Chapter 7.4, Problem 32E

a.

To determine

To solve the augmented matrix.

Answer to Problem 32E

The solution of the system of linear equation is $x=10$ , $y=-6$ and $z=4$ .

Explanation of Solution

Given information:

The augmented matrix is given as:

  $\left[\begin{array}{ccccc}7& 13& 1& ⋮& -4\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$

Calculation:

Perform row operation to solve the augmented matrix,

  $\begin{array}{l}\left[\begin{array}{ccccc}7& 13& 1& ⋮& -4\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]\\ \begin{array}{c}{R}_2+R_1\to \\ R_3\text{+}{R}_2\to \\ \end{array}\left[\begin{array}{ccccc}4& 8& 0& ⋮& -8\\ 0& 1& 0& ⋮& -6\\ 3& 6& 1& ⋮& -2\end{array}\right]\\ \begin{array}{c}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.R_1\to \\ \\ \end{array}\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ 3& 6& 1& ⋮& -2\end{array}\right]\\ \begin{array}{c}-2R_2+R_1\to \\ \\ -3R_1+R_3\to \end{array}\left[\begin{array}{ccccc}1& 0& 0& ⋮& 10\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]\end{array}$ [Perform row operation]

The matrix is in row-echelon form, and the corresponding system is,

  $\{\begin{array}{c}x=10\\ \begin{array}{l}y=-6\\ z=4\end{array}\end{array}$

Hence,

The solution of the system of linear equation is $x=10$ , $y=-6$ and $z=4$ .

b.

To determine

To solve the system of linear equation represented by the augmented matrix.

Answer to Problem 32E

The solution of the system of linear equation is $x=10$ , $y=-6$ and $z=4$ .

Explanation of Solution

Given information:

The augmented matrix is given as:

  $\left[\begin{array}{ccccc}7& 13& 1& ⋮& -4\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$

Calculation:

Convert the augmented matrix into system of linear equation,

  $\{\begin{array}{l}7x+13y+z=-4\\ -3x-5y-z=-4\\ 3x+6y+z=-2\end{array}$

Use back-substitution to solve the system of linear equation,

  $\begin{array}{l}7x+13y+z=-4\\ z=-7x-13y-4\end{array}$

Substitute, $z=-7x-13y-4$ on $-3x-5y-z=-4$ ,

  $\begin{array}{l}-3x-5y-\left(-7x-13y-4\right)=-4\\ -3x-5y+7x+13y+4=-4\\ 4x+8y=-4-4\\ y=\frac{-8-4x}{8}\end{array}$

Now, substitute, $z=-7x-13y-4$ and $y=\frac{-8-4x}{8}$ on $3x+6y+z=-2$ ,

  $\begin{array}{l}3x+6\left(\frac{-8-4x}{8}\right)+\left[-7x-13\left(\frac{-8-4x}{8}\right)-4\right]=-2\\ 3x+3\left(-2-x\right)+\left[-7x-13\left(\frac{-2-x}{2}\right)-4\right]=-2\\ 3x-6-3x-7x+13+\frac{13}{2}x-4=-2\\ 3x-3x-7x+\frac{13}{2}x=-2+6-13+4\\ \frac{-14x+13x}{2}=-5\\ -x=-10\\ x=10\end{array}$

Substitute back $x=10$ on $y=\frac{-8-4x}{8}$ ,

  $\begin{array}{l}y=\frac{-8-4\left(10\right)}{8}\\ y=\frac{-8-40}{8}\\ y=\frac{-48}{8}\\ y=-6\end{array}$

Again, substitute, $x=10$ and $y=-6$ on $z=-7x-13y-4$ ,

  $\begin{array}{l}z=-7\left(10\right)-13\left(-6\right)-4\\ z=-70+78-4\\ z=-74+78\\ z=4\end{array}$

Hence,

The solution of the system of linear equation is $x=10$ , $y=-6$ and $z=4$ .

c.

To determine

To compare the two-solution method.

Answer to Problem 32E

Method 1 is more preferable because it requires lesser time to solve the system of linear equation.

Explanation of Solution

Given information:

The augmented matrix is given as:

  $\left[\begin{array}{ccccc}7& 13& 1& ⋮& -4\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$

On comparing both the method of solving the system of linear equation, it can be concluded that both methods produce same solution that is the solution of the system of linear equation is $x=10$ , $y=-6$ and $z=4$ .

But method 1 is more preferable because it requires lesser time to solve the system of linear equation.