Solve Using a Matrix by Elimination 2a+b-g=4 , b+g+h=4 , a-d+2g=0

Problem

{[2*a+b−g=4],[b+g+h=4],[a−d+2*g=0])

Solution

1. Identify the variables and constants to set up the augmented matrix. The variables present are a,b,g,h,d To maintain a standard matrix form, we list them in alphabetical order: a,b,d,g,h

2. Construct the augmented matrix [A|B] where each row represents an equation and each column represents a variable coefficient or the constant term.

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

3. Swap the first and third rows ((R_1)↔(R_3) to get a 1 in the top-left pivot position.

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

4. Eliminate the a term in the third row by performing the row operation (R_3)−2*(R_1)→(R_3)

[[1,0,−1,2,0,0],[0,1,0,1,1,4],[0,1,2,−5,0,4]]

5. Eliminate the b term in the third row by performing the row operation (R_3)−(R_2)→(R_3)

[[1,0,−1,2,0,0],[0,1,0,1,1,4],[0,0,2,−6,−1,0]]

6. Normalize the third row by dividing by 2 ((R_3)/2→(R_3) to reach row-echelon form.

[[1,0,−1,2,0,0],[0,1,0,1,1,4],[0,0,1,−3,−0.5,0]]

7. Back-substitute to find the general solution. Since there are 3 equations and 5 variables, we assign parameters to the free variables g and h Let g=s() and h=t

8. Solve for the remaining variables using the rows of the matrix.

d−3*s−0.5*t=0⇒d=3*s+0.5*t

b+s+t=4⇒b=4−s−t

a−d+2*s=0⇒a−(3*s+0.5*t)+2*s=0⇒a=s+0.5*t

Final Answer

{[a=s+0.5*t],[b=4−s−t],[d=3*s+0.5*t],[g=s()],[h=t])

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