Chapter 7, Problem 43RE

To determine

To calculate: The solution of the system of linear equation.

Answer to Problem 43RE

The required values of $x,y\text{and}z$ are $-\frac{191}{16},-\frac{27}{16}\text{and}-\frac{11}{16}$ .

Explanation of Solution

Given information:

The given equation are

  $\{\begin{array}{c}x-2y+5z=-12\\ 2x+6z=-28\\ x+y+18z=-4\end{array}$

Calculation:

The given equation is $y=a{x}^2+bx+c$ .

Let,

  $x-2y+5z=-12$  ...... (1)

  $\begin{array}{l}2x+6z=-28\\ x+3z=-14\end{array}$  ...... (2)

  $x+y+18z=-4$  ...... (3)

Draw the graph.

  

Write in the matrix format.

  $\left[\begin{array}{ccc}1& -2& 5\\ 2& 0& 6\\ 1& 1& 18\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-12\\ -28\\ -4\end{array}\right]$

Now, use back substitution method to find $x,y\text{and}z$ .

  $\left[\begin{array}{ccc}1& -2& 5\\ 1& 0& 3\\ 1& 1& 18\end{array}\begin{array}{c}⋮-12\\ ⋮-14\\ ⋮-4\end{array}\right]$

Now, solve.

  $\begin{array}{r}\left[\begin{array}{ccc}1& -2& 5\\ 1& 0& 3\\ 1& 1& 18\end{array}\begin{array}{c}⋮-12\\ ⋮-14\\ ⋮-4\end{array}\right]\\ \begin{array}{c}\\ -R_1+R_2\\ -R_1+R_3\end{array}\left[\begin{array}{ccc}1& -2& 5\\ 0& 2& -2\\ 0& 3& 13\end{array}\begin{array}{c}⋮-12\\ ⋮-2\\ ⋮8\end{array}\right]\\ \begin{array}{c}\\ \\ -3R_2+2R_3\end{array}\left[\begin{array}{ccc}1& -2& 5\\ 0& 2& -2\\ 0& 0& 32\end{array}\begin{array}{c}⋮-12\\ ⋮-2\\ ⋮22\end{array}\right]\end{array}$

Thus, the equations are,

  $\{\begin{array}{c}x-2y+5z=-12\\ 2y-2z=-2\\ 32z=-22\end{array}$

Thus, $z=-\frac{11}{16}$ .

And, $\begin{array}{c}2y-2\left(-\frac{11}{16}\right)=-2\\ y+\frac{11}{16}=-1\\ y=-\frac{27}{16}\end{array}$

And $\begin{array}{c}x-2\left(-\frac{27}{16}\right)+5\left(-\frac{11}{16}\right)=-12\\ x=-\frac{191}{16}\end{array}$

Hence, the required values of $x,y\text{and}z$ are $-\frac{191}{16},-\frac{27}{16}\text{and}-\frac{11}{16}$ .