Solve Using a Matrix by Elimination 2a-2b=0 , d+3g=2 , 3a-3b=0 , a-b+2d+6g=4

Problem

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

Solution

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

[[2,−2,0,0,|,0],[0,0,1,3,|,2],[3,−3,0,0,|,0],[1,−1,2,6,|,4]]

2. Simplify the rows by dividing the first row by 2 and the third row by 3

[[1,−1,0,0,|,0],[0,0,1,3,|,2],[1,−1,0,0,|,0],[1,−1,2,6,|,4]]

3. Perform row operations to eliminate entries. Subtract the first row from the third row ((R_3)−(R_1)→(R_3) and the first row from the fourth row ((R_4)−(R_1)→(R_4).

[[1,−1,0,0,|,0],[0,0,1,3,|,2],[0,0,0,0,|,0],[0,0,2,6,|,4]]

4. Continue elimination by subtracting twice the second row from the fourth row ((R_4)−2*(R_2)→(R_4).

[[1,−1,0,0,|,0],[0,0,1,3,|,2],[0,0,0,0,|,0],[0,0,0,0,|,0]]

5. Identify the free variables and the pivot variables. The pivots are in the a and d columns. The variables b and g are free.

a−b=0⇒a=b

d+3*g=2⇒d=2−3*g

6. Express the general solution in terms of the free variables b and g

(a,b,d,g)=(b,b,2−3*g,g)

Final Answer

(a,b,d,g)=(b,b,2−3*g,g)* for any *b,g∈ℝ

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