linear Algebra -matricies with variables

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mlane

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Sep 6, 2005
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I am working on Matricies in a linear algebra college course.
I was assigned a homework problem and practice problems with variables.

the practice problem is
1 h 4
3 6 8

I need to put in reduced row echelon form to solve but I am having trouble. There are no examples in the book and my instructor did not discuss this type in class.
I began by multiplying -3 by row 1 + row 2 for a new row 2 which gives

1 h 4
0 -3h+6 -4 if I add them together for a new row 1 I get

1 -2h+6 0
0 -3h+6 -4

( -1R2 +R1 = R1)

1 h 4
0 -3h+6 -4
I am going in circles here.

I would greatly appreciate some help on this type of problem
I know the answer given in the book is h cannot =2. I just need help getting there.
Where should the h end up? And how can I get it there? Can h be a pivot?
 
mlane said:
I am working on Matricies in a linear algebra college course.
I was assigned a homework problem and practice problems with variables.

the practice problem is
1 h 4
3 6 8

I need to put in reduced row echelon form to solve but I am having trouble. There are no examples in the book and my instructor did not discuss this type in class.
I began by multiplying -3 by row 1 + row 2 for a new row 2 which gives

1 h 4
0 -3h+6 -4 if I add them together for a new row 1 I get

1 -2h+6 0
0 -3h+6 -4

( -1R2 +R1 = R1)

1 h 4
0 -3h+6 -4
I am going in circles here.

I would greatly appreciate some help on this type of problem
I know the answer given in the book is h cannot =2. I just need help getting there.
Where should the h end up? And how can I get it there? Can h be a pivot?
Click to expand...
A non-zero h could be a pivot, but continuing with your start:
Code: 
1     h       4
0  -3h+6     -4

1     h       4
0     1       -4/(-3h+6)       h not equal 2 to not divide by zero

1     0       4 + 4h/(-3h+6)   R1 - hR2
0     1       -4/(-3h+6)
 
Hello, mlane!

\(\displaystyle \L\begin{vmatrix}1 & h & | & 4 \\ 3 & 6 & | & 8 \end{vmatrix}\)
Click to expand...

Your first step is correct . . .

\(\displaystyle \begin{array}{ccccccc}. \\ \; \\ \; \\ \; \\ . \\ \; \\ R2-3\cdot R_1\end{array}\;
\L\begin{vmatrix}1 & h & | & 4 \\ 0 & -3h+6 & | & -4 \end{vmatrix}\)


We will not be able to avoid fractions. .We might as well introduce them now.

\(\displaystyle \begin{array}{cccccccc}. \\ \; \\ \; \\ . \\ \; \\ \; \\ . \\ R2\div(-3h+6)\end{array}\;
\L\begin{vmatrix}1 & h & | & 4 \\ 0 & 1 & | & \frac{4}{3h-6} \end{vmatrix}\)


\(\displaystyle \begin{array}{ccccccc}R1-h\cdot R2 \\ \; \\ \; \\ . \\ \; \\ \; \\ .\end{array}\;
\L\begin{vmatrix}1 & 0 & | &\frac{8h-24}{3h-6} \\ 0 & 1 & | & \frac{4}{3h-6}\end{vmatrix}\)


Edit: Too fast for me, JakeD!
 
1 h 4
3 6 8

the original equation was
x+hy=4
3x+6y=8
if h =2 then you do not have 2 equations with 2 independant variables,you have
x+2y=4
x+2y=8/3 no solution

=============================================
1 h 4
3 6 8
multiply 1 by -3 and add to 2nd

1 h 4
0 [6-3h] -4
multiply 2nd row by -h/[6-3h] and add to first

1 0 4+4/[6-3h]
0 [6-3h] -4
divide 2nd row by [6-3h]

1 0 [28-12h]/6-3h]
0 1 -4/[6-3h]

6-3h can't =0 for non trivial solution
h can't = 2 answer

for x=y then 28-12h=-4
h=3/8

please check math for algebra errors
Arthur
 
I am not sure I made this clear.
After row operations we came down to:
1 0 a/[6-3h]
0 1 b/[6-3h]

for a non trivial solution [x=y=infinity], 6-3h can't = 0 or h=2

for x to equal y set a=b
for x to equal ky set a=kb

I don't know if this will help make your algebra clearer, but I hope it does.
Arthur
 
you are more than welcome and I hope it helped some.
It is obvious from the matrix after it is reduced, that h can have any value except 2 for a non trivial solution.

I am afraid it has been 55 years since I studied matricies, so my language is poor, and you might not use the terms I do.
Arthur
 
It makes more sense now. I could see the relation but I thought I had to reduce them completely to solve. If I would not try to think so hard, maybe I would see that some solutions are simple. Anyhow, I appreciate every bit of help, sincerely.
Thanks again.
Mel
 
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