Question

Given, matrix $A=\left[\begin{array}{c}3\\ 2\end{array}\right]andB=\left[\begin{array}{c}-2\\ -1\end{array}\right],$ find the matrix X such that $X-A=B$

• A. $\left[\begin{array}{c}0\\ 0\end{array}\right]$
• B. $\left[\begin{array}{c}1\\ 1\end{array}\right]$
• C. $\left[\begin{array}{c}4\\ 0\end{array}\right]$
• D. $\left[\begin{array}{c}1\\ -1\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{c}1\\ 1\end{array}\right]$

Given, $A=\left[\begin{array}{c}3\\ 2\end{array}\right],B=\left[\begin{array}{c}-2\\ -1\end{array}\right],$ $X-A=B$

Let, $X=\left[\begin{array}{c}a\\ b\end{array}\right]$

$\therefore \left[\begin{array}{c}a\\ b\end{array}\right]-\left[\begin{array}{c}3\\ 2\end{array}\right]=\left[\begin{array}{c}-2\\ -1\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}a-3\\ b-2\end{array}\right]=\left[\begin{array}{c}-2\\ -1\end{array}\right]$

By equating above matrices, we have

$a-3=-2$ and $b-2=-1$

$\Rightarrow a=1$ and $b=1$

$\Rightarrow X=\left[\begin{array}{c}1\\ 1\end{array}\right]$

Hence, the answer is $X=\left[\begin{array}{c}1\\ 1\end{array}\right].$