Linear Algebra: Find the value/values of "a" for the system of equations when...

[hobomath]

Linear Algebra: Find the value/values of "a" for the system of equations when...

Hello, I've been stuck on this one since yesterday, so I appreciate the help!

For this system:
ax + y = 1
9x + ay = 3

where a ϵ R


1. find the row echelon form of the augmented matrix:
a 1 | 1
9 a | 3

L1/a -> L1

1 (1/a) | (1/a)
9 a | 3

L2 - (9*L1) -> L2

1 (1/a) | (1/a)
0 (a-(9/a)) | (3-(9/a))

**I'm not sure if what I did is good. It looks wrong and complicated :/
Anyway moving on... this is where I don't understand.

2. Find the value or values of "a" for this system when
a) it has an unique solution.
b) it has an infinite solution.
c) it has 0 solution.

Now I have 2 new equations.

x + (1/a)y = 1/a
(a-(9/a))y = (3-(9/a))

y = (3-(9/a)) / (a-(9/a))

x = 1/a - (1/a)y
x = 1/a - ( ( (1/a)*(3-(9/a)) ) / (a-(9/a)) )


I thought that a system could only have either a unique solution, an infinite of solutions OR no solution.
But in this case, I believe that "a" is much like a free variable, meaning it can take any value so this system has infinite solutions.

Could someone explain me how I'm suppose to solve this problem?
I did all the other problems where I have to solve the Systems of Linear Equations.
However, this one is confusing me because it's not really asking to solve the system, but to find the value of "a" so it respects the system.
Thank you so much !

 

[Ishuda]

First of all, treat the system as any other but just let a be a "given variable constant" so that it acts like a constant which is represented by the letter a. Now there may be particular values a can't have, so as you manipulate the augmented matrix, remember that in the back of your mind.

Start with the augmented matrix
\(\displaystyle \begin{pmatrix} a& 1& /\, 1\\ 9& a& /\, 3 \end{pmatrix}\)
Replace row 2 with a times the second row minus 9 times the first row. Now this sort of assumes a isn't zero since one normally doesn't multiply by zero in situations like this, so we might check that when we get finished. We now have
\(\displaystyle \begin{pmatrix} a& 1& /\, 1\\ 0& a^2-9& /\, 3a-9 \end{pmatrix}\)
and we can solve for x and y:
y = \(\displaystyle \frac{3a-9}{a^2-9}\, =\, \frac{3 (a-3)}{(a+3)(a-3)}\, =\, \frac{3}{a+3}\)
x = \(\displaystyle \frac{1-y}{a}\, =\, \frac{a + 3 - 3}{a (a+3)}\, =\, \frac{1}{a+3}\)

Now, given the above, how would you answer your three questions
2. Find the value or values of "a" for this system when
a) it has an unique solution.
b) it has an infinite solution.
c) it has 0 solution.

 

[hobomath]

First of all, treat the system as any other but just let a be a "given variable constant" so that it acts like a constant which is represented by the letter a. Now there may be particular values a can't have, so as you manipulate the augmented matrix, remember that in the back of your mind.

Start with the augmented matrix
\(\displaystyle \begin{pmatrix} a& 1& /\, 1\\ 9& a& /\, 3 \end{pmatrix}\)
Replace row 2 with a times the second row minus 9 times the first row. Now this sort of assumes a isn't zero since one normally doesn't multiply by zero in situations like this, so we might check that when we get finished. We now have
\(\displaystyle \begin{pmatrix} a& 1& /\, 1\\ 0& a^2-9& /\, 3a-9 \end{pmatrix}\)
and we can solve for x and y:
y = \(\displaystyle \frac{3a-9}{a^2-9}\, =\, \frac{3 (a-3)}{(a+3)(a-3)}\, =\, \frac{3}{a+3}\)
x = \(\displaystyle \frac{1-y}{a}\, =\, \frac{a + 3 - 3}{a (a+3)}\, =\, \frac{1}{a+3}\)

Now, given the above, how would you answer your three questions
2. Find the value or values of "a" for this system when
a) it has an unique solution.
b) it has an infinite solution.
c) it has 0 solution.

Hello, thank you for the reply.
I never thought about multiplying a row with "a"... such a clever trick :mrgreen: Ty.

a) I believe finding a unique solution is like finding the determinant of the matrix.
\(\displaystyle ad-bc ≠ 0 \)
\(\displaystyle a *(a^2-9) - (1 * 0) ≠ 0\)
\(\displaystyle a^2 - 9 ≠ 0\)
\(\displaystyle a^2 ≠ 9\)
\(\displaystyle a ≠ 3 \)

So if "a" is not equal to 3, then the system has a unique solution!

Now to find if it has an infinite or no solution... then a = 3.
Imma take a look on the new equations.

\(\displaystyle ax + y = 1 \)
\(\displaystyle (1/(3+3))*3 + (3/(3+3)) = 1 \)
\(\displaystyle 1=1 \)

\(\displaystyle (a^2-9)y = 3a-9 \)
\(\displaystyle (3^2-9)*(3/(3+3)) = 3*3 -9 \)
\(\displaystyle 0 = 0 \)


Edit: Ok ty I figured it out.