Solve Using a Matrix by Elimination 3^2a+3b+c=4 , 5^2a+5b+c=4 , 2^2a+2b+c=1

Problem

{[9*a+3*b+c=4],[25*a+5*b+c=4],[4*a+2*b+c=1])

Solution

1. Write the system of linear equations as an augmented matrix [A|B]

[[9,3,1,|,4],[25,5,1,|,4],[4,2,1,|,1]]

2. Swap the first and third rows to make the calculations easier.

(R_1)↔(R_3)⇒[[4,2,1,|,1],[25,5,1,|,4],[9,3,1,|,4]]

3. Eliminate the a terms in the second and third rows using the first row.

(R_2)−25/4*(R_1)→(R_2)⇒[[4,2,1,|,1],[0,−7.5,−5.25,|,−2.25],[9,3,1,|,4]]

(R_3)−9/4*(R_1)→(R_3)⇒[[4,2,1,|,1],[0,−7.5,−5.25,|,−2.25],[0,−1.5,−1.25,|,1.75]]

4. Eliminate the b term in the third row using the second row.

(R_3)−(−1.5)/(−7.5)*(R_2)→(R_3)⇒[[4,2,1,|,1],[0,−7.5,−5.25,|,−2.25],[0,0,−0.2,|,2.2]]

5. Solve for c using the third row.

−0.2*c=2.2⇒c=−11

6. Substitute c into the second row to solve for b

−7.5*b−5.25*(−11)=−2.25⇒−7.5*b+57.75=−2.25

−7.5*b=−60⇒b=8

7. Substitute b and c into the first row to solve for a

4*a+2*(8)+(−11)=1⇒4*a+16−11=1

4*a+5=1⇒4*a=−4⇒a=−1

Final Answer

(a,b,c)=(−1,8,−11)

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