## Precalculus (6th Edition) Blitzer

The given statement does not make any sense. To find ${{D}_{x}},{{D}_{y}},{{D}_{z}}\text{ or }{{D}_{n}}$ we need to apply the evaluation process for a $n\times n$ determinant $n$ more times. The values of ${{D}_{x}},{{D}_{y}},{{D}_{z}}\text{ or }{{D}_{n}}$ cannot be obtained from the number that occurs in the computation of D. Example: Given a linear system in three variables, \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 ${{a}_{1}},{{a}_{2}},{{a}_{3,}}{{b}_{1}}\text{,}{{\text{b}}_{2}}\text{,}{{\text{b}}_{3}}\text{,}{{\text{c}}_{1}}\text{,}{{\text{c}}_{2}}\text{and }{{\text{c}}_{3}}\text{ are coefficients and }{{\text{d}}_{1}},{{d}_{2}}\And {{c}_{3}}\text{ are constants}\text{.}$ 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. Hence, the provided statement does not make any sense.