Chapter 7, Problem 136CR

To determine

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

Answer to Problem 136CR

The required solution is $x=1$ , $y=1$ and $z=-3$ .

Explanation of Solution

Given: The given system of equations are-

  $\begin{array}{l}14x-21y-7z=14\\ -4x+2y-2z=4\\ 56x-21y+7z=14\end{array}$

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-

  $\begin{array}{l}14x-21y-7z=14\\ -4x+2y-2z=4\\ 56x-21y+7z=14\end{array}$

The associated coefficient matrix is $\left[\begin{array}{ccc}14& -21& -7\\ -4& 2& -2\\ 56& -21& 7\end{array}\right]$ .

Thus,

  $D=\left|\begin{array}{ccc}14& -21& -7\\ -4& 2& -2\\ 56& -21& 7\end{array}\right|$

  $=14\left|\begin{array}{cc}2& -2\\ -21& 7\end{array}\right|-(-21)\left|\begin{array}{cc}-4& -2\\ 56& 7\end{array}\right|+(-7)\left|\begin{array}{cc}-4& 2\\ 56& -21\end{array}\right|$ (Expanding along $R_1$ )

  $=14(14-42)+21(-28+112)-7(84-112)$

  $=-392+1764+196$

  $=1568$

  $D_x=\left|\begin{array}{ccc}14& -21& -7\\ 4& 2& -2\\ 14& -21& 7\end{array}\right|$

  $=14\left|\begin{array}{cc}2& -2\\ -21& 7\end{array}\right|-(-21)\left|\begin{array}{cc}4& -2\\ 14& 7\end{array}\right|+(-7)\left|\begin{array}{cc}4& 2\\ 14& -21\end{array}\right|$ (Expanding along $R_1$ )

  $=14(14-42)+21(28+28)-7(-84-28)$

  $=-392+1176+784$

  $=1568$

  $D_y=\left|\begin{array}{ccc}14& 14& -7\\ -4& 4& -2\\ 56& 14& 7\end{array}\right|$

  $=14\left|\begin{array}{cc}4& -2\\ 14& 7\end{array}\right|-14\left|\begin{array}{cc}-4& -2\\ 56& 7\end{array}\right|+(-7)\left|\begin{array}{cc}-4& 4\\ 56& 14\end{array}\right|$ (Expanding along $R_1$ )

  $=14(28+28)-14(-28+112)-7(-56-224)$

  $=784-1176+1960$

  $=1568$

  $D_z=\left|\begin{array}{ccc}14& -21& 14\\ -4& 2& 4\\ 56& -21& 14\end{array}\right|$

  $=14\left|\begin{array}{cc}2& 4\\ -21& 14\end{array}\right|-(-21)\left|\begin{array}{cc}-4& 4\\ 56& 14\end{array}\right|+14\left|\begin{array}{cc}-4& 2\\ 56& -21\end{array}\right|$ (Expanding along $R_1$ )

  $=14(28+84)+21(-56-224)+14(84-112)$

  $=1568-5880-392$

  $=-4704$

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

  $x=\frac{D_x}{D}$

  $=\frac{1568}{1568}$

  $=1$

  $y=\frac{D_y}{D}$

  $=\frac{1568}{1568}$

  $=1$

  $z=\frac{D_z}{D}$

  $=\frac{-4704}{1568}$

  $=-3$

Therefore the required solution is $x=1$ , $y=1$ and $z=-3$ .