Suppose that you are given the following system of linear equations

Question

[variables are x1, x2, x3 and x4]

x1 + 3x2 + 9x3 + 2x4 =5
x1 + 3x3 - 4x4 =5
x2 + 2x3 + 3x4 =-1
-2x1 + 3x2 + 5x4 =-1

Let b be the vector such that B = [A b] , and let a1, a2, a3 and a4 be the columns of A.

a) What is the general solution for the system?
i just want to double check my answers, please point out any mistakes.

the solution i got is
x1=1-3x3
x2=2-2x3
x3 is free
x4=-1

Let m be the number of linearly independent columns of A, let k be the number of parameters (free variables) in (a), and let n be the total number of columns in A. In our example above, n = 4.

b) Notice that m + k = n. Do you suppose that this relationship will be true for all systems of linear equations? Why or why not?

-i don't know how to approach this question, anybody mind explaining the answer to me? thanks.

x3 is zero, so if one puts that into what you got for x1, x2, you are right. I don't agree with your x3 is "free". It is zero. x3 is not a free parameter.

http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi

I don't understand the b) either.

but the row reduced matrix is the following:

1 0 3 0 1
0 1 2 0 2
0 0 0 1 -1
0 0 0 0 0

with pivots in columns 1, 2 and 4, does that make x3 a free variable?

Yes, I found the same solution. x3 is free so you have a "one dimensional solution space". The rank of the matrix is the number of independent rows (or columns, row rank and column rank are always the same) and this is 3.

what exactly is independent rows?

so as you said, m=3 and k=1
m+k=4, so this is true.

will this be true for all systems of linear equations (as the question asks)? How would you actually explain or proof it?

For b) see here:

http://en.wikipedia.org/wiki/Rank-nullity_theorem

If you write the equation as:

A x = b

if x and y are two solutions:

A x = b

A y = b

then

A (x-y) = 0

So, the difference is an element of the null space and all solutions are of the form:

x = n + p

Where p is one particular solution of:

A x = b

and n is an element of the null space.

The number of free parameters in te solution will thus be equal to the dimension of the null space. The equation m + k = n is thus a direct consequence of the Rank-nullity theorem.

Sketch of the proof here:

http://www.jiskha.com/display.cgi?id=1179363311.1179366947

Thanks, but i haven't learn Rank-nullity theorem or null space yet. We just got introduced to linearly independent and linearly dependent. Is there a way of explaining it with the limited knowledge i have?

Ok,

let's just start with the equation:

A x = b (1)

where A is an N by N matrix, and b is an N dimensional vector. Suppose that a solution exists, let's denote it by x.

Now let's consider the equation:

A y = 0 (2)

If a1 a2, a3, are the columns of A (considered as vectors) and y1, y2, y3, etc. are the components of the vector y, then this equation is simply:

a1 y1 + a2 y2 + a3 y3 + ...=0

Clearly if all the columns are independent then (by definition) you only have the trivial solution y = 0.

If the number of independent columns is k, then that means that you'll have a solution for y that depends on N - k independent parameters. I guess you just have to check in your lecture notes how the concept of "number of independent columns" or "rank" was explained to you.

You can now construct the general solution of (1) by adding to x a solution to (2):

A (x + y) = A x + A y = b + 0 = b

The number of free parameters in y is N - k. So all we have to show now is that there aren't any more degrees of freedom apart from the choice of y, or put differently, that all solutions are of the form x + y for one particular fixed x. But that's trivial. Suppose that z is any solution of (1) and that z is not equal to x. Then the difference of z and x is mapped by A to zero, so z - x would satisfy Eq. (2).

Answer 1

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