Solve Using a Matrix by Elimination x-4y+2z=8 , -2x+y-7z=0 , -7y-3z=16

Problem

{[x−4*y+2*z=8],[−2*x+y−7*z=0],[−7*y−3*z=16])

Solution

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

[[1,−4,2,|,8],[−2,1,−7,|,0],[0,−7,−3,|,16]]

2. Eliminate the x-term in the second row by adding 2 times the first row to the second row ((R_2)←(R_2)+2*(R_1).

[[1,−4,2,|,8],[0,−7,−3,|,16],[0,−7,−3,|,16]]

3. Eliminate the y-term in the third row by subtracting the second row from the third row ((R_3)←(R_3)−(R_2).

[[1,−4,2,|,8],[0,−7,−3,|,16],[0,0,0,|,0]]

4. Identify the system type based on the resulting row-echelon form. The third row 0 = 0$ indicates that the system is dependent and has infinitely many solutions.

5. Express the variables in terms of a free variable, z From the second row, solve for y

−7*y−3*z=16

−7*y=3*z+16

y=−(3*z+16)/7

6. Substitute the expression for y into the first row to solve for x

x−4*(−(3*z+16)/7)+2*z=8

x+(12*z+64)/7+(14*z)/7=56/7

x+(26*z+64)/7=56/7

x=(56−64−26*z)/7

x=(−26*z−8)/7

Final Answer

(x,y,z)=((−26*z−8)/7,(−3*z−16)/7,z)

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