1. ## Linear solutions

For which values of a will the following system have no solutions? one solution? or infinite solutions?

System
x + 2y - 3z = 4
3x - y + 5z = 2
4x + y + (a^2 - 14)z = a + 2

For which values of a will the following system have no solutions? one solution? or infinite solutions?

System
x + 2y - 3z = 4
3x - y + 5z = 2
4x + y + (a^2 - 14)z = a + 2
Multiply equation 1 by -3 and add it to equation 2 to get

$-7y+14z=-10$

Multiply eqaution 1 by -4 and add it to equation 3 to get

$-7y+(a^2-2)z=a-14$

Now multiply the first of these equations by -1 and add it to the other to get

$(a^2-16)z=a-4$

Now we factor the left hand side to get

$(a-4)(a+4)z=a-4$

Now lets check some values of a

Case I: a=4

we get 0=0 so the system is consistant dependant (has an infinite number of solutions.)

Case II: a=-4

We get 0=-8 So the system in inconsistant(has no solutions)

Case III: $a \ne 4 \mbox{ or } a \ne -4$

we get

$(a-4)(a+4)z=a-4 \iff z =\frac{1}{a+4}$

The system has exactly one solution and it is given above.
So the system is consistant independant.

I hope this clear it up.

Note this could have been done with a matrix and putting the matrix into eschelon form.