Suppose we have the matrix $A = \begin{bmatrix} 4 &6 \\ -8 &-12\\ \end{bmatrix} $ and the vector $b = \begin{bmatrix} 3\\ -6 \end{bmatrix} $. So I am tying to determine whether b lies in the column space of A. My answer for this problem is yes since the row 2 of the matrix and the vector is a multiple of -2 giving us infinitly many answers since it is parallel. First, is my answer correct? If so, what I am confused with is with 3x3 matrix. For the following matrix, $A = \begin{bmatrix} 2 &-6 &-6\\ 4 &2 &-8\\ 6 &-2 &-2\\ \end{bmatrix} $ and the following vector $b = \begin{bmatrix} -4 \\ -14\\ 4 \\ \end{bmatrix} $ , can someone show me how you decide whether b lies in the column space of A? For the above matrix, I made the problem up myself. So feel free to demonstrate you point in any matrix you feel like. What I am having a bit of trouble at the moment is grasping the mechanics of how in general this is determined. Any inutitive explanation would be much appreciated.
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$\begingroup$You're correct that $b$ in the first case is a multiple of the second column of your first matrix: $1/2$ Column 2 = b. So essentially $b = 0\cdot \begin{pmatrix} 4 \\ -8 \end{pmatrix} + \frac 12 \begin{pmatrix} 6 \\ -12\end{pmatrix}$. Indeed, we see that any vector of the form $\langle t, -2t\rangle $ is in the column space of the matrix.
Hint: Solve the augmented Matrix $[A \mid b]$ to see if a solution exists for the constants $c_1, c_2, c_3$. If so, you will know that $b$ is in the column space of $A$.
$$b = c_1\begin{pmatrix} 2 \\ 4 \\ 6 \\ \end{pmatrix} + c_2\begin{pmatrix} -6 \\ 2 \\ -2 \\ \end{pmatrix} + c_3\begin{pmatrix}-6 \\ -8 \\ -2 \\ \end{pmatrix}$$
$$[A\mid b] = \left[\begin{array}{rrr|r} 2 &-6 &-6 &-4 \\ 4 &2 &-8 &-14\\ 6 &-2 &-2 &4 \end{array}\right] $$
$\endgroup$ 6 $\begingroup$If $b$ is in Col$A = \text{Col}\begin{bmatrix} 2 &-6 &-6\\ 4 &2 &-8\\ 6 &-2 &-2\\ \end{bmatrix} $
then $b= c_1\begin{bmatrix} 2 \\ 4 \\ 6 \\ \end{bmatrix} + c_2\begin{bmatrix} -6 \\ 2 \\ -2 \\ \end{bmatrix} + c_3\begin{bmatrix} -6 \\ -8 \\ -2 \\ \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ c_3\\ \end{bmatrix}\begin{bmatrix} 2 &-6 &-6\\ 4 &2 &-8\\ 6 &-2 &-2\\ \end{bmatrix} $
For some $c_1, c_2,$ and $c_3$ so being in the column space is the same as having a solution to the system $Ax=b$
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