Q. 384 - Intermediate Algebra

Question

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

$\left\{\begin{array}{l}2y+3z=-1\\ 5x+3y=-6\\ 7x+z=1\end{array}\right.$ 

Step-by-Step Solution

Answer

The solution of the system of equations is:

$$\begin{gathered} x=0 \\ y=-2 \\ z=1 \end{gathered}$$

Step 1. Given information

The given system of equations is: 

$\left\{\begin{array}{l}2y+3z=-1\\ 5x+3y=-6\\ 7x+z=1\end{array}\right.$

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

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

$\left[\begin{array}{rrrr}0& 2& 3& -1\\ 5& 3& 0& -6\\ 7& 0& 1& 1\end{array}\right]$

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 $R_3\to 7R_2-5R_3$:

$\left[\begin{array}{rrrr}0& 2& 3& -1\\ 5& 3& 0& -6\\ 7& 0& 1& 1\end{array}\right]\stackrel{R_3\to 7R_2-5R_3}{⟶}\left[\begin{array}{cccc}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 21& -5& -47\end{array}\right]$

Performing row operation$R_3\to 5R_1+3R_3$:

$\left[\begin{array}{cccc}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 21& -5& -47\end{array}\right]\stackrel{R_3\to 5R_1+3R_3}{⟶}\left[\begin{array}{rrrl}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 73& 0& -146\end{array}\right]$

Dividing row 3 by $73$, we get:

$\left[\begin{array}{rrrl}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 73& 0& -146\end{array}\right]\stackrel{R_3\to \frac{1}{73}{R}_3}{\to }\left[\begin{array}{llll}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 1& 0& -2\end{array}\right]$

Step 4. Compare the equation by echelon form.

The echelon form of the matrix is: 

$\left[\begin{array}{llll}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 1& 0& -2\end{array}\right]$

The system of equation by the above matrix is: 

$$\begin{gathered} 2y+3z=-1 \\ 5x+3y=-6 \\ y=-2 \end{gathered}$$

Step 5. Find the value of x & z

Substituting $y=-2$ in the equation

$5x+3y=-6$ :

$$\begin{gathered} 5x+3(-2)=-6 \\ 5x-6=-6 \\ 5x=0 \\ x=0 \end{gathered}$$

Again substituting $y=-2$ in the equation $2y+3z=-1$:

$$\begin{gathered} 2(-2)+3z=-1 \\ -4+3z=-1 \\ 3z=-1+4 \\ 3z=3 \\ z=1 \end{gathered}$$

Thus, the solution of the system of equation is given as:

$$\begin{gathered} x=0 \\ y=-2 \\ z=1 \end{gathered}$$

step 6. Writing solution in the ordered triad

The solution of the system in the ordered triad is $\left(0,-2,1\right)$

Step 7. Check the solution.

Substitute the ordered triad $\left(0,-2,1\right)$ in the equation $2y + 3z = -1$:

$$\begin{gathered} 2y + 3z = -1 \\ 2(-2)+3\left(1\right)=-1 \\ -4+3=-1 \\ -1=-1 \end{gathered}$$

This is true

Substituting the ordered triad $\left(-2,5,2\right)$in the equation $5x + 3y = -6$:

$$\begin{gathered} 5x + 3y = -6 \\ 5\left(0\right)+3(-2)=-6 \\ -6=-6 \end{gathered}$$

This is true.

Substituting the ordered triad $\left(0,-2,1\right)$in the equation $7x + z = 1$:

$$\begin{gathered} 7x + z = 1 \\ 7\left(0\right)+1=1 \\ 1=1 \end{gathered}$$

This is true.

Thus the ordered triad $\left(0,-2,1\right)$ is the solution of the system of equations.