Chapter 11.2, Problem 48AYU

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

$\{\begin{array}{l}2x+y=-4\\ -2y+4z=0\\ 3x-2z=-11\end{array}$

To determine

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

Answer to Problem 48AYU

Solution:

$x=-3,y=2,z=1$

Consistent system.

Explanation of Solution

Given:

$2x+y=-4$

$-2y+4z=0$

$3x-2z=-11$

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:

$2x+y=-4$

$-2y+4z=0$

$3x-2z=-11$

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

$\left[\begin{array}{ccc}2& 1& 0\\ 0& -2& 4\\ 3& 0& -2\end{array}|\begin{array}{c}-4\\ 0\\ -11\end{array}\right]$

The element in (row1, column1) position is 2.

To get the (row1, column1) position as 1, divide row1 by 2:

$\left[\begin{array}{ccc}\frac{2}{2}& \frac{1}{2}& \frac{0}{2}\\ 0& -2& 4\\ 3& 0& -2\end{array}|\begin{array}{c}-\frac{4}{2}\\ 0\\ -11\end{array}\right]$

Simplify further:

$\left[\begin{array}{ccc}1& \frac{1}{2}& 0\\ 0& -2& 4\\ 3& 0& -2\end{array}|\begin{array}{c}-2\\ 0\\ -11\end{array}\right]$

We see that the (row2, column1) position is zero.

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

$\left[\begin{array}{ccc}1& \frac{1}{2}& 0\\ \frac{0}{-2}& \frac{-2}{-2}& \frac{4}{-2}\\ 3& 0& -2\end{array}|\begin{array}{c}-2\\ \frac{0}{-2}\\ -11\end{array}\right]$

Simplify further:

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

(row3, column1) position is 3.

To get the (row3, column 1) position as zero, multiply row 1 by $-3$ and add to row3:

$\left[\begin{array}{ccc}1& \frac{1}{2}& 0\\ 0& 1& -2\\ -3\left(1\right)+3& -3\left(\frac{1}{2}\right)+0& -3\left(0\right)-2\end{array}|\begin{array}{c}-2\\ 0\\ -3\left(-2\right)-11\end{array}\right]$

Simplify further:

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

To get the (row3, column 2) position as zero, multiply row 2 by $\frac{3}{2}$ and add to row3:

$\left[\begin{array}{ccc}1& \frac{1}{2}& 0\\ 0& 1& -2\\ 0& \frac{3}{2}\left(1\right)-\frac{3}{2}& \frac{3}{2}\left(-2\right)-2\end{array}|\begin{array}{c}-2\\ 0\\ \frac{3}{2}\left(0\right)-5\end{array}\right]$

Simplify further:

$\left[\begin{array}{ccc}1& \frac{1}{2}& 0\\ 0& 1& -2\\ 0& 0& -5\end{array}|\begin{array}{c}-2\\ 0\\ -5\end{array}\right]$

To get the (row3, column3) position as 1, divide row 3 by $-5$:

$\left[\begin{array}{ccc}1& \frac{1}{2}& 0\\ 0& 1& -2\\ \frac{0}{-5}& \frac{0}{-5}& \frac{-5}{-5}\end{array}|\begin{array}{c}-2\\ 0\\ \frac{-5}{-5}\end{array}\right]$

Simplify:

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

The above matrix corresponds to the system of equations:

$x+\frac{1}{2}y=-2$

$y-2z=0$

$z=1$

Now substitute $z=1$ in the second equation $y-2z=0$, to get the value of $y$.

$y-2z=0$

$y-2\left(1\right)=0$

$y=2$

Now substitute $y=2$ in the first equation $x+\frac{1}{2}y=-2$, to get the value of $x$.

$x+\frac{1}{2}y=-2$

$x+\frac{1}{2}\left(2\right)=-2$

$x+1=-2$

$x=-2-1$

$x=-3$

Thus the solutions to the given system of equations are: $x=-3,y=2;z=1$.