The following equations have infinitely many solutions. 8 x + 9 y + 4 z = -58 -5

Question

The following equations have infinitely many solutions.
8 x + 9 y + 4 z = -58 -5 x - 5 y - 3 z = 37 2 x - 4 y + 6 z = -22 Give the right hand side of the vector form of the general solution, using a parameter such as s or t. (Any lowercase letter will do as a parameter, so long as it is not x, y or z.) For example, for the equations
x = y + 1, y = z + 1, z = x - 2 one correct answer is
[x, y, z] = [0, -1, -2] + t [1, 1, 1] and you would enter the right hand side of this equation in the space provided:
[0, -1, -2] + t * [1, 1, 1] or, if you prefer:
[t, t - 1, t - 2]
[x, y, z] = For example, for the equations
x = y + 1, y = z + 1, z = x - 2 one correct answer is
[x, y, z] = [0, -1, -2] + t [1, 1, 1] and you would enter the right hand side of this equation in the space provided:
[0, -1, -2] + t * [1, 1, 1] or, if you prefer:
[t, t - 1, t - 2] 8 x + 9 y + 4 z = -58 -5 x - 5 y - 3 z = 37 2 x - 4 y + 6 z = -22

Explanation / Answer

First, we must row reduce our matrix: 8   9    4    -58 -5 -5   -3    37 2 -4    6    -22 to obtain: 5   0   7   -43 0 5 -4     6 0   0   0     0 One way to obtain this would be the following sequence of elementary row operations: 1/2 R3 -> R3 R1 <--> R3 5R1 + R2 -> R2 -8R1 + R3 -> R3 -1/3 R2 -> R2 1/5 R3 -> R3 R3 - R2 -> R3 5R1 + 2R2 -> R1. Next we find the free variables. A free variable will not have a pivot in its column. Here, the only free variable is z. We can assign any variable to z, let's assign t. Then, from the second row, we see 5y - 4z = 6. Plugging in z=t, and solving this for y gives y = 6/5 + 4/5 t. The first row says 5x + 7z = -43. Plugging in z=t, and solving for x yields x = -43/5 - 7/5 t. This means our solution is [x, y, z] = [-43/5 - 7/5 t, 6/5 + 4/5 t, t]

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