# Thread: [SOLVED] Solving if a matrix is infinite

1. ## [SOLVED] Solving if a matrix is infinite

x + y + z = 4
2x + 3y + z = 8
3x + (3-p)y + 2z = 13-(p^2)

I've reduced the system of equations to:

x + y + z = 4
y - z = 0
(p+1)y = (p^2) - 1

It then says:

"For each of p = 1 and p = -1 indicate how many solutions there are to the system of equations. I there is a unique solution, give that solution. If there are infinitely many solutions give the resulting unique solution when
(i) z = 1
(ii) z = -1

I've found that p=1 give unique and p = -1 give infinite.

PS. My elimination is fine I have the answers.
(i) z = 1, y = 1, x = 6
(ii) z = -1, y = -1, x = 6

I just don't know how to solve if it is infinite.

2. Originally Posted by Evales
x + y + z = 4
2x + 3y + z = 8
3x + (3-p)y + 2z = 13-(p^2)

I've reduced the system of equations to:

x + y + z = 4
y - z = 0
(p+1)y = (p^2) - 1

It then says:

"For each of p = 1 and p = -1 indicate how many solutions there are to the system of equations. I there is a unique solution, give that solution. If there are infinitely many solutions give the resulting unique solution when
(i) z = 1
(ii) z = -1

I've found that p=1 give unique and p = -1 give infinite.

PS. My elimination is fine I have the answers.
(i) z = 1, y = 1, x = 6
(ii) z = -1, y = -1, x = 6

I just don't know how to solve if it is infinite.
If p=-1

The last equation turns into 0=0 That is true for all values of y.

so let y=t

subbing into the equation above it you get
$t-z=0 \iff z=t$

subbing both of these back into the first equation we get

$x+t+t=4 \iff x=-2t+4$ then the general solution is

$(-2t+4,t,t) \\\ t \in \mathbb{R}$

I hope this helps.

3. Yup tis all good thanks!