If A = [ * 20 c 1&2 3& - 5 ], then A^ - 1 = - Vidyadip

MCQ

If $A = \left[ {\begin{array}{*{20}{c}}1&2\\3&{ - 5}\end{array}} \right]$, then ${A^{ - 1}}$=

A
$\left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 2}\\{ - 3}&1\end{array}} \right]$
✓
$\left[ {\begin{array}{*{20}{c}}{\frac{5}{{11}}}&{\frac{2}{{11}}}\\{\frac{3}{{11}}}&{ - \frac{1}{{11}}}\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}{ - \frac{5}{{11}}}&{ - \frac{2}{{11}}}\\{ - \frac{3}{{11}}}&{ - \frac{1}{{11}}}\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}5&2\\3&{ - 1}\end{array}} \right]$

✓

Answer

Correct option: B.

$\left[ {\begin{array}{*{20}{c}}{\frac{5}{{11}}}&{\frac{2}{{11}}}\\{\frac{3}{{11}}}&{ - \frac{1}{{11}}}\end{array}} \right]$

b
(b) $A = \left[ {\,\begin{array}{*{20}{c}}1&2\\3&{ - 5}\end{array}\,} \right]$

$adj\,A = \left[ {\,\begin{array}{*{20}{c}}{ - 5}&{ - 2}\\{ - 3}&{ + 1}\end{array}\,} \right]$

$|A| = \left| {\,\begin{array}{*{20}{c}}1&2\\3&{ - 5}\end{array}\,} \right| = - 11$

$\therefore \,\,{A^{ - 1}} = \frac{1}{{11}}\left[ {\,\begin{array}{*{20}{c}}5&2\\3&{ - 1}\end{array}\,} \right] = \left[ {\,\begin{array}{*{20}{c}}{5/11}&{2/11}\\{3/11}&{ - 1/11}\end{array}\,} \right]$.

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