\sin\theta = b\sin\theta \implies (a - b)\sin\theta = 0

The equation (a−b)sin⁡θ=0(a - b)\sin\theta = 0 is satisfied if and only if:

  • a−b=0  ⟹  a=ba - b = 0 \implies a = b, or
  • sin⁡θ=0  ⟹  θ=nπ\sin\theta = 0 \implies \theta = n\pi (where nn is an integer).

Thus, the matrices commute if a=ba = b or θ=nπ\theta = n\pi, which corresponds to Option A.

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