Q. 382 - Intermediate Algebra

Question

In the following exercises, solve each system of equations using a matrix.

$\left\{\begin{array}{l}4x + y = 6\\ x - y = 4\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system of equations is:

$$\begin{gathered} x=2 \\ y=-2 \end{gathered}$$

Step 1. Given information

The given system of equations is:

$\left\{\begin{array}{l}4x + y = 6\\ x - y = 4\end{array}\right.$

Step 2. Convert the system of equations into an augmented matrix.

The augmented matrix of the given system of equations is given as:

$\left[\begin{array}{rrr}4& 1& 6\\ 1& -1& 4\end{array}\right]$

In the augmented matrix, the first equation gives the first row and the second equation gives the second row. The vertical line replaces the equal signs.

Step 3. Convert the augmented matrix into echelon form

Interchanging the row 1 and $2$, we get:

$\left[\begin{array}{rrr}4& 1& 6\\ 1& -1& 4\end{array}\right]\stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{rrr}1& -1& 4\\ 4& 1& 6\end{array}\right]$

Then we multiply the row $1$ by $-4$ and add it to row $2$, we get:

$$\begin{gathered} \left[\begin{array}{rrr}1& -1& 4\\ 4& 1& 6\end{array}\right]\stackrel{-4R_1+R_2}{⟶}\left[\begin{array}{lll}1& -1& 4\\ (-4·1)+4& (-4(-1\left)\right)+1& (-4·4)+6\end{array}\right] \\ =\left[\begin{array}{rrr}1& -1& 4\\ 0& 5& -10\end{array}\right] \end{gathered}$$

Dividing row $2$ by $5$ , we get:

$\left[\begin{array}{rrr}1& -1& 4\\ 0& 5& -10\end{array}\right]\stackrel{\frac{R_2}{5}}{\to }\left[\begin{array}{rrr}1& -1& 4\\ 0& 1& -2\end{array}\right]$

Step 4. Compare the equation by echelon form.

The echelon form of the matrix is:

$\left[\begin{array}{rrr}1& -1& 4\\ 0& 1& -2\end{array}\right]$

The system of equation by the above matrix is:

$$\begin{gathered} x-y=4 \\ y=-2 \end{gathered}$$

Step 5. Find the value of x

Substituting the value $y=-2$ in the equation

$$\begin{gathered} x-y=4 \\ x-(-2)=4 \\ x+2=4 \\ x=4-2 \\ x=2 \end{gathered}$$

So, the solution of the given system of equation is :

$$\begin{gathered} x=2 \\ y=-2 \end{gathered}$$

Step 6. Check the solution for equation 4 x   +   y   =   6 .

Substituting $$\begin{gathered} x=2 \\ y=-2 \end{gathered}$$ in the equation, we get:

$$\begin{gathered} 4x + y = 6 \\ 4\left(2\right)-2=6 \\ 6=6 \end{gathered}$$

This is true.

Step 7. Check the solution for equation x   −   y   =   4  

Substituting $$\begin{gathered} x=2 \\ y=-2 \end{gathered}$$ in the equation, we get:

$$\begin{gathered} x - y = 4 \\ 2-(-2)=4 \\ 4=4 \end{gathered}$$

This is true.

So the ordered pair $\left(2,2\right)$ is the solution.