Solve Using a Matrix by Elimination

Problem

{[3*a+5*b+2*d=0],[4*a+7*b+5*d=0],[a+b−4*d=0],[2*a+9*b+6*d=0])

Solution

1. Write the augmented matrix for the system of linear equations. Since all equations are equal to zero, this is a homogeneous system.

[[3,5,2,0],[4,7,5,0],[1,1,−4,0],[2,9,6,0]]

2. Swap rows to place a 1 in the top-left corner (pivot position). Swap (R_1) and (R_3)

[[1,1,−4,0],[4,7,5,0],[3,5,2,0],[2,9,6,0]]

3. Eliminate entries below the first pivot by performing row operations: (R_2)→(R_2)−4*(R_1) (R_3)→(R_3)−3*(R_1) and (R_4)→(R_4)−2*(R_1)

[[1,1,−4,0],[0,3,21,0],[0,2,14,0],[0,7,14,0]]

4. Normalize the second row by dividing (R_2) by 3 to create a pivot of 1.

[[1,1,−4,0],[0,1,7,0],[0,2,14,0],[0,7,14,0]]

5. Eliminate entries below the second pivot by performing (R_3)→(R_3)−2*(R_2) and (R_4)→(R_4)−7*(R_2)

[[1,1,−4,0],[0,1,7,0],[0,0,0,0],[0,0,−35,0]]

6. Simplify the matrix by swapping (R_3) and (R_4) and then dividing the new (R_3) by −35

[[1,1,−4,0],[0,1,7,0],[0,0,1,0],[0,0,0,0]]

7. Perform back-substitution or further row reduction to reach reduced row echelon form. Use (R_3) to clear entries above it: (R_2)→(R_2)−7*(R_3) and (R_1)→(R_1)+4*(R_3)

[[1,1,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,0]]

8. Finalize the reduction by performing (R_1)→(R_1)−(R_2)

[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,0]]

9. Interpret the result as a system of equations. The matrix indicates a=0 b=0 and d=0

Final Answer

a=0,b=0,d=0

Want more problems? Check here!