Chapter 11.2, Problem 38AYU

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

$\{\begin{array}{l}x+2y=5\\ x+y=3\end{array}$

To determine

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

Answer to Problem 38AYU

Solution:

$x=1,y=2$

Consistent system.

Explanation of Solution

Given:

$x+2y=5$

$x+y=3$

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 $a$, $b$, $c$ are constants.

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

Calculation:

$x+2y=5$

$x+y=3$

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

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

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

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

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

Simplify further:

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

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

Divide row2 by $-1$:

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

Simplify further:

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

The above matrix corresponds to the system of equations:

$x+2y=5$

$y=2$

Substitute $y=2$ in the first equation, $x+2y=5$.

$x+2\left(2\right)=5$

$x=5-4$

$x=1$

Thus the solutions of the given system of equations are: $x=1,y=2$.