Chapter 10.3, Problem 27E

To determine

The solution of given system of linear equations having unique solutions using Gaussian elimination or gauss-Jordan elimination.

Answer to Problem 27E

The solution for given system of linear equation is.

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}2x-3y-z=13\\ -x+2y-5z=6\\ 5x-y-z=49\end{array}$

Concept Used:

Calculation:

Consider first the given equations,

  $\begin{array}{l}2x-3y-z=13\\ -x+2y-5z=6\\ 5x-y-z=49\end{array}$

Step1:

In the first step first convert the given, equation into augmented matrix.

  $\left[\begin{array}{cccc}2& -3& -1& 13\\ -1& 2& -5& 6\\ 5& -1& -1& 49\end{array}\right]$

Step 2:

In the second step convert the augmented matrix into row echelon form.

For this row operation is performed. In the first row operation,

Perform,

  $Row1/2=NewRow1$

  $Row1/2=\left[\begin{array}{cccc}1& -1.5& -0.5& 6.5\\ -1& 2& -5& 6\\ 5& -1& -1& 49\end{array}\right]$

  $Row2+Row1=NewRow2$

  $Row2+Row1=\left[\begin{array}{cccc}1& -1.5& -0.5& 6.5\\ 0& 0.5& -5.5& 12.5\\ 5& 1& 1& 49\end{array}\right]$

Now the next row operation to be performed is,

  $Row2*2=NewRow2$

  $Row2*2=$ $\left[\begin{array}{cccc}1& 1.5& -0.5& 6.5\\ 0& 0.5& 11& 25\\ 5& -1& 1& 49\end{array}\right]$

next row operation to be performed is,

  $Row3-5*Row1=NewRow3$

  $\left[\begin{array}{cccc}5& -7.5& -2.5& 32.5\\ 0& 1& 11& 25\\ 5& 1& 1& 49\end{array}\right]$

  $15*Row1=\left[\begin{array}{cccc}15& 15& -30& 90\\ 15& 20& 30& -25\\ -5& 30& -10& 45\end{array}\right]$ and

  $Row2-15*Row1=\left[\begin{array}{cccc}1& 1& 2& 6\\ 0& 5& 60& -115\\ -5& 30& -10& 45\end{array}\right]$

  $=NewRow2$

Step3:

In the step 3 again perform row operations,

First perform,

  $Row3+5*Row1$

  $=NewRow3$

  $Row3+5*Row1=\left[\begin{array}{cccc}5& 5& -10& 30\\ 0& 5& 60& -115\\ -5& 30& -10& 45\end{array}\right]$

And,

  $Row3+5*Row1=\left[\begin{array}{cccc}1& 1& -2& 6\\ 0& 5& 60& 115\\ 0& 35& 20& 75\end{array}\right]$

  $=NewRow3$

Step 4:

The next step for row operation is,

  $Row3-7*Row2=NewRow3$

  $7*Row2=\left[\begin{array}{cccc}1& 1& -2& 6\\ 0& 35& 420& 805\\ 0& 35& -20& 755\end{array}\right]$

And,

  $Row3-7*Row2=\left[\begin{array}{cccc}1& 1& -2& 6\\ 0& 5& 60& -805\\ 0& 0& -440& 880\end{array}\right]$ = $NewRow3$

Step4:

In this the matrix is in row echloen form,

Divide

  $Row3/-440$

And,

  $Row2/5$

  $=\left[\begin{array}{cccc}1& 1& -2& 6\\ 0& 1& 12& -23\\ 0& 0& 1& -2\end{array}\right]$

Now the matrix is obtained in the row echelon form , write the equations from the matrix,

  $x+y-2z=6$ ….. $\left(1\right)$

  $y+12z=-23$ ….. $\left(2\right)$

  $z=-2$ ..... $\left(3\right)$

Step5:

In the step 5 values of $x,y,z$ can be found using back substitution,From equation

Conclusion:

Hence, the solution for given system is $\left(3,1,2\right)$