Question

Consider the following system of equation:

x + y − z =2,
x + 2y + z =3,
x + y + (a^2−5)z =a,

where a ∈ R. Find all values of a for which the resulting linear system has

(a) no solution,
(b) a unique solution,
(c) infinitely many solutions.

Answer 1

OMG it's Tori Black... LOL

Answer 2

For a unique solution, the determinant of these equation needs to be non zero. For an infinite number of solutions, we must reduce to 2 equations with 3 unknowns. For no solutions, one or more of these planes would be parallel.

Answer 3

do you know about determinants?

Answer 4

no

Answer 5

Ok then.

For a unique solution, all rows should be independent. Choose an "a" that makes this true.

Answer 6

i solved the system and got a unique solution of \[\frac{ 1 }{ 2 }(1\pm \sqrt(17) \]

Answer 7

so how do i find no solutions and infinitely many

Answer 8

You have 4 unknowns, x,y, z and a but only 3 equations. There are no unique solutions until "a" is fixed. For certain "a" any of these three possibilities are possible (unique, infinite,none).

Answer 9

yeah i found "a" for a unique solution

Answer 10

could you solve it?

Answer 11

yes

Answer 12

If a=2, then last row would be equal to the 1st row, right? What doe that mean?

Answer 13

What does that imply about the independence of these rows? Are they dependent or independent? When a=2.

Answer 14

...or -2.

Answer 15

correct.

Answer 16

On the other hand, a = 2 is not as catastrophic as a = -2.

Answer 17

With a = 2, you find that there is infinitely many solutions. So how do we find one "a" that has no solution. We need to choose an a that makes the last equation "parallel" to any of the other two equations.

Answer 18

so any value other than the unique solution value and a=+-2 should ake it parallel?

Answer 19

when a=2 there are NOT infinitely many solutions

Answer 20

My line of thinking is this.

Each of the coefficients of this equation represents a vector that is perpendicular to the planes each equation represents. If two equations have the same perpendicular vector then they are also parallel to each other. Does this make sense?

Answer 21

When we used a=2, we found the last row equal to the first row, right? But what if a = -2?

Answer 22

yes sorry made an error

Answer 23

With a=-2 we have two planes that are parallel, but not the same plane. This means...?

Answer 24

This means we found a plane parallel to another but doesn't intersect it. That is, no solution.

Answer 25

what is the value of "a" when there are no solutions?

Answer 26

a=-2 is one possibility that I see

Answer 27

but there could be many values?
how would i right that?

Answer 28

a=2 is infinite solution

Answer 29

something like: all values excluding a=-1 and a=2 will give no solutions?

Answer 30

Look's like a=-1 would give a unique solution

Answer 31

and values that would give no solutions?

Answer 32

a=-2 would give no solutions. I don't see another possibility for no solutions

Answer 33

What I'm looking for is an a that makes the last row a linear combination of one of the 1st two but the constants are not equal

Answer 34

\[\left[\begin{matrix}1 & 0 & 0 & \dfrac{a+5}{a+2} \\ 0 & 1 & 0 & \dfrac{a}{a+2}? \\ 0 & 0 & 1 & \dfrac{1}{a+2}\end{matrix}\right]\]

Only a = -2 is that big a problem.

Answer 35

thanks @tkhunny

Answer 36

Oops, I forgot to delete a question mark and I forgot to say that was Reduced Row Echelon form.

Darn that I.E. I had to switch to Firefox just to see what I was typing!

Answer 37

lol. In any case, without some tools in linear algebra this is probably the best I can do to describe. I hope this helped