Question

If $x\left[\begin{array}{c}-3\\ 4\end{array}\right]+y\left[\begin{array}{c}4\\ 3\end{array}\right]=\left[\begin{array}{c}10\\ -5\end{array}\right]$, then

• A. $x=-2,y=1$
• B. $x=-9,y=10$
• C. $x=22,y=1$
• D. $x=2,y=-1$

Answer 1

The correct option is A

$x=-2,y=1$

Explanation for the correct option.

Matrix addition operation:

$\begin{array}{rcl}x\left[\begin{array}{c}-3\\ 4\end{array}\right]+y\left[\begin{array}{c}4\\ 3\end{array}\right]& =& \left[\begin{array}{c}10\\ -5\end{array}\right]\\ & \Rightarrow & \left[\begin{array}{c}-3x\\ 4x\end{array}\right]+\left[\begin{array}{c}4y\\ 3y\end{array}\right]=\left[\begin{array}{c}10\\ -5\end{array}\right]\\ \Rightarrow -3x+4y& =& 10...\left(1\right)\\ and4x+3y& =& -5...\left(2\right)\end{array}$

From $\left(1\right)$, we get

$y=\frac{10+3x}{4}$

By substituting the value of $y$ in $\left(2\right)$, we get

$\begin{array}{rcl}4x+3\left(\frac{10+3x}{4}\right)& =& -5\\ & \Rightarrow & 16x+30+9x=-20\\ \Rightarrow x& =& -2\end{array}$

By substituting the value of $x$ in $y$, we get

$y=\frac{10+3\left(-2\right)}{4}=1$

Hence, option A is correct.