Q. 386 - Intermediate Algebra

Question

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

$\left\{\begin{array}{l}x+y-3z=-1\\ y-z=0\\ -x+2y=1\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system of equations is:

$$\begin{gathered} x=-1+4z \\ y=-z \end{gathered}$$

Where $z$ is any real number.

Step 1. Given information

The given system of equations is: 

$\left\{\begin{array}{l}x+y-3z=-1\\ y-z=0\\ -x+2y=1\end{array}\right.$

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

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

$\left[\begin{array}{rrrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ -1& 2& 0& 1\end{array}\right]$

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

Perform row operation $R_3\to R_1+R_3$

$\left[\begin{array}{rrrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ -1& 2& 0& 1\end{array}\right]\stackrel{R_3\to R_1+R_3}{⟶}\left[\begin{array}{rrrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ 0& 3& -3& 0\end{array}\right]$

Perform row operation $R_3\to \frac{1}{3}{R}_3$

$\left[\begin{array}{rrrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ 0& 3& -3& 0\end{array}\right]\stackrel{R_3\to \frac{1}{3}{R}_3}{\to }\left[\begin{array}{llrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ 0& 1& -1& 0\end{array}\right]$

Perform a row operation $R_3\to R_2-R_3$

$\left[\begin{array}{llrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ 0& 1& -1& 0\end{array}\right]\stackrel{R_1\to R_2-R_3}{\to }\left[\begin{array}{rrrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ 0& 0& 0& 0\end{array}\right]$

Step 4. Compare the equation by echelon form.

The echelon form of the matrix is:  

$\left[\begin{array}{rrrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ 0& 0& 0& 0\end{array}\right]$

The system of equation by the above matrix is:  

$$\begin{gathered} x+y-3z=-1 \\ -y-z=0 \\ 0=0 \end{gathered}$$

So, this system has infinitely many solutions. So we this system of equation in terms of $z$.

Step 5. Find the value of x and y in terms of z .

Expressing $y$ in terms of $z$, we get:

$$\begin{gathered} -y-z=0 \\ y=-z \end{gathered}$$

Now, substituting $y=-z$ in the equation $x+y-3z=-1$, we get

 $$\begin{gathered} x+y-3z=-1 \\ x-z-3z=-1 \\ x-4z=-1 \\ x=-1+4z \end{gathered}$$

So the solution of the given system of equation is:

$$\begin{gathered} x=-1+4z \\ y=-z \end{gathered}$$

Where $z$ is any real number.