## Precalculus (6th Edition) Blitzer

The solution of the system is $\left( x,y,z \right)=\left( 1,3,2 \right)$.
We know that according to Cramer’s rule, $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$. Where, $D$ is the determinant of the matrix that consists of the coefficients of the variables in the equations $D=$ $\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 4 & -7 \\ 2 & -1 & 3 \\ \end{matrix} \right|$ ${{D}_{x}}$, acts as determinant in the numerator which is obtained by replacing the x coefficient in D with the right side of the equation having the constants. ${{D}_{x}}=$ $\left| \begin{matrix} 6 & 1 & 1 \\ 1 & 4 & -7 \\ 5 & -1 & 3 \\ \end{matrix} \right|$ ${{D}_{y}}$, acts as determinant in the numerator which is obtained by replacing the y coefficient in D with the right side of the equation having the constants. ${{D}_{y}}=$ $\left| \begin{matrix} 1 & 6 & 1 \\ 3 & 1 & -7 \\ 2 & 5 & 3 \\ \end{matrix} \right|$ ${{D}_{z}}$, acts as determinant in the numerator which is obtained by replacing the z coefficient in D with the right side of the equation having the constants. ${{D}_{z}}=$ $\left| \begin{matrix} 1 & 1 & 6 \\ 3 & 4 & 1 \\ 2 & -1 & 5 \\ \end{matrix} \right|$. Solve the four determinants and compute the value of D as given below: \begin{align} & D=\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 4 & -7 \\ 2 & -1 & 3 \\ \end{matrix} \right| \\ & =1\left\{ 4\left( 3 \right)-\left( \left( -1 \right)\left( -7 \right) \right) \right\}-1\left\{ 3\left( 3 \right)-2\left( -7 \right) \right\}+1\left\{ 3\left( -1 \right)-2\left( 4 \right) \right\} \\ & =\left( 12-7 \right)-\left( 9+14 \right)+\left( -3-8 \right) \\ & =5-23-11 \\ & =-29 \end{align} Compute the value of ${{D}_{x}}$ as given below: \begin{align} & {{D}_{x}}=\left| \begin{matrix} 6 & 1 & 1 \\ 1 & 4 & -7 \\ 5 & -1 & 3 \\ \end{matrix} \right| \\ & =6\left\{ 4\left( 3 \right)-\left( -1 \right)\left( -7 \right) \right\}-1\left( 1\left( 3 \right)-5\left( -7 \right) \right)+1\left( 1\left( -1 \right)-5\left( 4 \right) \right) \\ & =6\left( 12-7 \right)-\left( 3+35 \right)+\left( -1-20 \right) \\ & =30-38-21 \\ & =-29 \end{align} Compute the value of ${{D}_{y}}$ as given below: \begin{align} & {{D}_{y}}=\left| \begin{matrix} 1 & 6 & 1 \\ 3 & 1 & -7 \\ 2 & 5 & 3 \\ \end{matrix} \right| \\ & =1\left\{ 1\left( 3 \right)-5\left( -7 \right) \right\}-6\left\{ 3\left( 3 \right)-2\left( -7 \right) \right\}+1\left( 3\left( 5 \right)-2\left( 1 \right) \right) \\ & =\left( 3+35 \right)-6\left( 9+14 \right)+\left( 15-2 \right) \\ & =38-138+13 \\ & =-87 \end{align} Compute the value of ${{D}_{z}}$ as given below: \begin{align} & {{D}_{z}}=\left| \begin{matrix} 1 & 1 & 6 \\ 3 & 4 & 1 \\ 2 & -1 & 5 \\ \end{matrix} \right| \\ & =1\left\{ 4\left( 5 \right)-\left( -1 \right)\left( 1 \right) \right\}-1\left\{ 3\left( 5 \right)-2\left( 1 \right) \right\}+6\left\{ 3\left( -1 \right)-2\left( 4 \right) \right\} \\ & =\left( 20+1 \right)-\left( 15-2 \right)+6\left( -3-8 \right) \\ & =21-13-66 \\ & =-58 \end{align} Put the values in the formula $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$ to solve for x, y, and z as given below: \begin{align} & x=\frac{{{D}_{x}}}{D} \\ & =\frac{-29}{-29} \\ & =1 \end{align} \begin{align} & y=\frac{{{D}_{y}}}{D} \\ & =\frac{-87}{-29} \\ & =3 \end{align} Then, \begin{align} & z=\frac{{{D}_{z}}}{D} \\ & =\frac{-58}{-29} \\ & =2 \end{align} Thus, the solution of the system of equations is $\left( x,y,z \right)=\left( 1,3,2 \right)$.