Question

If $x\left[\begin{array}{c}2\\ 1\end{array}\right]+y\left[\begin{array}{c}3\\ 5\end{array}\right]+\left[\begin{array}{c}-8\\ -11\end{array}\right]=O,$ then

• A. $x=1$
• B. $x=2$
• C. $y=2$
• D. $y=1$

Answer 1

Answer: (A), (C)

$y=2$

Given: $x\left[\begin{array}{c}2\\ 1\end{array}\right]+y\left[\begin{array}{c}3\\ 5\end{array}\right]+\left[\begin{array}{c}-8\\ -11\end{array}\right]=O$
$\Rightarrow \left[\begin{array}{c}2x\\ x\end{array}\right]+\left[\begin{array}{c}3y\\ 5y\end{array}\right]+\left[\begin{array}{c}-8\\ -11\end{array}\right]=O$

So,
$2x+3y-8=0\Rightarrow 2x+3y=8\cdots \left(1\right)$
$x+5y-11=0\Rightarrow x+5y=11\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ the values are, $x=1,y=2$.