Linear algebra help!

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prokrastinera

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May 26, 2020
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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
 
Reduce the matrix. Have you tried that? To receive help, you have to show us your work.
 
HINT \(\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&a \\ 1&a&2 \end{array}} \right| = 2a - {a^2}\)
 
prokrastinera said:
I dont know where to start reducinng the matrix...
Click to expand...
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!
 
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.
 
prokrastinera said:
I dont know where to start reducinng the matrix...
Click to expand...
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
 
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