Chapter 8.3, Problem 57E

To determine

The system of linear equations by using inverse matrix.

Answer to Problem 57E

  $\left(\begin{array}{c}x=\frac{13}{16}+\frac{5}{16}z\\ y=\frac{11}{16}+\frac{19}{16}z\\ z\end{array}\right)$ .

Explanation of Solution

Given information:

The given system of equation is:

  $\{\begin{array}{l}5x-3y+2z=2\\ 2x+2y-3z=3\\ x-7y+8z=-4\end{array}$

Calculation:

The given system of equation is:

  $\{\begin{array}{l}5x-3y+2z=2\\ 2x+2y-3z=3\\ x-7y+8z=-4\end{array}$

Convert the augmented matrix into the row-echelon form:

Divide the first row by five:

  ${\left[\begin{array}{ccc}5& -3& 2\\ 2& 2& -3\\ 1& -7& 8\end{array}|\begin{array}{c}2\\ 3\\ -4\end{array}\right]}^{\times \left(\frac{1}{5}\right)}\Rightarrow \left[\begin{array}{ccc}1& -3/5& 2/5\\ 2& 2& -3\\ 1& -7& 8\end{array}|\begin{array}{c}2/5\\ 3\\ -4\end{array}\right]$

Multiply the first row by two and after that subtract the first row from the second:

  $\left[\begin{array}{ccc}1& -3/5& 2/5\\ 0& 16/5& -19/5\\ 1& -7& 8\end{array}|\begin{array}{c}2/5\\ 11/5\\ -4\end{array}\right]$

Subtract one from the row first and multiply the resultant matrix by first row:

  $\left[\begin{array}{ccc}1& -3/5& 2/5\\ 0& 16/5& -19/5\\ 0& -32/5& 38/5\end{array}|\begin{array}{c}2/5\\ 11/5\\ -22/5\end{array}\right]$

Divide the second row by $\frac{16}{5}$ .

  $\left[\begin{array}{ccc}1& -3/5& 2/5\\ 0& 1& -19/16\\ 0& -32/5& 38/5\end{array}|\begin{array}{c}2/5\\ 11/16\\ -22/5\end{array}\right]$

Multiply the second row by $\frac{-32}{5}$ after that subtract the second row by third row:

  $\left[\begin{array}{ccc}1& -3/5& 2/5\\ 0& 1& -19/16\\ 0& 0& 0\end{array}|\begin{array}{c}2/5\\ 11/16\\ 0\end{array}\right]$

Multiply the second row by $\frac{-3}{5}$ and after that subtract the resultant from the row first:

  $\left[\begin{array}{ccc}1& 0& -5/16\\ 0& 1& -19/16\\ 0& 0& 0\end{array}|\begin{array}{c}13/16\\ 11/16\\ 0\end{array}\right]$

Now find the variable $x$ from the equation first of the system:

  $x=\frac{13}{16}+\frac{5}{16}z$

Now find the variable $y$ from the equation second of the system:

  $y=\frac{11}{16}+\frac{19}{16}z$

Therefore, the solutions are $\left(\begin{array}{c}x=\frac{13}{16}+\frac{5}{16}z\\ y=\frac{11}{16}+\frac{19}{16}z\\ z\end{array}\right)$