If A = [ * 20 c 2&2 - 3 &2 ] and B = [ * 20 c 0& - 1 1&0 ], then … - Vidyadip

MCQ

If $A = \left[ {\begin{array}{*{20}{c}}2&2\\{ - 3}&2\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right],$ then ${({B^{ - 1}}{A^{ - 1}})^{ - 1}}$=

✓
$\left[ {\begin{array}{*{20}{c}}2&{ - 2}\\2&3\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}3&{ - 2}\\2&2\end{array}} \right]$
C
$\frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}2&2\\{ - 2}&3\end{array}} \right]$
D
$\frac{1}{{10}}\left[ {\begin{array}{*{20}{c}}3&2\\{ - 2}&2\end{array}} \right]$

✓

Answer

Correct option: A.

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

a
(a) ${({B^{ - 1}}{A^{ - 1}})^{ - 1}} = {({A^{ - 1}})^{ - 1}}{({B^{ - 1}})^{ - 1}} = AB$

(Reversal law of inverses)

$ = \,\left[ {\,\begin{array}{*{20}{c}}2&2\\{ - 3}&2\end{array}\,} \right]\,\left[ {\,\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}\,} \right] = \left[ {\,\begin{array}{*{20}{c}}2&{ - 2}\\2&3\end{array}\,} \right]$.

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