Chapter 8.1, Problem 39E

a.

To determine

Solve the augmented matrix.

Answer to Problem 39E

  $\boxed{\left[\begin{array}{cccc}3& 0& :& -6\\ 6& -4& :& -28\end{array}\right]}$ , $\boxed{\left[\begin{array}{cccc}3& 0& :& -6\\ 0& -4& :& -16\end{array}\right]}$ , $\boxed{\left[\begin{array}{cccc}3& 0& :& -6\\ 0& 1& :& -4\end{array}\right]}$ and $\boxed{\left[\begin{array}{cccc}1& 0& :& -2\\ 0& 1& :& 4\end{array}\right]}$

Explanation of Solution

Given information:

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

Add $R_2$ to $R_1$

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

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

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

Calculation:

Consider the augmented matrix

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

Perform the below row operations

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 resultant matrix is $\boxed{\left[\begin{array}{cccc}3& 0& :& -6\\ 6& -4& :& -28\end{array}\right]}$

Add $-2$ times $R_1$ to $R_2$ that is apply $-2R_1+R_2$

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

  $\to \left[\begin{array}{cccc}3& 0& :& -6\\ 0& -4& :& -16\end{array}\right]$

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

Multiply $R_2$ by $-\frac{1}{4}$ that is apply $-\frac{1}{4}$

  $R_2$

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

  $\to \left[\begin{array}{cccc}3& 0& :& -6\\ 0& 1& :& 4\end{array}\right]$

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

Multiply $R_1$ by $\frac{1}{3}$ that is apply $\frac{1}{3}$

  $R_3$

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

  $\to \left[\begin{array}{cccc}1& 0& :& -2\\ 0& 1& :& 4\end{array}\right]$

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

b.

To determine

Write and solve the system of linear equations

Answer to Problem 39E

  $\boxed{x=-2,y=4}$

Explanation of Solution

Given information:

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

Calculation:

The system of equations from the augmented matrix is

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

Solve these equations

Multiply the 1^st equation by 2 and then add it to the 2^nd equation

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

Substitute $y=4$ in the equation $-3x+4y=22$

  $\begin{array}{l}-3x+4(4)=22\\ -3x=22-16\\ -3x=6\\ x=-2\end{array}$

Hence, the solution is $\boxed{x=-2,y=4}$

This is same as in (a).

c.

To determine

Compare the two solution methods.

Answer to Problem 39E

Using row operations is easy method

Explanation of Solution

Given information:

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

Calculation:

Hence, using row operations is easy method.