Solve Using a Matrix by Row Operations x+y+4z=1 , 3x+2y+11z=1 , x+3z=-1

Problem

{[x+y+4*z=1],[3*x+2*y+11*z=1],[x+3*z=−1])

Solution

1. Write the augmented matrix representing the system of linear equations.

[[1,1,4,|,1],[3,2,11,|,1],[1,0,3,|,−1]]

2. Eliminate the x-terms in the second and third rows by performing row operations (R_2)−3*(R_1)→(R_2) and (R_3)−(R_1)→(R_3)

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

3. Simplify the second row by multiplying by −1 (−1*(R_2)→(R_2).

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

4. Eliminate the y-term in the third row by performing (R_3)+(R_2)→(R_3)

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

5. Identify the system type as having infinitely many solutions because the third row is all zeros, indicating a dependent system.

6. Express the variables in terms of a parameter z=t From the second row, y+z=2 so y=2−t From the first row, x+y+4*z=1

7. Substitute and solve for x by plugging in y=2−t and z=t

x+(2−t)+4*t=1

x+2+3*t=1

x=−1−3*t

Final Answer

(x,y,z)=(−1−3*t,2−t,t)

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