Matrices & some matlab?

Question

Hi guys and thanks for the help in advance :)

1.(a) Find out when the following system of linear equations has no solution, a unique solution, or infinitely many solutions.

x + -3z = -3
2x + ky - z = -2
x + 2y + kz = 1

{*the sheet gave me instructions to use matlab to reduce it to this form:
[1 , 0 , -3 , -3 ]
[0 , 1 , 5/k , 4/k ]
[0 , 0 , k+3-10/k , 4-8/k ]
what is the answer?*}
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{*The question sheet then tells me to use the >>rref(aug) command which will give me this output:
[1 , 0 , 0 , -3*(k+1)/(k+5) ]
[0 , 1 , 0 , 4/(k+5) ]
[0 , 0 , 1 , 4/(k+5) ] *}
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(b) What do you notice about the difference in the answers?
Try to write down what you think has happened in Mat lab.

Answer 1

Look at the last line: [0 , 0 , k+3-10/k , 4-8/k ]
We need to solve for k when k + 3 - 10/k = 0; multiply through by k to get
k^2 + 3k - 10 = 0 => (k+5)(k-2) = 0
When k = 2 the last line is [0, 0, 0, 0]; when k = -5 it is [0, 0, 0, 28/5].
The last line isn't defined when k = 0, so we have to treat this case separately: in this case the system of equations is x + -3z = -3, 2x - z = -2, x + 2y = 1 which has a unique solution (the first two equations are l.i. and will give us a unique solution for x and z, and we can get y from the third equation).
So when k = 2 there are infinitely many solutions; for k = -5 there are no solutions; and for any other k there is a unique solution.

After using rref to get the reduced row-echelon form, you can see that there is a single answer for all k ≠ -5. In particular, k = 2 gives a unique solution in this system, whereas in the original system there were infinitely many solutions.

Matlab has presumably divided (4-8/k) by (k + 3 - 10/k) and cancelled out the common factor of (k-2).