Find y, if\[left( \begin{matrix} 5 & -6 \\ 2 & 7 \\\end{matrix} \right)\left( \begin{align} & x \\ & y \\\end{align} \right)=\left( \begin{align} & 7 \\ & -11 \\\end{align} \right)\]
3
2
8
5
\[\begin{align} & \left( \begin{matrix} 5 & -6 \\ 2 & 7 \\\end{matrix} \right)\left( \begin{align} & x \\ & y \\\end{align} \right)=\left( \begin{align} & 7 \\ & -11 \\\end{align} \right) \\ & \text{Expanding the matrix} \\ & 5x-6y=7----(i) \\ & 2x-7y=-11---(ii) \\ & \text{Mulitply (i) by }2 \\ & \text{Multiply (ii) by 5} \\ & 10x-12y=14----(iii) \\ & 10x-35x=-55----(iv) \\ & \text{Subtract (iv) from (iii)} \\ & 23y=69 \\ & y=3 \\ & Method2 \\ & \text{Using inverse method} \\ & \left( \begin{matrix} 5 & -6 \\ 2 & -7 \\\end{matrix} \right)\left( \begin{align} & x \\ & y \\\end{align} \right)=\left( \begin{align} & 7 \\ & -11 \\\end{align} \right) \\ & \text{ }A\text{ }X\text{ }B \\ & {{A}^{-1}}=\frac{1}{-35+12}\left( \begin{matrix} -7 & 6 \\ -2 & 5 \\\end{matrix} \right)=\frac{1}{-23}\left( \begin{matrix} -7 & 6 \\ -2 & 5 \\\end{matrix} \right) \\ & X={{A}^{-1}}B \\ & \left( \begin{align} & x \\ & y \\\end{align} \right)=\frac{1}{-23}\left( \begin{matrix} -7 & 6 \\ -2 & 5 \\\end{matrix} \right)\left( \begin{align} & 7 \\ & -11 \\\end{align} \right) \\ & \left( \begin{align} & x \\ & y \\\end{align} \right)=\frac{1}{-23}\left( \begin{align} & -49-66 \\ & -14-55 \\\end{align} \right)=-\frac{1}{23}\left( \begin{align} & -115 \\ & -69 \\\end{align} \right) \\ & \left( \begin{align} & x \\ & y \\\end{align} \right)=\left( \begin{align} & \frac{-115}{-23} \\ & \frac{-69}{-23} \\\end{align} \right)=\left( \begin{align} & 5 \\ & 3 \\\end{align} \right) \\ & x=5,\text{ }y=3 \\\end{align}\]