## Precalculus (6th Edition) Blitzer

A linear system in three variables is: \begin{align} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}} \\ \end{align} Where ${{a}_{1}},{{a}_{2}},{{a}_{3,}}{{b}_{1}}\text{,}{{\text{b}}_{2}}\text{,}{{\text{b}}_{3}}\text{,}{{\text{c}}_{1}}\text{,}{{\text{c}}_{2}}\text{and }{{\text{c}}_{3}}\text{ are coffecients and }{{\text{d}}_{1}},{{d}_{2}}\And {{c}_{3}}\text{ are constants}\text{.}$ Then, \begin{align} & x=\frac{{{D}_{x}}}{D},y=\frac{{{D}_{y}}}{D},z=\frac{{{D}_{z}}}{D},\text{ where D}\ne \text{0} \\ & \\ \end{align} Where, $D=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|\text{ }$ These are the coefficients of variables x, y, z. ${{D}_{x}}=\left| \begin{matrix} {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|$ Replace $x$ -coefficients in $D$ with the constants on the right of the three equations. ${{D}_{y}}=\left| \begin{matrix} {{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\ \end{matrix} \right|$ Replace $y$ -coefficients in $D$ with the constants on the right of the three equations. ${{D}_{z}}=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\ \end{matrix} \right|$ Replace $z$ -coefficients in $D$ with the constants on the right of the three equations. Example, consider the given linear system \begin{align} & 2x+1y+1z=3 \\ & 1x-1y-1z=0 \\ & 1x+2y+1z=0 \\ \end{align} So, $D=\left| \begin{matrix} 2 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & 2 & 1 \\ \end{matrix} \right|$ Now, we will find the variable x ${{D}_{x}}=\left| \begin{matrix} 3 & 1 & 1 \\ 0 & -1 & -1 \\ 0 & 2 & 1 \\ \end{matrix} \right|$ Solve for $D$ along row 1, \begin{align} & \text{D=2}\left| \begin{matrix} -1 & -1 \\ 2 & 1 \\ \end{matrix} \right|-1\left| \begin{matrix} 1 & -1 \\ 1 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} 1 & -1 \\ 1 & 2 \\ \end{matrix} \right| \\ & D=2\left\{ -1\cdot 1-(-1)2 \right\}-1\left\{ 1\cdot 1-(-1)1 \right\}+1\left\{ 1\cdot 2-(-1)1 \right\} \\ & D=2(-1+2)-1(1+1)+1(2+1) \\ & D=2-2+3=3 \end{align} Similarly, ${{D}_{x}}=\left| \begin{matrix} 3 & 1 & 1 \\ 0 & -1 & -1 \\ 0 & 2 & 1 \\ \end{matrix} \right|$ Expand along column 1, \begin{align} & {{D}_{x}}=3\left| \begin{matrix} -1 & -1 \\ 2 & 1 \\ \end{matrix} \right|-0\left| \begin{matrix} 1 & 1 \\ -1 & 1 \\ \end{matrix} \right|+0\left| \begin{matrix} 1 & 1 \\ -1 & -1 \\ \end{matrix} \right| \\ & {{D}_{x}}=3\left\{ -1\cdot 1-(-1)(2) \right\}-0+0 \\ & {{D}_{x}}=3(-1+2)=3 \end{align} Therefore, \begin{align} & x=\frac{{{D}_{x}}}{D} \\ & x=\frac{3}{3}=1 \\ \end{align} By this method, one can easily evaluate the determinant in a faster way. Cramer’s rule is a faster method for solve a determinant.