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The given matrix is $$A=\begin{bmatrix} 1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\5 & -6 & 10 & 7 \end{bmatrix}$$ Row reducing yields $$B=\begin{bmatrix} 1 & 0 & -1 & 5 \\ 0 & 1 & -\frac{5}{2} & 3 \\0 & 0 & 0 & 0 \end{bmatrix}$$ By the number of free variables and pivots the Nullity $= 2$ and Rank $=2$ respectively.

The basis of the $Col(A)=\begin{bmatrix}1 \\-1\\5 \end{bmatrix} \begin{bmatrix} -4\\2\\-6 \end{bmatrix}$

The basis of the $Nul(A)=\begin{bmatrix}1 \\-\frac{5}{2}\\1\\0 \end{bmatrix} \begin{bmatrix} -5\\-3\\0\\1 \end{bmatrix}$

Then I am given a vector $x=\begin{bmatrix} 6\\-5\\23 \end{bmatrix}$ and asked to determine if it is in the $Col(A)$, so I set up the augmented matrix and row reduce. $$\begin{bmatrix} 1 &-4 & |6 \\-1 & 2 &|-5 \\ 5 &-6 & |23 \end{bmatrix}\to^{rref} \begin{bmatrix}1 &0 & |4\\0 & 1 & |-\frac{1}{2}\\ 0&0&|0 \end{bmatrix} $$

From this I find $x=4, y=-\frac{1}{2}$ so the vector is in the column space of A. Then I am asked to determine the coordinates of $x$ relative to the basis above but how can I if the basis has 4 columns and my vector has only 3 rows?

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