Chapter 8.1, Problem 42E

a.

To determine

Solve the augmented matrix by comparing Linear Systems.

Answer to Problem 42E

1. $\left[\begin{array}{ccccc}4& 8& 0& ⋮& -8\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$ 2. $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$ 3. $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ 3& 6& 1& ⋮& -2\end{array}\right]$

4. $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]$ 5. $\left[\begin{array}{ccccc}1& 0& 0& ⋮& 10\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]$

Explanation of Solution

Given information:

Perform the row operations to solve the augmented matrix by comparing Linear Systems.

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

1. Add $R_2$ to $R_1$

2. Multiply $R_2$ by $\frac{1}{4}$

3. Add $R_3$ to $R_2$

4. Add $-3$ times $R_1$ to $R_3$

5. Add $-2$ times $R_2$ to $R_1$

Calculation:

a) The augmented matrix

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

1. Add $R_2$ to $R_1$

  $\left[\begin{array}{ccccc}7+(-3)& 13+(-5)& 1+(-1)& ⋮& -4+(-4)\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$ = $\left[\begin{array}{ccccc}4& 8& 0& ⋮& -8\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$

Hence the new matrix is $\left[\begin{array}{ccccc}4& 8& 0& ⋮& -8\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$

2. Multiply $R_2$ by $\frac{1}{4}$

Apply $\frac{1}{4}{R}_1$ operation on new matrix.

  $\left[\begin{array}{ccccc}4& 8& 0& ⋮& -8\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$ ⇒ $\frac{1}{4}{R}_1$

  $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$

Hence the resultant matrix is $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$

3. Add $R_3$ to $R_2$

Apply $R_2\to R_3+R_2$ operation on resultant matrix

  $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ -3& -5& -1& ⋮& -4\\ 3& 6& 1& ⋮& -2\end{array}\right]$ ⇒ $R_2\to R_3+R_2$

  $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ -3+3& -5+6& -1+1& ⋮& -4-2\\ 3& 6& 1& ⋮& -2\end{array}\right]$ ⇒ $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ 3& 6& 1& ⋮& -2\end{array}\right]$

Hence the resultant matrix is $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ 3& 6& 1& ⋮& -2\end{array}\right]$

4. Add $-3$ times $R_1$ to $R_3$

Apply $R_3\to (-3R_1+R_3)$ operation on resultant matrix.

  $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ 3& 6& 1& ⋮& -2\end{array}\right]$ ⇒ $R_3\to (-3R_1+R_3)$

  $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ -3+3& -3\times 2+6& -3\times 0+1& ⋮& (-3\times -2+)-2\end{array}\right]$ ⇒ $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]$

Hence the resultant matrix is $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]$

5. Add $-2$ times $R_2$ to $R_1$

  $\left[\begin{array}{ccccc}1& 2& 0& ⋮& -2\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]$

  $R_1\Rightarrow -2R_2+R_1$

  $\left[\begin{array}{ccccc}1-0& 2-2& 0-0& ⋮& -2+(-2\times -6)\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]$ ⇒ $\left[\begin{array}{ccccc}1& 0& 0& ⋮& 10\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]$

Hence the resultant matrix is $\left[\begin{array}{ccccc}1& 0& 0& ⋮& 10\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]$

b.

To determine

Solve and write the system of linear equations represented by the augmented matrix,

Answer to Problem 42E

  $x=10$ , $y=-6$ , $z=4$

Explanation of Solution

Given information:

Write and solve the system of linear equations represented by the augmented matrix,

Calculation:

The equation system of augmented matrix is

System ⇒ $\{\begin{array}{ccccc}7x& +13y& +z& ⋮& -4\\ -3x& -5y& -z& ⋮& -4\\ 3x& +6y& +z& ⋮& -2\end{array}$

Now eliminate $x$ by adding $eq2\&eq3$ .

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

Now eliminate $z$ by adding $eq1\&eq2$ .

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

Substitute $y=-6$ in equation $4x+8y=-8$

  $\begin{array}{l}4x+8(-6)=-8\\ 4x=-8+48\\ x=10\end{array}$

Substitute $y=-6$ , $x=10$ in equation $-3x-5y-z=-4$

  $\begin{array}{l}-3\times 10-5\times -6-z=-4\\ -30+30-z=-4\\ z=4\end{array}$

Hence the answer is $x=10$ , $y=-6$ , $z=4$ same as part (a)

c.

To determine

Compare the both solution methods.

Answer to Problem 42E

  $x=10$ , $y=-6$ , $z=4$

Explanation of Solution

Given information:

Compare the both solution methods. Which do you prefer⇒

Calculation:

To apply row method is preferable.