Question

Let $A=\left[ \begin{matrix} 1 & 0 & 0 \\\ 2 & 1 & 0 \\\ 3 & 2 & 1 \\\ \end{matrix} \right]$ . If ${{u}_{1}}$ and ${{u}_{2}}$ are column matrices such that $A{{u}_{1}}=\left[ \begin{matrix} 1 \\\ 0 \\\ 0 \\\ \end{matrix} \right]$ and $A{{u}_{2}}=\left[ \begin{matrix} 0 \\\ 1 \\\ 0 \\\ \end{matrix} \right]$ then ${{u}_{1}}+{{u}_{2}}$ is equal to.
(a) $\left[ \begin{matrix} -1 \\\ 1 \\\ 0 \\\ \end{matrix} \right]$
(b) $\left[ \begin{matrix} 1 \\\ -1 \\\ -1 \\\ \end{matrix} \right]$
(c) $\left[ \begin{matrix} -1 \\\ -1 \\\ 0 \\\ \end{matrix} \right]$
(d) $\left[ \begin{matrix} -1 \\\ 1 \\\ -1 \\\ \end{matrix} \right]$

Answer 1

Hint: For solving this problem we should know how to multiply any two matrices. First, we will assume the unknown matrices and then solve some linear equations to find the elements of these matrices and ultimately we will find ${{u}_{1}}$ and ${{u}_{2}}$ . Then, we will solve for ${{u}_{1}}+{{u}_{2}}$ .

Complete step-by-step answer:
$A=\left[ \begin{matrix} 1 & 0 & 0 \\\ 2 & 1 & 0 \\\ 3 & 2 & 1 \\\ \end{matrix} \right]$ and $A{{u}_{1}}=\left[ \begin{matrix} 1 \\\ 0 \\\ 0 \\\ \end{matrix} \right]$ , $A{{u}_{2}}=\left[ \begin{matrix} 0 \\\ 1 \\\ 0 \\\ \end{matrix} \right]$ where ${{u}_{1}}$ and ${{u}_{2}}$ are column matrices.
Now, as $A$ has 3 rows and 3 columns then, order of $A$ is $3\times 3$ .So, ${{u}_{1}}$ and ${{u}_{2}}$ will have 3 rows and only one column. Then, order of ${{u}_{1}}$ and ${{u}_{2}}$ will be $3\times 1$ .
Let, ${{u}_{1}}=\left[ \begin{matrix} {{a}_{1}} \\\ {{b}_{1}} \\\ {{c}_{1}} \\\ \end{matrix} \right]$ . Then,
$\begin{aligned} & A{{u}_{1}}=\left[ \begin{matrix} 1 & 0 & 0 \\\ 2 & 1 & 0 \\\ 3 & 2 & 1 \\\ \end{matrix} \right]\times \left[ \begin{matrix} {{a}_{1}} \\\ {{b}_{1}} \\\ {{c}_{1}} \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ 0 \\\ 0 \\\ \end{matrix} \right] \\\ & \Rightarrow A{{u}_{1}}=\left[ \begin{matrix} {{a}_{1}} \\\ 2{{a}_{1}}+{{b}_{1}} \\\ 3{{a}_{1}}+2{{b}_{1}}+{{c}_{1}} \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ 0 \\\ 0 \\\ \end{matrix} \right] \\\ \end{aligned}$
Now, in the above equation, we will equate each term of the two matrices. Then,
$\begin{aligned} & \left[ \begin{matrix} {{a}_{1}} \\\ 2{{a}_{1}}+{{b}_{1}} \\\ 3{{a}_{1}}+2{{b}_{1}}+{{c}_{1}} \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ 0 \\\ 0 \\\ \end{matrix} \right] \\\ & \Rightarrow {{a}_{1}}=1.............\left( 1 \right) \\\ & \Rightarrow 2{{a}_{1}}+{{b}_{1}}=0..............\left( 2 \right) \\\ & \Rightarrow 3{{a}_{1}}+2{{b}_{1}}+{{c}_{1}}=0................\left( 3 \right) \\\ \end{aligned}$
Now, put ${{a}_{1}}=1$ from (1) in equation (2) to get the value of ${{b}_{1}}$ . Then,
$\begin{aligned} & 2{{a}_{1}}+{{b}_{1}}=0 \\\ & \Rightarrow 2\times 1+{{b}_{1}}=0 \\\ & \Rightarrow {{b}_{1}}=-2..............\left( 4 \right) \\\ \end{aligned}$
Now, put ${{a}_{1}}=1$ from (1) and ${{b}_{1}}=-2$ from (4) to get the value of ${{c}_{1}}$ . Then,
$\begin{aligned} & 3{{a}_{1}}+2{{b}_{1}}+{{c}_{1}}=0 \\\ & \Rightarrow 3\times 1+2\times \left( -2 \right)+{{c}_{1}}=0 \\\ & \Rightarrow 3-4+{{c}_{1}}=0 \\\ & \Rightarrow {{c}_{1}}=1 \\\ \end{aligned}$
Now, we have the value of ${{a}_{1}}=1$ , ${{b}_{1}}=-2$ and ${{c}_{1}}=1$ . Then,
$\begin{aligned} & {{u}_{1}}=\left[ \begin{matrix} {{a}_{1}} \\\ {{b}_{1}} \\\ {{c}_{1}} \\\ \end{matrix} \right] \\\ & \Rightarrow {{u}_{1}}=\left[ \begin{matrix} 1 \\\ -2 \\\ 1 \\\ \end{matrix} \right].....................\left( 5 \right) \\\ \end{aligned}$
Similarly, we can get ${{u}_{2}}$ . Let ${{u}_{2}}=\left[ \begin{matrix} {{a}_{2}} \\\ {{b}_{2}} \\\ {{c}_{2}} \\\ \end{matrix} \right]$ . Then,
$\begin{aligned} & A{{u}_{2}}=\left[ \begin{matrix} 1 & 0 & 0 \\\ 2 & 1 & 0 \\\ 3 & 2 & 1 \\\ \end{matrix} \right]\times \left[ \begin{matrix} {{a}_{2}} \\\ {{b}_{2}} \\\ {{c}_{2}} \\\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\\ 1 \\\ 0 \\\ \end{matrix} \right] \\\ & \Rightarrow A{{u}_{2}}=\left[ \begin{matrix} {{a}_{2}} \\\ 2{{a}_{2}}+{{b}_{2}} \\\ 3{{a}_{2}}+2{{b}_{2}}+{{c}_{2}} \\\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\\ 1 \\\ 0 \\\ \end{matrix} \right] \\\ \end{aligned}$
Now, in the above equation, we will equate each term of the two matrices. Then,
$\begin{aligned} & \left[ \begin{matrix} {{a}_{2}} \\\ 2{{a}_{2}}+{{b}_{2}} \\\ 3{{a}_{2}}+2{{b}_{2}}+{{c}_{2}} \\\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\\ 1 \\\ 0 \\\ \end{matrix} \right] \\\ & \Rightarrow {{a}_{2}}=0.............\left( 6 \right) \\\ & \Rightarrow 2{{a}_{2}}+{{b}_{2}}=1..............\left( 7 \right) \\\ & \Rightarrow 3{{a}_{2}}+2{{b}_{2}}+{{c}_{2}}=0................\left( 8 \right) \\\ \end{aligned}$
Now, put ${{a}_{2}}=0$ from (6) in equation (7) to get the value of ${{b}_{2}}$ . Then,
$\begin{aligned} & 2{{a}_{2}}+{{b}_{2}}=1 \\\ & \Rightarrow 2\times 0+{{b}_{2}}=1 \\\ & \Rightarrow {{b}_{2}}=1..............\left( 9 \right) \\\ \end{aligned}$
Now, put ${{a}_{2}}=0$ from (6) and ${{b}_{2}}=1$ from (9) to get the value of ${{c}_{2}}$ . Then,
$\begin{aligned} & 3{{a}_{2}}+2{{b}_{2}}+{{c}_{2}}=0 \\\ & \Rightarrow 3\times 0+2\times 1+{{c}_{2}}=0 \\\ & \Rightarrow 0+2+{{c}_{2}}=0 \\\ & \Rightarrow {{c}_{2}}=-2 \\\ \end{aligned}$
Now, we have the value of ${{a}_{2}}=0$ , ${{b}_{2}}=1$ and ${{c}_{2}}=-2$ . Then,
$\begin{aligned} & {{u}_{2}}=\left[ \begin{matrix} {{a}_{2}} \\\ {{b}_{2}} \\\ {{c}_{2}} \\\ \end{matrix} \right] \\\ & \Rightarrow {{u}_{2}}=\left[ \begin{matrix} 0 \\\ 1 \\\ -2 \\\ \end{matrix} \right].....................\left( 10 \right) \\\ \end{aligned}$
Now, adding equation (5) and (10). Then,

& {{u}_{1}}+{{u}_{2}}=\left[ \begin{matrix} 1 \\\ -2 \\\ 1 \\\ \end{matrix} \right]+\left[ \begin{matrix} 0 \\\ 1 \\\ -2 \\\ \end{matrix} \right] \\\ & \Rightarrow {{u}_{1}}+{{u}_{2}}=\left[ \begin{matrix} 1+0 \\\ -2+1 \\\ 1-2 \\\ \end{matrix} \right] \\\ & \Rightarrow {{u}_{1}}+{{u}_{2}}=\left[ \begin{matrix} 1 \\\ -1 \\\ -1 \\\ \end{matrix} \right] \\\ \end{aligned}$$ Thus, we got $${{u}_{1}}+{{u}_{2}}=\left[ \begin{matrix} 1 \\\ -1 \\\ -1 \\\ \end{matrix} \right]$$ . Hence, (b) is the correct option. Note: We can solve this question by another approach in which first we write $A{{u}_{1}}+A{{u}_{2}}$ by simply adding the given data and then pre multiplying the result with ${{A}^{-1}}$ for which we have to find the matrix ${{A}^{-1}}$ from the given data. After pre multiplying the result with ${{A}^{-1}}$ we will directly get ${{u}_{1}}+{{u}_{2}}$ . Moreover, the student must take care of calculation mistakes while solving the question.