*** Simultaneous Equations 1?

Question

i) for which values of k does the following set of equations have:
a) no solution
b) many solutions
c) one solution?

x + ky = 5
kx + y = 5

ii) choose a value for k to illustrate each of the above in a sketch. what would this look like?

try using matrices!!!!!!

Answer 1

x + ky = 5
kx + y = 5

First, let's put both of these in slope-intercept form,
y = mx + b.

x + ky = 5 implies ky = -x + 5, which implies y = (-1/k)x + (5/k)

kx + y = 5 implies y = -kx + 5.

y = (-1/k)x + (5/k)
y = -kx + 5

a) In order for this set of equations to have no solutions, they must be parallel lines with a distinct y-intercept. In the first equation, the slope is equal to -1/k. In the second equation, the slope is equal to -k. Parallel lines have the same slope, so equating the slopes, we get

-1/k = -k. Solve for k.
1 = k^2; therefore, k = +/- 1.

To obtain a distinct y-intercept, note that in the first equation, the y-intercept is equal to 5/k, and in the second equation, the y-intercept is equal to 5. We want these values NOT to be equal; therefore,

5/k = 5 implies k = 1, so we want k NOT to equal 1. Reject
k = 1 as a solution, and so our final answer that makes the set of equations have no solution would be k = -1. That is

x - y = 5
-x + y = 5

Will have no solution.

b) In order to have infinitely many solutions, our set of equations must have the same slope AND the same y-intercept (i.e. they have to be the same line). We've just solved for equating the slopes and equating the y-intercepts in the previous part; the difference is we WANT k to equal 1.
Therefore, k = 1 gives many solutions; i.e.

x + y = 5
x + y = 5

has infinitely many solutions.

c) In order to obtain one solution, the slopes must NOT be equal. Since the slopes are equal when k = +/- 1, it follows that we obtain a unique solution for all other values.
As long as k is not equal to 1 and k is not equal to -1, we have one solution.

******
Matrices? I wasn't aware of your math level. In that case, the augmented matrix is represented by this:

[1 k | 5]
[k 1 | 5]

Put the four entries in reduced row-echelon form.

R2 -> R2 - k(R1)

[1 k | 5]
[0 (1 - k^2) | 5 - 5k]

In order for this system to have no solution, we must have
1 - k^2 = 0 and 5 - 5k ≠ 0.

1 - k^2 = 0 when k = {-1, 1}.

When k = -1, 5 - 5k = 5 - 5(-1) = 10, which is ≠ 0.
Therefore, k = -1 yields no solution.

To have many solutions, 1 - k^2 = 0 and 5 - 5k = 0.
When k = 1, 5 - 5k = 5 - 5(1) = 0.
Therefore, k = 1 yields infinitely many solutions.

One solution is obtained from any other value of k.

Answer 2

C and 1