1. ## Linear Systems

This is to check some of my work, and to get some additional help.

I will denote a matrix as I would in Maple A = ([1,2],[3,4],[4,5]), for example, will mean a 3x2 matrix, with the first row being [1,2].

1.) Given that
Matrix A = ([1,3],[2,1],[7,2]) and that
vector b = ([1], [2], [3])

a.) Is A*x = b consistent or inconsistent?

I row reduced it to echelon form, and got:

([1,3,1],[0,-5,0],[0,0,-4])

Therefore, since there is a pivot in the augmented column, the solution is inconsistent, that is, has no solutions.

b.) Draw a sketch of the vectors in R^3 that you'll illustrate your answer in part a. Note: this doesn't have to be a 'to scale' graph, instead, just a general picture of what's happening, with a paragraph or two. (HINT: think in terms of the span of columns of A).

MY THOUGHTS:

Inconsistent? Graph exists?

c.) Now, draw a sketch of the lines in R^2 that illustrates the answer in part a. And since since you'll be drawing this in R^2, this should be easier to graph and should be done 'more to scale'. (HINT: Think in terms of orig. system of lin. equations.)

MY THOUGHTS:

Again, if it's inconsistent, how are you able to draw a sketch of part a? I was thinking along the lines of there being a free variable, that is the R generated, in R^3, but this isn't the case.

2.) Let g be the transformation from R^3 to R^3 defined as:
g : R^3 -> R^2 with
vector ([x_1],[x_2],[x_3]) |-> matrix ([x_1 + x_2 + x_3],[x_1 - 2])

Is g a linear transformation? If it is, show why it is. If it is not, show why it is not.

MY THOUGHTS:

I believe this involves some of the principles T(u + v) = T(u) + T(v) or something to that extent, though I'm not too sure how to go about this.

2. Let me show you how I did it,
The matrix,
A is 3 rows 2 colums.
Matrix (vector) b is 2 rows 1 colums
Thus,
X is 1 row 2 coloms.
That means,
X=[x,y]
For some numbers x and y.

But,
AX=([x +3y],[2x+y],[7x+2y])
Thus, by equality,
x+3y=1
2x+y=2
7x+2y=3
Solving the first two equations we have,
y=0 and x=1
But,
That does not solve the third equations.
There there is no such coefficient matrix X.

3. Indeed; I think row reducing is faster though . But I see what you did. Any idea how to graph them, and maybe how to do the linear transformation. Time to view the definitions and try BS my way through it, haha.

4. Originally Posted by ThePerfectHacker
Let me show you how I did it,
The matrix,
A is 3 rows 2 colums.
Matrix (vector) b is 2 rows 1 colums
Thus,
X is 1 row 2 coloms.
That means,
X=[x,y]
For some numbers x and y.

But,
AX=([x +3y],[2x+y],[7x+2y])
Thus, by equality,
x+3y=1
2x+y=2
7x+2y=3
Solving the first two equations we have,
y=0 and x=1
But,
That does not solve the third equations.
There there is no such coefficient matrix X.

"Matrix (vector) b is 2 rows 1 colums"; I think you mean 3 rows, 1 column.

5. I'm confused on how to graph this. The echelon form of the matrix is a 3x3 matrix with:

([1,3,1],[0,-5,0],[0,0,-4]);

Since there is a pivot in every row (and consequently in every column), don't the rows of A span b, and therefore SHOULD be consistent, which I just proved it was not (as did Perfect). I can't even say whether its one to one, onto or linearly independent (which if it was lin. ind. would mean its onto).

6. The power of Maple!

I figured it out, thanks though.