Solve Using a Matrix by Elimination a+2b+g=7 , a+b+d-g=3 , 3a+b+5d-7g=1

Problem

{[a+2*b+g=7],[a+b+d−g=3],[3*a+b+5*d−7*g=1])

Solution

1. Write the augmented matrix for the system of linear equations, using the variables a b d and g as columns.

[[1,2,0,1,7],[1,1,1,−1,3],[3,1,5,−7,1]]

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

[[1,2,0,1,7],[0,−1,1,−2,−4],[0,−5,5,−10,−20]]

3. Eliminate the second column entry below the second pivot by performing the row operation (R_3)−5*(R_2)→(R_3)

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

4. Identify the free variables by observing that the third row is all zeros, indicating a dependent system. The variables d and g correspond to columns without pivots.

d=s()

g=t

5. Solve for the leading variables a and b in terms of the parameters s() and t using back-substitution.

−b+s−2*t=−4⇒b=s−2*t+4

a+2*(s−2*t+4)+t=7⇒a+2*s−4*t+8+t=7⇒a=−2*s+3*t−1

Final Answer

(a,b,d,g)=(−2*s+3*t−1,s−2*t+4,s,t)

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