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Let $A=\begin{bmatrix} 1 & 1 & 1 & 1 \\ 3 & 2 & 4 & 1 \\ 1 & 5 & 4 & 2 \\ \end{bmatrix}$. By attempting to solve the system $Ax=b$ in terms of $b=\begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ \end{bmatrix}$ either show that $col A=R^3$, or find a vecotr in $R^3$ that is NOT in col $A$.

When attempting to solve this I got $$ \left[ \begin{array}{cccc|c} 1&1&1&1&b_1\\ 0&-1&1&-2&b_2-3b_1\\ 0&0&7&-7&(b_3-b_1)-4(b_2-3b_1)\\ \end{array} \right] $$ Since there are three pivots that means this matrix has at least three columns that are not redundant and therefore it fills up all of $R^3$ so col $A=R^3$ correct?

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Yes, this is correct. Notice that you didn't even need to do this with the vector $b$ on the right side, you could claim that the column space is $\mathbb{R}^3$ by the number of pivot positions.

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