ä-Hi folks,

we are asked to show that the only values of t for which the 3 equations

x + (t - 4)y +2z = 0

2x - 6y + (t + 3)z = 0

(t - 5)x + 12y -8z = 0

have a solution in which x,y and z are not all zero are t = 1 and t = 4.

Putting the equations into a 3 x 3 matrix A, we know that if $det A \ne 0$ then we have a unique solution.

If det A = 0 then we do not. In this case A is singular and we either have no solutions (inconsistent equations) or an infinite number of solutions. i.e. a common line or 3 coincident planes.

So, I evaluate $det A \ne 0$ to get an equation in t that I can solve. This should enable me to find the values of t.

My confusion is to do with the inequality. I am not dealing with det A = 0.

We are told that (t - 1) and (t - 4) are solutions and therefore factors of the equation and when I do the calculations I end up with:

$(t - 1) (t - 1) (t - 4) \ne 0$

I would be fine with this if det A = 0, for then I would get t = 1 and t = 4 as required. But I cannot interpret what the factors not equal to zero mean.

Any advice?

Thanks