LinAlg: Determine the value(s) of h such that the matrix....

[Velouria555]

Homework Statement
Determine the values of h such that the matrix is the augmented matrix of a consistent linear system.

1 4 -2
3 h -6

The attempt at a solution
The answer I got differs from the back of the book.

I tried solving it by adding R1(4) to R2

1 3 -2
-4 h 8

becomes

1 3 -2
0 h-12 0

I thought that h had to equal 12 so that (0)x₁ + (0)x₂ = 0, but the book says the answer is all h. I don't know what I'm missing

 

[Mark44]

Homework Statement
Determine the values of h such that the matrix is the augmented matrix of a consistent linear system.

1 4 -2
3 h -6

The attempt at a solution
The answer I got differs from the back of the book.

I tried solving it by adding R1(4) to R2

1 3 -2
-4 h 8

I don't understand what you did to get this matrix. Also, I don't understand your notation -- R1(4) -- does that mean you are adding 4 times R1 to R2? If so, R1 shouldn't change. How did the row 1, column 2 entry change from 4 to 3?

becomes

1 3 -2
0 h-12 0

I thought that h had to equal 12 so that (0)x₁ + (0)x₂ = 0, but the book says the answer is all h. I don't know what I'm missing

If h = 12 the two equations are dependent, meaning that they both represent the same line, so there are an infinite number of solutions. For this problem, an inconsistent system would be two parallel lines (no intersection).

 

[SammyS]

Homework Statement
Determine the values of h such that the matrix is the augmented matrix of a consistent linear system.

1 4 -2
3 h -6

The attempt at a solution
The answer I got differs from the back of the book.

I tried solving it by adding R1(4) to R2

1 3 -2
-4 h 8

becomes

1 3 -2
0 h-12 0

I thought that h had to equal 12 so that (0)x₁ + (0)x₂ = 0, but the book says the answer is all h. I don't know what I'm missing

$\begin{bmatrix} 1 & 3 & | & -2\\ 0 & (h-12) & | & 0 \end{bmatrix}$

Depending upon the value of h, this system may be consistent and dependent, or it may be consistent and independent.

(0)x₁ + (h - 12)x₂ = 0

simplifies to: (h - 12)x₂ = 0 .

Either
h - 12 = 0, leaving x₂ arbitrary.

or
x₂ = 0 .

 

[Velouria555]

Oh man, sorry people. I was confusing two different problems as I typed this one out... been a long, fartbrain day.

What I meant to say is:

1 4 -2
3 h -6

R₁(-3)+R₂=

1 4 -2
0 h-12 0

After seeing both responses, I think I might be missing something very fundamental and obvious about this problem. My thought process is that to make this system consistent, R₂ is (0)x₁+(0)x₂=0 so (h-12) must equal 0 and the only way that can occur is if h = 12. If, for example, h = 1, then R₂ implies that 11x₂=0. Actually... I guess that makes sense because no matter what h is, when you solve for x₂, it will equal 0/h which is always 0. Is this correct?

 

[SammyS]

Oh man, sorry people. I was confusing two different problems as I typed this one out... been a long, fartbrain day.

What I meant to say is:

1 4 -2
3 h -6

R₁(-3)+R₂=

1 4 -2
0 h-12 0

After seeing both responses, I think I might be missing something very fundamental and obvious about this problem. My thought process is that to make this system consistent, R₂ is (0)x₁+(0)x₂=0 so (h-12) must equal 0 and the only way that can occur is if h = 12. If, for example, h = 1, then R₂ implies that 11x₂=0. Actually... I guess that makes sense because no matter what h is, when you solve for x₂, it will equal 0/h which is always 0. Is this correct?

If h ≠ 12, then we must have x₂ = 0. The solution for x₁ may also be determined.

If h = 12, then x₂ can be any value. The value of x₁ will the depend upon the value chosen for x₂ .

 

[Mark44]

1 4 -2
0 h-12 0

After seeing both responses, I think I might be missing something very fundamental and obvious about this problem. My thought process is that to make this system consistent, R₂ is (0)x₁+(0)x₂=0 so (h-12) must equal 0 and the only way that can occur is if h = 12. If, for example, h = 1, then R₂ implies that 11x₂=0. Actually... I guess that makes sense because no matter what h is, when you solve for x₂, it will equal 0/h which is always 0. Is this correct?

To expand on what SammyS said, your final augmented matrix represents this system of equations:
1x + 4y = -2
... (h - 12)y = 0

For simplicity I'm going to use x and y instead of subscripts on x.
Obviously (??) you can solve for y in the second equation, which yields y = 0 if h ≠ 12, and y is arbitrary if h = 12. In the latter case (h = 12), x's value will depend on y.

To summarize, if h ≠ 12, the single solution is x = 0 and y = 0. If h = 12 there are an infinite number of solutions, as both equations represent the same line. Every solution of the system is a point on the line x + 4y = -2.

 

[Ray Vickson]

Oh man, sorry people. I was confusing two different problems as I typed this one out... been a long, fartbrain day.

What I meant to say is:

1 4 -2
3 h -6

R₁(-3)+R₂=

1 4 -2
0 h-12 0

After seeing both responses, I think I might be missing something very fundamental and obvious about this problem. My thought process is that to make this system consistent, R₂ is (0)x₁+(0)x₂=0 so (h-12) must equal 0 and the only way that can occur is if h = 12. If, for example, h = 1, then R₂ implies that 11x₂=0. Actually... I guess that makes sense because no matter what h is, when you solve for x₂, it will equal 0/h which is always 0. Is this correct?

What is your answer to the original problem (which asked for values of h such that the system is consistent)? Is it consistent if $h \neq 12$? Is it consistent if $h = 12$?

 

[Velouria555]

Got it! For some reason, I was under the impression that the bottom row could not be [0 * 0] (as * represents a non-zero number), but that is totally wrong because such a row would still make a system consistent and is quite common, actually since * would be the pivot position. And [0 0 0] is also consistent and essentially means that the row is 'free' as my textbook refers to it.

Thanks everyone, for your patience and great explanations!

 

[SammyS]

@Velouria555 ,

We neglected welcoming you to PF ... so Welcome to PF !