Chapter 11.2, Problem 47AYU

In problems 39-74, solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent.

$\{\begin{array}{l}x-y=6\\ 2x-3z=16\\ 2y+z=4\end{array}$

To determine

To find: The solution to the given system of equations using matrices.

Answer to Problem 47AYU

Solution:

$x=8,y=2,z=0$

Consistent system.

Explanation of Solution

Given:

$x-y=6$

$2x-3z=16$

$2y+z=4$

Formula used:

To solve a system of three equations in $x,y$ and $z$ using matrices:

Step 1: Write the corresponding matrix associated with the system of equations.

Step 2: Use elementary row operations to get equivalent matrix of the form:

$\left[\begin{array}{ccc}1& a& b\\ 0& 1& c\\ 0& 0& 1\end{array}|\begin{array}{c}d\\ e\\ f\end{array}\right]$; where $a,b,c,d,e,f$ are constants.

Step 3: Solve for $x,y$ and $z$.

Calculation:

$x-y=6$

$2x-3z=16$

$2y+z=4$

The corresponding matrix associated with the above system of equations is:

$\left[\begin{array}{ccc}1& -1& 0\\ 2& 0& -3\\ 0& 2& 1\end{array}|\begin{array}{c}6\\ 16\\ 4\end{array}\right]$

The element in (row1, column1) position is already 1.

To get the (row2, column1) position as zero, multiply row1 by $-2$ and add to row2:

$\left[\begin{array}{ccc}1& -1& 0\\ -2\left(1\right)+2& -2\left(-1\right)+0& -2\left(0\right)-3\\ 0& 2& 1\end{array}|\begin{array}{c}6\\ -2\left(6\right)+16\\ 4\end{array}\right]$

Simplify:

$\left[\begin{array}{ccc}1& -1& 0\\ 0& 2& -3\\ 0& 2& 1\end{array}|\begin{array}{c}6\\ 4\\ 4\end{array}\right]$

To get the (row2, column2) position as 1, divide row2 by 2:

$\left[\begin{array}{ccc}1& -1& 0\\ \frac{0}{2}& \frac{2}{2}& -\frac{3}{2}\\ 0& 2& 1\end{array}|\begin{array}{c}6\\ \frac{4}{2}\\ 4\end{array}\right]$

Simplify further:

$\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -\frac{3}{2}\\ 0& 2& 1\end{array}|\begin{array}{c}6\\ 2\\ 4\end{array}\right]$

Now the (row3, column1) position is already zero.

To get the (row3, column 2) position as zero, multiply row2 by $-2$ and add to row3:

$\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -\frac{3}{2}\\ 0& -2\left(1\right)+2& -2\left(-\frac{3}{2}\right)+1\end{array}|\begin{array}{c}6\\ 2\\ -2\left(2\right)+4\end{array}\right]$

Simplify further:

$\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -\frac{3}{2}\\ 0& 0& 4\end{array}|\begin{array}{c}6\\ 2\\ 0\end{array}\right]$

To get the (row3, column3) position as 1, divide row3 by 4:

$\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -\frac{3}{2}\\ \frac{0}{4}& \frac{0}{4}& \frac{4}{4}\end{array}|\begin{array}{c}6\\ 2\\ \frac{0}{4}\end{array}\right]$

Simplify:

$\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -\frac{3}{2}\\ 0& 0& 1\end{array}|\begin{array}{c}6\\ 2\\ 0\end{array}\right]$

The above matrix corresponds to the system of equations:

$x-y=6$

$y-\frac{3}{2}z=2$

$z=0$

Now substitute $z=0$ in the second equation $y-\frac{3}{2}z=2$, to get the value of $y$.

$y-\frac{3}{2}z=2$

$y-\frac{3}{2}\left(0\right)=2$

$y-0=2$

$y=2$

Now substitute $y=2$ in the first equation $x-y=6$, to get the value of $x$.

$x-y=6$

$x-2=6$

$x=6+2$

$x=8$

Thus the solutions to the given system of equations are: $x=8,y=2;z=0$.