Chapter 8.1, Problem 40E

a.

To determine

Perform the row operations to solve the augmented matrix.

Answer to Problem 40E

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

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

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

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

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

Explanation of Solution

Given information:

Perform the row operations to solve the augmented matrix.

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

Add $R_2$ to $R_1$ .

Multiply $R_1$ by $1/4$ .

Add $R_3$ to $R_2$ .

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

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

Calculation:

Add $R_2$ to $R_1$ .

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

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

Multiply $R_1$ by $1/4$ .

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

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

Add $R_3$ to $R_2$ .

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

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

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

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

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

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

  $\left[\begin{array}{ccccc}1-0& 2-2& 0-0& ⋮& -2+12\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right] \Rightarrow \left[\begin{array}{ccccc}1& 0& 0& ⋮& 10\\ 0& 1& 0& ⋮& -6\\ 0& 0& 1& ⋮& 4\end{array}\right]$

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

Write and solve the system of linear equations.

Answer to Problem 40E

  $\boxed{x=10,y=-6,z=4}$

Explanation of Solution

Given information:

Write and solve the system of linear equations (in variables $x,y$ and $z$ , if applicable) represented by the augmented matrix.

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

Calculation:

Solve the equations,

  $\begin{array}{l}7x+13y+z=-\mathrm{4.....}\left(1\right)\\ -3x-5y-z=-\mathrm{4......}\left(2\right)\\ 3x+6y+z=-\mathrm{2.....}\left(3\right)\end{array}$

Eliminate $x$ by adding equation $\left(2\right)$ and $\left(3\right)$ ,

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

Now eliminate $z$ .

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

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

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

Now put $y=-6$ and $x=10$ in equation $\left(3\right)$ we get,

  $\begin{array}{l}3x+6y+z=-2\\ 3\left(10\right)+6\left(-6\right)+z=-2\\ 30-36+z=-2\\ z=-2+6\\ z=4\end{array}$

Hence, the solution is $\boxed{x=10,y=-6,z=4}$ .

c.

To determine

Identify the best method.

Answer to Problem 40E

  $\boxed{\text{Row opration method}}$

Explanation of Solution

Given information:

Compare the two solution methods. Which do you prefer?

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

Calculation:

Both the methods are best applicable, but row opration is the easy metod,

Hence we prefer $\boxed{\text{Row opration method}}$ .