# Matrices

• Mar 5th 2007, 02:09 PM
buckaroobill
Matrices
So I got this problem in which I had to determine whether a vector [1,2,5,-1] was in the row space of the following matrix:

First row: 2, -1, 0, 3
2nd row: 7, -1, 5, 8

I know that in order to do this problem, you have to set up the matrix:

2 7 |1
-1 -1 | 2
0 5 | 5
3 8 | - 1

I row reduced it using row-reduced echelon form, and I was wondering if I got a correct answer (my graphic calculator doesn"t work for some reason when I create matrices in which the number of rows exceed the number of columns so I couldn't use it to check my work).

Using Type I (what you use to turn a pivot entry to 1) and Type II (what you use to turn a number into 0, which means taking the negative of the desired entry, multiplying it by the pivot row, and adding the row with the desired entry) row operations, I finally wound up with the matrix:
1 0 | 5/3
0 1 | -1/3
0 0 | 20/3
0 0 | -10/3
• Mar 5th 2007, 04:40 PM
ThePerfectHacker
Quote:

Originally Posted by buckaroobill
So I got this problem in which I had to determine whether a vector [1,2,5,-1] was in the row space of the following matrix:

First row: 2, -1, 0, 3
2nd row: 7, -1, 5, 8

You need to show whether there exists A and B such that,
A[2 -1 0 3]+B[7 -1 5 8]=[1 2 5 -1]
That just leads to linear systems.
• Mar 5th 2007, 05:13 PM
buckaroobill
Okay, thanks. But don't you find A and B by first finding the row-reduced echelon form of the matrix (which is what I did in my earlier post)?
• Mar 5th 2007, 05:24 PM
ThePerfectHacker
Quote:

Originally Posted by buckaroobill
Okay, thanks. But don't you find A and B by first finding the row-reduced echelon form of the matrix (which is what I did in my earlier post)?

I have been thinking and I have another way of doing this.

First note that the two given vectors are linearly independent (why?). Thus, they form a real vector space of dimension 2.

Now, [1,2,5,-1] is in the row space of two two vectors if and only if the row space of all three is the same.

Meaning, show that,
[1 2 5 -1]
[2 -1 0 3]
[7 -1 5 8]
Has rank = 2.
• Mar 5th 2007, 06:31 PM
buckaroobill
okay, so i realized there was an error in calculating my matrix in my first post.

when i put the matrix
{2, 7 | 1}
{-1, -1 | 2}
{0, 5 | 5}
{3, 8 | -1}

in row reduced echelon form, I got

{1, 0 | 3/2}
{0, 1 | 3/5}
{0, 0 | 2 }
{0, 0 | -1}

Therefore, the rank of this matrix is 2 because there are two rows with nonzero pivot entries. But how does one go about finding the A and the B for linear combination?
• Mar 5th 2007, 07:36 PM
ThePerfectHacker
Quote:

Originally Posted by buckaroobill
okay, so i realized there was an error in calculating my matrix in my first post.

when i put the matrix
{2, 7 | 1}
{-1, -1 | 2}
{0, 5 | 5}
{3, 8 | -1}

in row reduced echelon form, I got

{1, 0 | 3/2}
{0, 1 | 3/5}
{0, 0 | 2 }
{0, 0 | -1}

Therefore, the rank of this matrix is 2 because there are two rows with nonzero pivot entries. But how does one go about finding the A and the B for linear combination?

Since you where working with the coloum spaces you can see that that actual vectors are the corresponding colums in the original matrix.

By looking note that the third coloum is 3/5 the second and 3/2 the first.

That means those constants express that vector in terms of those ones (in the corresponding matrix).
• Mar 5th 2007, 08:19 PM
buckaroobill
Okay, so A is equal to 3/2 and B is equal to 3/5?

If this is true, then shouldn't 3/2{2, -1, 0, 3) + 3/5 (7, -1, 5, 8) yield the the vector (1, 2, 5, -1)?

Because when I do this out, I get a different answer.
• Mar 5th 2007, 08:48 PM
qpmathelp
Quote:

Originally Posted by buckaroobill
Okay, so A is equal to 3/2 and B is equal to 3/5?

If this is true, then shouldn't 3/2{2, -1, 0, 3) + 3/5 (7, -1, 5, 8) yield the the vector (1, 2, 5, -1)?

Because when I do this out, I get a different answer.

1 0 -3

0 1 1

0 0 0

0 0 0

A= -3

B = 1
• Mar 5th 2007, 08:58 PM
buckaroobill
Ah, got it. I messed up the matrix.

Thanks for the help!