Chapter 11.2, Problem 45AYU

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

$\{\begin{array}{l}3x-5y=3\\ 15x+5y=21\end{array}$

To determine

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

Answer to Problem 45AYU

Solution:

$x=\frac{4}{3},y=\frac{1}{5}$

Consistent system.

Explanation of Solution

Given:

$3x-5y=3$

$15x+5y=21$

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:

$3x-5y=3$

$15x+5y=21$

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

$\left[\begin{array}{cc}3& -5\\ 15& 5\end{array}|\begin{array}{c}3\\ 21\end{array}\right]$

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

To get the (row2, column1) position as 1, divide row1 by 3.

$\left[\begin{array}{cc}\frac{3}{3}& -\frac{5}{3}\\ 15& 5\end{array}|\begin{array}{c}\frac{3}{3}\\ 21\end{array}\right]$

Simplify further:

$\left[\begin{array}{cc}1& -\frac{5}{3}\\ 15& 5\end{array}|\begin{array}{c}1\\ 21\end{array}\right]$

To get the (row2, column1) position as zero, multiply row1 by $-15$ and add to row2:

$\left[\begin{array}{cc}1& -\frac{5}{3}\\ -15\left(1\right)+15& -15\left(-\frac{5}{3}\right)+5\end{array}|\begin{array}{c}1\\ -15\left(1\right)+21\end{array}\right]$

Simplify further:

$\left[\begin{array}{cc}1& -\frac{5}{3}\\ 0& 30\end{array}|\begin{array}{c}1\\ 6\end{array}\right]$

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

Divide row2 by 30:

$\left[\begin{array}{cc}1& -\frac{5}{3}\\ \frac{0}{30}& \frac{30}{30}\end{array}|\begin{array}{c}1\\ \frac{6}{30}\end{array}\right]$

Simplify further:

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

The above matrix corresponds to the system of equations:

$x-\frac{5}{3}y=1$

$y=\frac{1}{5}$

Substitute $y=\frac{1}{5}$ in the first equation, $x-\frac{5}{3}y=1$.

$x-\frac{5}{3}\left(\frac{1}{5}\right)=1$

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

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

$x=\frac{4}{3}$

Thus the solutions of the given system of equations are: $x=\frac{4}{3};y=\frac{1}{5}$.