Chapter 11.2, Problem 46AYU

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

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

To determine

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

Answer to Problem 46AYU

Solution:

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

Consistent system.

Explanation of Solution

Given:

$2x-y=-1$

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

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 $a,b,c$ are constants.

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

Calculation:

$2x-y=-1$

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

Rewrite the equations,

$2x-y=-1$

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

Or

$2x-y=-1$

$2x+y=3$

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

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

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

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

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

Simplify further:

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

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

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

Simplify further:

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

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

Divide row2 by 2:

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

Simplify further:

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

The above matrix corresponds to the system of equations:

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

$y=2$

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

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

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

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

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

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