Chapter 11.2, Problem 43AYU

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

$\{\begin{array}{l}2x+3y=6\\ x-y=\frac{1}{2}\end{array}$

To determine

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

Answer to Problem 43AYU

Solution:

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

Consistent system.

Explanation of Solution

Given:

$2x+3y=6$

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

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:

$2x+3y=6$

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

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

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

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

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

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

Simplify further:

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

To get the (row2, column1) position as zero, multiply row1 by -1 and add to row2:

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

Simplify further:

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

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

Divide row2 by $-\frac{5}{2}$:

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

Simplify further:

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

The above matrix corresponds to the system of equations:

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

$y=1$

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

$x+\frac{3}{2}\left(1\right)=3$

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

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

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