The inverse of the matrix [ * 20 c 3& - 2 1&4 ]is - Vidyadip

MCQ

The inverse of the matrix $\left[ {\begin{array}{*{20}{c}}3&{ - 2}\\1&4\end{array}} \right]$is

✓
$\left[ {\begin{array}{*{20}{c}}{\frac{4}{{14}}}&{\frac{2}{{14}}}\\{\frac{{ - 1}}{{14}}}&{\frac{3}{{14}}}\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}{\frac{3}{{14}}}&{\frac{{ - 2}}{{14}}}\\{\frac{1}{{14}}}&{\frac{4}{{14}}}\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}{\frac{4}{{14}}}&{\frac{{ - 2}}{{14}}}\\{\frac{1}{{14}}}&{\frac{3}{{14}}}\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}{\frac{3}{{14}}}&{\frac{2}{{14}}}\\{\frac{1}{{14}}}&{\frac{4}{{14}}}\end{array}} \right]$

✓

Answer

Correct option: A.

$\left[ {\begin{array}{*{20}{c}}{\frac{4}{{14}}}&{\frac{2}{{14}}}\\{\frac{{ - 1}}{{14}}}&{\frac{3}{{14}}}\end{array}} \right]$

a
(a) Let $A = \left[ {\begin{array}{*{20}{c}}3&{ - 2}\\1&4\end{array}} \right]\, \Rightarrow |A| = 14$
$\therefore $ $adj\,A = \left[ {\begin{array}{*{20}{c}}4&2\\{ - 1}&3\end{array}} \right]$ $ \Rightarrow $ ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\frac{4}{{14}}}&{\frac{2}{{14}}}\\{\frac{{ - 1}}{{14}}}&{\frac{3}{{14}}}\end{array}} \right]$.

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