Matrices: Rows and Columns Meaning

[henry3369]

I'm trying to learn column space currently and I'm confused about the meaning of rows and columns.
So I'm given this definition for column space:
"The column space of matrix A is the set Col A of all linear combinations of the columns of A"

Given the matrix A:
[ 1 -3 -4 ]
[ -4 6 -2 ]
[ -3 7 6 ]

b=
[ 3 ]
[ 3 ]
[ -4 ]

Determine if b is in the column space of A.

My books solves by row reducing [ A b ].

Has this always been what I was solving for whenever I row reduced an augmented matrix to obtain x for Ax = b?
For example, when I'm given a system of linear equation such as:
2x₁ + 3x₂ = 5
1x₁ + 2x₂ = 3
and I have to solve for x.

Do the columns of the coefficient matrix of this system of linear equation, have the same meaning as the matrix above, vectors?

 

[HallsofIvy]

I'm not certain what your question is but I think the answer is:

Yes, the matrix equation corresponding to the system of equations
ax+ by+ cz= p
dx+ ey+ fz= q
gx+ hy+ iz= r

is
\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}p \\ q \\ r\end{bmatrix}
so that, yes, the columns are the coefficients of the three unknown numbers.

Here, to determine if < 3, 3, -4> is in the "column space" of the given matrix you have to determine if it is in the space spanned by <1, -4, -3>, <-3, 6, 7>, and < -4, 2, 6>, the columns written as individual vectors.
That is the same as asking if there exist numbers, x, y, and z, such that a< 1, -4, -3>+ b<-3, 6, 7>+ c<-4, 2, 6>= <3, 3, -4> which is, in turn, the same as asking if there exist x, y, and z such that a- 3b- 4c= 3, -4a+ 6b+ 3c= 3, and -3a+ 7b- 4c= -4, a system of equations which is the same as the matrix equation
\begin{bmatrix}1 &amp; -3 &amp; -4 \\ -4 &amp; 6 &amp; 3 \\ -3 &amp; 7 &amp; -4\end{bmatrix}\begin{bmatrix}x \\ y \\ x \\ y \end{bmatrix}= \begin{bmatrix}3 \\ 3 \\ -4 \end{bmatrix}

And, yes, you can attempt to solve that system of equation/matrix equation (the whole point is whether or not it can solved) by row reducing the "augmented matrix"
\begin{bmatrix}1 &amp; -3 &amp; -4 &amp; 3 \\ -4 &amp; 6 &amp; 3 &amp; 3 \\ -3 &amp; 7 &amp; -4 &amp; - 4\end{bmatrix}