Solve Using a Matrix by Elimination 3y+2z=2 , 2x-y-3z=2 , 2x+2y-z=4

Problem

{[3*y+2*z=2],[2*x−y−3*z=2],[2*x+2*y−z=4])

Solution

1. Write the augmented matrix representing the system of linear equations, ensuring variables are aligned in x,y,z order.

[[0,3,2,|,2],[2,−1,−3,|,2],[2,2,−1,|,4]]

2. Swap rows to place a non-zero leading entry in the first row, first column.

(R_1)↔(R_2)

[[2,−1,−3,|,2],[0,3,2,|,2],[2,2,−1,|,4]]

3. Eliminate the x term in the third row by subtracting the first row from the third row.

(R_3)−(R_1)⇒(R_3)

[[2,−1,−3,|,2],[0,3,2,|,2],[0,3,2,|,2]]

4. Eliminate the y term in the third row by subtracting the second row from the third row.

(R_3)−(R_2)⇒(R_3)

[[2,−1,−3,|,2],[0,3,2,|,2],[0,0,0,|,0]]

5. Identify the nature of the system by observing the row of zeros, which indicates that the equations are dependent and there are infinitely many solutions.

0=0

6. Express the variables in terms of a free variable, let z=t

3*y+2*t=2⇒y=(2−2*t)/3

7. Substitute back into the first equation to solve for x

2*x−(2−2*t)/3−3*t=2

2*x=2+(2−2*t)/3+3*t

2*x=(6+2−2*t+9*t)/3

2*x=(8+7*t)/3⇒x=(8+7*t)/6

Final Answer

{[x=(8+7*t)/6],[y=(2−2*t)/3],[z=t])

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