Chapter 11.2, Problem 40AYU

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

$\{\begin{array}{l}3x+3y=3\\ 4x+2y=\frac{8}{3}\end{array}$

To determine

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

Answer to Problem 40AYU

Solution:

$x=\frac{1}{3}\text{, }y=\frac{2}{3}$

Consistent system.

Explanation of Solution

Given:

$3x+3y=3$

$4x+2y=\frac{8}{3}$

Formula used:

To solve a system of two equations in $x$ and $y$ 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:

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

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

Calculation:

$3x+3y=3$

$4x+2y=\frac{8}{3}$

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

$\left[\begin{array}{cc}3& 3\\ 4& 2\end{array}|\begin{array}{c}3\\ \frac{8}{3}\end{array}\right]$

The element in (row1, column1) position is 3.

To get the (row1, column1) position as 1, divide row1 by $3$:

$\left[\begin{array}{cc}\frac{3}{3}& \frac{3}{3}\\ 4& 2\end{array}|\begin{array}{c}\frac{3}{3}\\ \frac{8}{3}\end{array}\right]$

Simplify further:

$\left[\begin{array}{cc}1& 1\\ 4& 2\end{array}|\begin{array}{c}1\\ \frac{8}{3}\end{array}\right]$

To get the (row2, column1) position as zero, multiply row1 by $-4$ and add to row2:

$\left[\begin{array}{cc}1& 1\\ -4\left(1\right)+4& -4\left(1\right)+2\end{array}|\begin{array}{c}1\\ -4\left(1\right)+\frac{8}{3}\end{array}\right]$

Simplify further:

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

Use elementary row operation to get the (row2, column2) position as 1.

Divide row2 by $-2$:

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

Simplify further:

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

The above matrix corresponds to the system of equations:

$x+y=1$

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

Substitute $y=\frac{2}{3}$ in the first equation, $x+y=1$.

$x+\frac{2}{3}=1$

$x=1-\frac{2}{3}$

$x=\frac{1}{3}$

Thus the solutions of the given system of equations are: $x=\frac{1}{3}\text{; }y=\frac{2}{3}$.