InterviewSolution
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InterviewSolution
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The inverse of the matrix \(\rm \begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}\)is1. \( \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & -2+3i \end{bmatrix}\)2. \( \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\)3. \( \frac{1}{12}\begin{bmatrix} 2+3i & i\\ -i & 2-3i \end{bmatrix}\)4. \( \frac{1}{12}\begin{bmatrix} 2-3i & -i\\ i & 2+3i \end{bmatrix}\) |
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Answer» Correct Answer - Option 2 : \( \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\) Concept: For a 2 x 2 matrix there is a short-cut formula for inverse as given below \(\rm {\begin{bmatrix} a & & b\\ c& & d \end{bmatrix}}^{-1} = \frac{1}{(ad-bc)}\begin{bmatrix} d & & -b\\ -c& & a \end{bmatrix}\) Calculation: A we know that inverse of matrix \(\rm {\begin{bmatrix} a & & b\\ c& & d \end{bmatrix}}^{-1} = \frac{1}{(ad-bc)}\begin{bmatrix} d & & -b\\ -c& & a \end{bmatrix}\) ⇒ \({\begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}}^{-1}= \frac{1}{(2+3i)(2-3i)-1}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\) ⇒ \({\begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}}^{-1}= \frac{1}{4+9-1}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\) ⇒ \({\begin{bmatrix} 2+3i & -i\\ i & 2-3i \end{bmatrix}}^{-1}= \frac{1}{12}\begin{bmatrix} 2-3i & i\\ -i & 2+3i \end{bmatrix}\) Hence, option 2 is correct. |
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