## Precalculus (6th Edition) Blitzer

Using Cramer's rule to solve \left\{ \begin{align} & 3x+y+4z=-8 \\ & 2x+3y-2z=11 \\ & x-3y-2z=4 \\ \end{align} \right. for $y$, we obtain $y=\frac{\left| \begin{matrix} \underline{3} & \underline{-8} & \underline{4} \\ \underline{2} & \underline{11} & \underline{-2} \\ \underline{1} & \underline{4} & \underline{-2} \\ \end{matrix} \right|}{\left| \begin{matrix} \underline{3} & \underline{1} & \underline{4} \\ \underline{2} & \underline{3} & \underline{-2} \\ \underline{1} & \underline{-3} & \underline{-2} \\ \end{matrix} \right|}$
Cramer's rule states that for any linear equations: \begin{align} & {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} \\ & {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} \end{align} The value for $x,y$ can be calculated as here: $x=\frac{\left| \begin{matrix} {{c}_{1}} & {{b}_{1}} \\ {{c}_{2}} & {{b}_{2}} \\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|}$ $y=\frac{\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{c}_{2}} \\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|}$ Therefore, the values of $y$ for the given linear equations are: $y=\frac{\left| \begin{matrix} 3 & -8 & 4 \\ 2 & 11 & -2 \\ 1 & 4 & -2 \\ \end{matrix} \right|}{\left| \begin{matrix} 3 & 1 & 4 \\ 2 & 3 & -2 \\ 1 & -3 & -2 \\ \end{matrix} \right|}$