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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
Inverse of the matrix $\left( {\begin{array}{*{20}{c}}1&{ - 2}\\3&4\end{array}} \right)$ is
  • ✓
    $\frac{1}{{10}}\left( {\begin{array}{*{20}{c}}4&2\\{ - 3}&1\end{array}} \right)$
  • B
    $\frac{1}{{10}}\left( {\begin{array}{*{20}{c}}1&{ - 2}\\3&4\end{array}} \right)$
  • C
    $\frac{1}{{10}}\left( {\begin{array}{*{20}{c}}4&2\\{  3}&1\end{array}} \right)$
  • D
    $\left( {\begin{array}{*{20}{c}}4&2\\{ - 3}&1\end{array}} \right)$
✓

Answer

Correct option: A.
$\frac{1}{{10}}\left( {\begin{array}{*{20}{c}}4&2\\{ - 3}&1\end{array}} \right)$
a
(a) Let $A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\3&4\end{array}} \right]$
$|A|\,\, = 4 + 6 = 10 \ne 0$

Now, ${A_{11}} = 4$, ${A_{12}}=-3$, ${A_{21}} = - ( - 2) = 2$, ${A_{22}} = 1$

$\therefore $ $adj\,(A) = \left[ {\begin{array}{*{20}{c}}4&2\\{ - 3}&1\end{array}} \right]$

$\therefore $ ${A^{ - 1}} = \frac{{adj\,(A)}}{{|A|}} = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}4&2\\{ - 3}&1\end{array}} \right]$.

Trick : Check from the options $A{A^{ - 1}} = I\,$

==> $A{A^{ - 1}}$=$\left[ {\begin{array}{*{20}{c}}1&{ - 2}\\3&4\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{\frac{4}{{10}}}&{\frac{2}{{10}}}\\{\frac{{ - 3}}{{10}}}&{\frac{1}{{10}}}\end{array}} \right]$= $\left[ {\begin{array}{*{20}{c}}{\frac{{10}}{{10}}}&0\\0&{\frac{{10}}{{10}}}\end{array}} \right]$

= $A{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I$.

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