Chapter 8 - Test - Page 951: 10

The value of $x=2$.

Consider the given system of equations $\begin{align} & 3x+y-2z=-3 \\ & 2z+7y+3z=9 \\ & 4x-3y-z=7 \end{align}$ Therefore, in matrix form, the system of equations can be written as below: $AX=b$ Where $A=\left[ \begin{array}{*{35}{r}} 3 & 1 & -2 \\ 2 & 7 & 3 \\ 4 & -3 & -1 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} -3 \\ 9 \\ 7 \\ \end{array} \right];X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]$ Therefore, the solution of the system of equations is given by: $X={{A}^{-1}}b$ Therefore, using Cramer’s rule, the solution of the system of the equations is $x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|},y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|},z=\frac{\left| {{A}_{z}} \right|}{\left| A \right|}$ Where ${{A}_{x}}=\left[ \begin{array}{*{35}{r}} -3 & 1 & -2 \\ 9 & 7 & 3 \\ 7 & -3 & -1 \\ \end{array} \right],{{A}_{y}}=\left[ \begin{array}{*{35}{r}} 3 & -3 & -2 \\ 2 & 9 & 3 \\ 4 & 7 & -1 \\ \end{array} \right],{{A}_{z}}=\left[ \begin{array}{*{35}{r}} 3 & 1 & -3 \\ 2 & 7 & 9 \\ 4 & -3 & 7 \\ \end{array} \right]$ Consider the determinant of the matrix $\begin{align} & \left| A \right|=\left| \begin{array}{*{35}{r}} 3 & 1 & -2 \\ 2 & 7 & 3 \\ 4 & -3 & -1 \\ \end{array} \right| \\ & =3\left( -7+9 \right)-\left( -2-12 \right)-2\left( -6-28 \right) \\ & =6+14+68 \\ & =88 \end{align}$ $\begin{align} & \left| {{A}_{x}} \right|=\left| \begin{array}{*{35}{r}} -3 & 1 & -2 \\ 9 & 7 & 3 \\ 7 & -3 & -1 \\ \end{array} \right| \\ & =-3\left( -7+9 \right)-1\left( -9-21 \right)-2\left( -27-49 \right) \\ & =-6+30+152 \\ & =176 \end{align}$ $\begin{align} & \left| {{A}_{y}} \right|=\left| \begin{array}{*{35}{r}} 3 & -3 & -2 \\ 2 & 9 & 3 \\ 4 & 7 & -1 \\ \end{array} \right| \\ & =3\left( -9-21 \right)+3\left( -2-12 \right)-2\left( 14-36 \right) \\ & =-90-42+44 \\ & =-88 \end{align}$ $\begin{align} & \left| {{A}_{z}} \right|=\left| \begin{array}{*{35}{r}} 3 & 1 & -3 \\ 2 & 7 & 9 \\ 4 & -3 & 7 \\ \end{array} \right| \\ & =3\left( 49+27 \right)-\left( 14-36 \right)-3\left( -6-28 \right) \\ & =228+22+102 \\ & =352 \end{align}$ Therefore, the solution of the system of equations is given by: $\begin{align} & x=\frac{176}{88}=2 \\ & y=\frac{-88}{88}=-1 \\ & z=\frac{352}{88}=4 \end{align}$ The value of x is $x=2$.