Question
Let \[A = \left( \begin{array}{l}
1\,\,\,\,\,\,0\,\,\,\,\,\,0\\
2\,\,\,\,\,\,1\,\,\,\,\,\,0\\
3\,\,\,\,\,\,2\,\,\,\,\,\,1
\end{array} \right).\] If $${u_1}$$ and $${u_2}$$ are column matrices such that \[A{u_1} = \left( \begin{array}{l}
1\\
0\\
0
\end{array} \right){\rm{and }}\,\,A{u_2} = \left( \begin{array}{l}
0\\
1\\
0
\end{array} \right),\] then $${u_1} + {u_2}$$ is equal to:
A.
\[\left( \begin{array}{l}
- 1\\
\,\,\,\,\,1\\
\,\,\,\,\,0
\end{array} \right)\]
B.
\[\left( \begin{array}{l}
- 1\\
\,\,\,\,\,1\\
- 1
\end{array} \right)\]
C.
\[\left( \begin{array}{l}
- 1\\
- 1\\
\,\,\,\,\,0
\end{array} \right)\]
D.
\[\left( \begin{array}{l}
\,\,\,\,\,1\\
- 1\\
- 1
\end{array} \right)\]
Answer :
\[\left( \begin{array}{l}
\,\,\,\,\,1\\
- 1\\
- 1
\end{array} \right)\]
Solution :
Let \[A{u_1} = \left( \begin{array}{l}
1\\
0\\
0
\end{array} \right){\rm{and }}\,\,A{u_2} = \left( \begin{array}{l}
0\\
1\\
0
\end{array} \right),\]
\[{\rm{Then, }}\,\,A{u_1} + A{u_2} = \left( \begin{array}{l}
1\\
0\\
0
\end{array} \right)\, + \left( \begin{array}{l}
0\\
1\\
0
\end{array} \right)\]
\[ \Rightarrow \,\,A\left( {{u_1} + {u_2}} \right) = \left( \begin{array}{l}
1\\
1\\
0
\end{array} \right)\,\,\,\,\,\,\,\,\,....\left( 1 \right)\]
\[{\rm{Also, }}\,\,A = \left( \begin{array}{l}
1\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\
2\,\,\,\,\,\,\,1\,\,\,\,\,\,\,0\\
3\,\,\,\,\,\,2\,\,\,\,\,\,\,\,1
\end{array} \right)\]
$$ \Rightarrow \,\,\left| A \right| = 1\left( 1 \right) - 0\left( 2 \right) + 0\left( {4 - 3} \right) = 1$$
We know,
$$\eqalign{
& {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj\,A \cr
& \Rightarrow \,\,{A^{ - 1}} = adj\left( A \right)\,\,\,\,\,\,\left( {\because \left| A \right| = 1} \right) \cr} $$
Now, from equation (1), we have
\[{u_1} + {u_2} = {A^{ - 1}}\left( \begin{array}{l}
1\\
1\\
0
\end{array} \right)\]
$$ \Rightarrow \,\,{u_1} + {u_2} = {A^{ - 1}}$$
\[ = \left[ \begin{array}{l}
\,\,1\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,0\\
- 2\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,0\\
\,\,1\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,1
\end{array} \right]\left( \begin{array}{l}
1\\
1\\
0
\end{array} \right) = \left[ \begin{array}{l}
\,\,\,1\\
- 1\\
- 1
\end{array} \right]\]