Not sure how to go about this?
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
$\displaystyle -7y+14z=-10$
Multiply eqaution 1 by -4 and add it to equation 3 to get
$\displaystyle -7y+(a^2-2)z=a-14$
Now multiply the first of these equations by -1 and add it to the other to get
$\displaystyle (a^2-16)z=a-4$
Now we factor the left hand side to get
$\displaystyle (a-4)(a+4)z=a-4$
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: $\displaystyle a \ne 4 \mbox{ or } a \ne -4$
we get
$\displaystyle (a-4)(a+4)z=a-4 \iff z =\frac{1}{a+4}$
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.