# Thread: Help in solving a following matrix

1. ## Help in solving a following matrix

Hello,

I have a question that's asking me if a set of four 3x1 matrices are a spanning set of R^3 (the entire 3-D space)..

So I have to figure out how to represent them in a linear combination with 4 different variables that I call alpha1 - alpha4. If the system's inconsistent, then I know that the set of 4 matrices is not a spanning set of R^3. I'm stuck at the following matrix:

1 0 0 0 -- a + b - c
0 1 0 2 -- b
0 0 1 1 -- -b + c

Where $[a,b,c]^T$ represents the idea that any 3x1 matrix can be written as a linear combination of the 4 matrices that were given in the problem.

My concern is that I'm stuck at the matrix I wrote above, and am not sure at how to continue. It's not in reduced row echelon form, and I'm not sure how to get it in that kind of form.

2. Originally Posted by Lord Darkin
Hello,

I have a question that's asking me if a set of four 3x1 matrices are a spanning set of R^3 (the entire 3-D space)..

So I have to figure out how to represent them in a linear combination with 4 different variables that I call alpha1 - alpha4. If the system's inconsistent, then I know that the set of 4 matrices is not a spanning set of R^3. I'm stuck at the following matrix:

1 0 0 0 -- a + b - c
0 1 0 2 -- b
0 0 1 1 -- -b + c

Where $[a,b,c]^T$ represents the idea that any 3x1 matrix can be written as a linear combination of the 4 matrices that were given in the problem.

My concern is that I'm stuck at the matrix I wrote above, and am not sure at how to continue. It's not in reduced row echelon form, and I'm not sure how to get it in that kind of form.
Hint: the set is dependant what is the span of

$\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

3. Oh, do you mean that in the second row, I can subtract 2 from both sides to leave alpha2 = b-2, and subtract 1 from both sides of the 3rd row, leading to alpha3 = -1 -b + c?

4. Originally Posted by Lord Darkin
Oh, do you mean that in the second row, I can subtract 2 from both sides to leave alpha2 = b-2, and subtract 1 from both sides of the 3rd row, leading to alpha3 = -1 -b + c?
Not quite your have 4 vectors in a 3 dimensional space.

For now just ignore the 4th vector what I am saying is that any point

$\begin{bmatrix} a \\ b \\ c \end{bmatrix}= a\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}+b\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} +c \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}+0\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}$

The forth vector is dependent on the other 3 and is not needed.