Question
Let \[A = \left[ {\begin{array}{*{20}{c}}
{x + y}&y\\
{2x}&{x - y}
\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}
2\\
{ - 1}
\end{array}} \right]\] and \[C = \left[ {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right].\] If $$AB =C,$$ then what is $$A^2$$ equal to ?
A.
\[\left[ {\begin{array}{*{20}{c}}
6&{ - 10}\\
4&{26}
\end{array}} \right]\]
B.
\[\left[ {\begin{array}{*{20}{c}}
{ - 10}&{5}\\
4&{24}
\end{array}} \right]\]
C.
\[\left[ {\begin{array}{*{20}{c}}
{ - 5}&{ - 6}\\
{ - 4}&{ - 20}
\end{array}} \right]\]
D.
\[\left[ {\begin{array}{*{20}{c}}
{- 5}&{ - 7}\\
{- 5}&{20}
\end{array}} \right]\]
Answer :
\[\left[ {\begin{array}{*{20}{c}}
6&{ - 10}\\
4&{26}
\end{array}} \right]\]
Solution :
\[A = \left[ {\begin{array}{*{20}{c}}
{x + y}&y\\
{2x}&{x - y}
\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}
2\\
{ - 1}
\end{array}} \right]{\rm{ and }}\,\,C = \left[ {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right]\]
Here, $$AB = C$$
\[\begin{array}{l}
\therefore \left[ {\begin{array}{*{20}{c}}
{x + y}&y\\
{2x}&{x - y}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2\\
{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right]\\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
{2\left( {x + y} \right)}&{ - y}\\
{4x}&{ - x + y}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right]
\end{array}\]
$$\eqalign{
& 2x + y = 3\,\,\,.....\left( {\text{i}} \right) \cr
& 3x + y = 2\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
From equations (i) and (ii), we get
$$x = - 1$$ and $$y = 5$$
\[\begin{array}{l}
\therefore A = \left[ {\begin{array}{*{20}{c}}
4&5\\
{ - 2}&{ - 6}
\end{array}} \right]\\
{\rm{Now, }}\,\,{A^2} = \left[ {\begin{array}{*{20}{c}}
4&5\\
{ - 2}&{ - 6}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
4&5\\
{ - 2}&{ - 6}
\end{array}} \right]\\
= \left[ {\begin{array}{*{20}{c}}
{16 - 10}&{20 - 30}\\
{ - 8 + 12}&{ - 10 + 36}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
6&{ - 10}\\
4&{26}
\end{array}} \right]
\end{array}\]