Linear algebra help!

[prokrastinera]

Hi! I need help with this question:

Determine all the values of the constant a so that the following equation system lacks solutions:

x + y + z = 0
x +2y + az = 1 + 2a
x + ay + 2z = -1 + a

 

[tkhunny]

Have you considered the Determinant of the Coefficients?

 

[Steven G]

Reduce the matrix. Have you tried that? To receive help, you have to show us your work.

 

[pka]

HINT \(\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&a \\ 1&a&2 \end{array}} \right| = 2a - {a^2}\)

 

[prokrastinera]

I dont know where to start reducinng the matrix...

 

[tkhunny]

I dont know where to start reducinng the matrix...

Ah. Paralysis. That's never good.

The real answer is ANYWHERE! Just jump it. It's a matrix, you won't hurt it.

What would you do if it were this: [MATH] \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&5 \\ 1&6&2 \end{array}} \right][/MATH] ?

Of course, you may have specific requirements or demands that would suggest you proceed in a certain way or from a certain place. Please note that if this is the case, it didn't do you any good. Seriously, just start!

 

[Steven G]

I was going to say exactly pka said. I think that it is worth repeating. You can start reducing from anywhere! There really does not have to be any formal rules to reduce a matrix other than the basic operations. You can interchange any two rows. You can multiply any row by any non-zero number and you can add a multiply of one row to another row. Just be sure in the end that the matrix is in the form that you want. Treat the letters just like numbers.

 

[Shiloh]

I dont know where to start reducinng the matrix...

The equations are
x+ y+ z= 0
x+ 2y+ az= 1+ 2a
x+ ay + 2z= -1+ a

The first thing I notice is that the coefficients of x are all 1.
Subtracting the first equation from the second eliminates x:
y+ (a- 1)z= 1+ 2a

Subtracting the first equation from the third also eliminates x:
(a- 1)y+ z= -1

Multiply the previous equation by a- 1:
(a- 1)y+ (a- 1)^2z= (a- 1)(1+ 2a)
and subtract (a- 1)y+ z= -1 from that
(a- 1)^2z- z= (a- 1)(1+ 2a)+ 1= a^2+ a

z= (a^2+ a)/(a- 1)^2

That has a solution for all a except a= 1