Answer: Choice B) Infinitely many solutions
- one solution: x = 8, y = -7/2, z = 0
- another solution: x = -12, y = 13/2, z = 10
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Explanation:
Here's the starting original augmented matrix.
[tex]\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\-4 & 0 & -8 & -32\\\end{array}\right][/tex]
We'll multiply everything in row 3 (abbreviated R3) by the value -1/4 or -0.25, which will make that -4 in the first column turn into a 1.
We use this notation to indicate what's going on: [tex](-1/4)*R3 \to R3[/tex]
That notation says "multiply everything in R3 by -1/4, then replace the old R3 with the new corresponding values".
So we have this next step:
[tex]\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\1 & 0 & 2 & 8\\\end{array}\right]\begin{array}{l} \ \\\ \\(-1/4)*R3 \to R3\\\end{array}[/tex]
Notice that the new R3 is perfectly identical to R1.
So we can subtract rows R1 and R3, and replace R3 with the result of nothing but 0's
[tex]\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l} \ \\\ \\R3-R1 \to R3\\\end{array}[/tex]
Whenever you get an entire row of 0's, it always means there are infinitely many solutions.
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Now let's handle the second row. That 5 needs to turn into a 0. We can multiply R1 by 5, and subtract that from R2.
So we need to compute 5*R1-R2 and have that replace R2.
[tex]\left[\begin{array}{ccc|c} 1 & 0 & 2 & 8\\0 & 1 & -1 & -7/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l} \ \\5*R1-R2 \to R2\ \\\ \\\end{array}[/tex]
Notice that in the third column of R2, we have 9-5*2 = 9-10 = -1. So we have -1 replace the 9. In the fourth column of R2, we have 73/2 - 5*8 = -7/2. So the -7/2 replaces the 73/2.
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At this point, the augmented matrix is in RREF form. RREF stands for Reduced Row Echelon Form. It seems a bit odd that the "F" of "RREF" stands for "form" even though we say "form" right after "RREF", but I digress.
Because the matrix is in RREF form, this means R1 and R2 lead to these equations:
[tex]R1 : 1x+0y+2z = 8\\ R2: 0z+1y-1z = -7/2[/tex]
which simplify to
[tex]R1: x+2z = 8\\R2: y-z = -7/2[/tex]
Let's get the z terms to each side like so:
[tex]x+2z = 8\\x = -2z+8\\\text{ and }\\y-z = -7/2\\y = z-7/2\\[/tex]
Therefore, all of the solutions are of the form [tex](x,y,z) = (-2z+8, z-7/2, z)[/tex] where z is any real number.
If z is allowed to be any real number, then we can simply pick any number we want to replace it. We consider z to be the "free variable", in that it's free to be whatever it wants. The values of x and y will depend on what we pick for z.
So the concept of "infinitely many solutions" doesn't exactly mean we can pick just any triple for x,y,z (admittedly it would be nice to randomly pick any 3 numbers off the top of my head and be done right away). Instead, we can pick anything we want for z, and whatever we picked, will directly determine x and y. The x and y are locked into place so to speak.
Let's say we picked z = 0.
That would lead to...
[tex]x = -2z+8\\x = -2(0)+8\\x = 8\\\text{ and }\\y = z-7/2\\y = 0-7/2\\y = -7/2\\[/tex]
So z = 0 would lead to x = 8 and y = -7/2
Rearranging the items in alphabetical order gets us:
x = 8, y = -7/2, z = 0
We have one solution of (x,y,z) = (8, -7/2, 0)
Now let's say we picked z = 10
[tex]x = -2z+8\\x = -2(10)+8\\x = -12\\\text{ and }\\y = z-7/2\\y = 10-7/2\\y = 13/2\\[/tex]
So we have x = -12, y = -13/2, z = 10
Another solution is (x,y,z) = (-12, 13/2, 10)
There's nothing special about z = 0 or z = 10. You can pick any two real numbers you want for z. Just be sure to recalculate the x and y values of course.
To verify each solution, you'll need to plug them back into the original equations formed by the original augmented matrix. After simplifying, you should get the same thing on both sides.
