Chapter 7, Problem 45RE

To determine

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

Answer to Problem 45RE

The required values of $x,y\text{and}z$ are 0.

Explanation of Solution

Given information:

The given equation are

  $\{\begin{array}{c}x-2y+z=-6\\ 2x-3y=-7\\ -x+3y-3z=11\end{array}$

Calculation:

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

Let,

  $x-2y+z=-6$  ...... (1)

  $2x-3y=-7$  ...... (2)

  $-x+3y-3z=11$  ...... (3)

Write in the matrix format.

  $\left[\begin{array}{ccc}1& -2& 1\\ 2& -3& 0\\ -1& 3& -3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-6\\ -7\\ 11\end{array}\right]$

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

  $\left[\begin{array}{ccc}1& -2& 1\\ 2& -3& 0\\ -1& 3& -3\end{array}\begin{array}{c}⋮-6\\ ⋮-7\\ ⋮11\end{array}\right]$

Now, solve.

  $\begin{array}{r}\begin{array}{c}\\ -2R_1+R_2\\ R_1+R_3\end{array}\left[\begin{array}{ccc}1& -2& 1\\ 2& -3& 0\\ -1& 3& -3\end{array}\begin{array}{c}⋮-6\\ ⋮-7\\ ⋮11\end{array}\right]\\ \left[\begin{array}{ccc}1& -2& 1\\ 0& 1& -2\\ 0& 1& -2\end{array}\begin{array}{c}⋮-6\\ ⋮5\\ ⋮5\end{array}\right]\end{array}$

Since, second row and the third row values are same thus,

Hence, the required values of $x,y\text{and}z$ are $0$ .