Chapter 8.1, Problem 41E

a.

To determine

Solve the augmented matrix by comparing Linear Systems.

Answer to Problem 41E

1. $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 6& -4& ⋮& -28\end{array}\right]$ 2. $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 0& -4& ⋮& -16\end{array}\right]$ 3. $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 0& 1& ⋮& 4\end{array}\right]$ 4. $\left[\begin{array}{cccc}1& 0& ⋮& -2\\ 0& 1& ⋮& 4\end{array}\right]$

Explanation of Solution

Given information:

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

  $(a)$ $\left[\begin{array}{cccc}-3& 4& ⋮& 22\\ 6& -4& ⋮& -28\end{array}\right]$

1. Add $R_2$ to $R_1$

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

3. Multiply $R_2$ by $-\frac{1}{4}$

4. Multiply $R_1$ by $\frac{1}{3}$

Calculation:

a) The augmented matrix

  $\left[\begin{array}{cccc}-3& 4& ⋮& 22\\ 6& -4& ⋮& -28\end{array}\right]$

1.Add $R_2$ to $R_1$

  $\left[\begin{array}{cccc}-3+6& 4+(-4)& ⋮& 22+(-28)\\ 6& -4& ⋮& -28\end{array}\right]$

  $=\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 6& -4& ⋮& -28\end{array}\right]$

Hence the new matrix is $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 6& -4& ⋮& -28\end{array}\right]$

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

Apply $-2R_1+R_2\to R_2$ operation on new matrix.

  $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 6& -4& ⋮& -28\end{array}\right]$

  $-2R_1+R_2\to R_2$

  $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ -2(3)+6& -2(0)-4& ⋮& -2(-6)-28\end{array}\right]$ ⇒ $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 0& -4& ⋮& -16\end{array}\right]$

Hence the resultant matrix is $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 0& -4& ⋮& -16\end{array}\right]$

3. Multiply $R_2$ by $-\frac{1}{4}$ .

Apply $-\frac{1}{4}{R}_2$ operation on resultant matrix

  $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 0& -4& ⋮& -16\end{array}\right]$ ⇒ $-\frac{1}{4}{R}_2$

  $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 0& -4/-4& ⋮& -16/-4\end{array}\right]$ ⇒ $\left[\begin{array}{cccc}3& 0& ⋮& -6\\ 0& 1& ⋮& 4\end{array}\right]$

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

4. Multiply $R_1$ by $\frac{1}{3}$

Apply $\frac{1}{3}{R}_1$ operation on resultant matrix

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

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

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

b.

To determine

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

Answer to Problem 41E

  $\boxed{x=-2,y=4}$

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 ⇒ $\{\begin{array}{cccc}-3x& +4y& =& 22\\ 6x& -4y& =& -28\end{array}$

Now multiply $eq1$ by $2$ and add both the equations.

  $\frac{\begin{array}{cccc}-6x& +8y& =& 44\\ 6x& -4y& =& -28\end{array}}{\begin{array}{l}4y=16\\ y=4\end{array}}$

Substitute $y=4$ in first equation of system we get,

  $\begin{array}{l}-3x+4(4)=22\\ -3x=22-16\\ x=-2,y=4\end{array}$

Hence the answer is $\boxed{x=-2,y=4}$ same as part (a)

c.

To determine

Compare the both solution methods.

Answer to Problem 41E

  $\boxed{x=-2,y=4}$

Explanation of Solution

Given information:

Compare the both solution methods. Which do you prefer?

Calculation:

To apply row method is preferable.