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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
If $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&i\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}0&{ - i}\\{ - i}&0\end{array}} \right]$, then $(A + B)(A - B)$ is equal to
  • ✓
    ${A^2} - {B^2}$
  • B
    ${A^2} + {B^2}$
  • C
    ${A^2} - {B^2} + BA + AB$
  • D
    None of these
✓

Answer

Correct option: A.
${A^2} - {B^2}$
a
(a) Here $AB = \left[ {\begin{array}{*{20}{c}}i&0\\0&i\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}0&{ - i}\\{ - i}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]$

and $BA = \left[ {\begin{array}{*{20}{c}}0&{ - i}\\{ - i}&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}i&0\\0&i\end{array}} \right]\, = \left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]$

Since $AB = BA,$ therefore $(A + B)(A - B) = {A^2} - {B^2}$.

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