Chapter 7, Problem 135CR

To determine

To find: The solution of the given system of equations.

Answer to Problem 135CR

The required solution is $x=0$ , $y=-3$ and $z=-3$ .

Explanation of Solution

Given: The given system of equations are-

  $\begin{array}{l}x-3y+2z=3\\ 2x+2y-3z=3\\ x-7y+8z=-3\end{array}$

Concept Used: Consider the system of equations-

  $a_{1}x+b_{1}y+c_{1}z=d_1$

  $a_{2}x+b_{2}y+c_{2}z=d_2$

  $a_{3}x+b_{3}y+c_{3}z=d_3$

The associated coefficient matrix is-

  $A=\left[\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right]$

Consider the following determinants-

  $D=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|$ (Determinant of the coefficient matrix)

  $D_x=\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|$ (Determinant of the coefficient matrix where the coefficients of x are replaced by the constant terms)

  $D_y=\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|$ (Determinant of the coefficient matrix where the coefficients of y are replaced by the constant terms)

  $D_z=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|$ (Determinant of the coefficient matrix where the coefficients of z are replaced by the constant terms)

Hence, according to Cramer’s rule the solution of the given system of equations is given by-

  $x=\frac{D_x}{D}$ ; $y=\frac{D_y}{D}$ ; $z=\frac{D_z}{D}$ provided $D\ne 0$ .

If however $D=0$ then the given system is inconsistent and Cramer’s rule can’t be applied to find a solution.

Calculations: The given system of equations are-

  $\begin{array}{l}x-3y+2z=3\\ 2x+2y-3z=3\\ x-7y+8z=-3\end{array}$

The associated coefficient matrix is $\left[\begin{array}{ccc}1& -3& 2\\ 2& 2& -3\\ 1& -7& 8\end{array}\right]$ .

Thus,

  $D=\left|\begin{array}{ccc}1& -3& 2\\ 2& 2& -3\\ 1& -7& 8\end{array}\right|$

  $=1\left|\begin{array}{cc}2& -3\\ -7& 8\end{array}\right|-(-3)\left|\begin{array}{cc}2& -3\\ 1& 8\end{array}\right|+2\left|\begin{array}{cc}2& 2\\ 1& -7\end{array}\right|$ (Expanding along $R_1$ )

  $=1(16-21)+3(16+3)+2(-14-2)$

  $=-5+57-32$

  $=20$

  $D_x=\left|\begin{array}{ccc}3& -3& 2\\ 3& 2& -3\\ -3& -7& 8\end{array}\right|$

  $=3\left|\begin{array}{cc}2& -3\\ -7& 8\end{array}\right|-(-3)\left|\begin{array}{cc}3& -3\\ -3& 8\end{array}\right|+2\left|\begin{array}{cc}3& 2\\ -3& -7\end{array}\right|$ (Expanding along $R_1$ )

  $=3(16-21)+3(24-9)+2(-21+6)$

  $=-15+45-30$

  $=0$

  $D_y=\left|\begin{array}{ccc}1& 3& 2\\ 2& 3& -3\\ 1& -3& 8\end{array}\right|$

  $=1\left|\begin{array}{cc}3& -3\\ -3& 8\end{array}\right|-3\left|\begin{array}{cc}2& -3\\ 1& 8\end{array}\right|+2\left|\begin{array}{cc}2& 3\\ 1& -3\end{array}\right|$ (Expanding along $R_1$ )

  $=1(24-9)-3(16+3)+2(-6-3)$

  $=15-57-18$

  $=-60$

  $D_z=\left|\begin{array}{ccc}1& -3& 3\\ 2& 2& 3\\ 1& -7& -3\end{array}\right|$

  $=1\left|\begin{array}{cc}2& 3\\ -7& -3\end{array}\right|-(-3)\left|\begin{array}{cc}2& 3\\ 1& -3\end{array}\right|+3\left|\begin{array}{cc}2& 2\\ 1& -7\end{array}\right|$ (Expanding along $R_1$ )

  $=1(-6+21)+3(-6-3)+3(-14-2)$

  $=15-27-48$

  $=-60$

Therefore, using Cramer’s rule, the required solution of the given system of equations is-

  $x=\frac{D_x}{D}$

  $=\frac{0}{20}$

  $=0$

  $y=\frac{D_y}{D}$

  $=\frac{-60}{20}$

  $=-3$

  $z=\frac{D_z}{D}$

  $=\frac{-60}{20}$

  $=-3$

Therefore the required solution is $x=0$ , $y=-3$ and $z=-3$ .