Solve Using a Matrix by Elimination 52f+48b+36m=67936 , f-b-m=398 , b-3m=453

Problem

{[52*ƒ+48*b+36*m=67936],[ƒ−b−m=398],[b−3*m=453])

Solution

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

[[52,48,36,|,67936],[1,−1,−1,|,398],[0,1,−3,|,453]]

2. Swap rows to place a 1 in the top-left corner for easier calculation.

(R_1)↔(R_2)

[[1,−1,−1,|,398],[52,48,36,|,67936],[0,1,−3,|,453]]

3. Eliminate the first entry in the second row by subtracting 52 times the first row from the second row.

(R_2)−52*(R_1)→(R_2)

[[1,−1,−1,|,398],[0,100,88,|,47240],[0,1,−3,|,453]]

4. Swap rows again to get a 1 in the second pivot position.

(R_2)↔(R_3)

[[1,−1,−1,|,398],[0,1,−3,|,453],[0,100,88,|,47240]]

5. Eliminate the entry below the second pivot by subtracting 100 times the second row from the third row.

(R_3)−100*(R_2)→(R_3)

[[1,−1,−1,|,398],[0,1,−3,|,453],[0,0,388,|,1940]]

6. Solve for m by dividing the third row by 388.

(R_3)/388→(R_3)

[[1,−1,−1,|,398],[0,1,−3,|,453],[0,0,1,|,5]]

m=5

7. Solve for b using back-substitution into the second row equation.

b−3*(5)=453

b−15=453

b=468

8. Solve for f using back-substitution into the first row equation.

ƒ−468−5=398

ƒ−473=398

ƒ=871

Final Answer

(ƒ,b,m)=(871,468,5)

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