I believe that the answer to Part 1 is correct.
For Part 2, setting x4 = r = 0 and x5 = s = 0, we get x1 = -5 - 7(0) - 5(0) = -5, x2 = 2 + 4(0) + 3(0) = 2, and
x3 = 4 - 6(0) + 2(0) = 4.
So, a solution is < -5, 2, 4, 0, 0 >
John L.
asked 09/02/21Part 1: Verify that the following augmented matrices are in rref(row echelon form), identify the leading and the free
variables, then find all the solutions, if there are any, to the systems of equations
corresponding to the rref(row echelon form). Assume that the variables are x1, x2, . . . , xn. Use the letters r,
s, t, and u for values of the free variables. Write your solutions in vector form.
[ 1 0 0 7 5 l -5 ]
0 1 0 -4 -3 l 2
0 0 1 6 -2 l 4
I got <x1,x2, x3, x4, x5> = <-5 -7r -5s, 2+ 4r + 3s, 4 - 6r +2s, r, s > as the answer.
Part 2: Give the simplest solution for the system (i.e. where all the free variables are set to zero), and find the
particular solution which fits the given description/s, if it is possible to do so.
x1 = −3 and x3 = 12.
I believe that the answer to Part 1 is correct.
For Part 2, setting x4 = r = 0 and x5 = s = 0, we get x1 = -5 - 7(0) - 5(0) = -5, x2 = 2 + 4(0) + 3(0) = 2, and
x3 = 4 - 6(0) + 2(0) = 4.
So, a solution is < -5, 2, 4, 0, 0 >
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