Consider the matrix $A=\begin{bmatrix}5&5&1&1\\-3&-2&-21&5\\4&6&4&1\end{bmatrix}$. Which set is a basis for the column space of $A$? (1) $\left\{\begin{bmatrix}1\\5\\1\end{bmatrix},\begin{bmatrix}5\\-3\\4\end{bmatrix},\begin{bmatrix}5\\-2\\6\end{bmatrix}\right\}$ (2) $\left\{\begin{bmatrix}1\\5\\1\end{bmatrix},\begin{bmatrix}5\\-3\\4\end{bmatrix}\right\}$ (3) $\left\{\begin{bmatrix}5\\-3\\4\end{bmatrix},\begin{bmatrix}5\\-2\\6\end{bmatrix},\begin{bmatrix}10\\-7\\6\end{bmatrix}\right\}$ (4) $\left\{\begin{bmatrix}1\\5\\1\end{bmatrix},\begin{bmatrix}5\\-2\\6\end{bmatrix}\right\}$

What we are trying to find

A basis for the column space is a smallest set of columns of $A$ that (1) spans all columns of $A$ and (2) is linearly independent.

Name the columns

Let $$ \begin{aligned} c_1&=\begin{bmatrix}5\\-3\\4\end{bmatrix},\quad c_2=\begin{bmatrix}5\\-2\\6\end{bmatrix},\quad c_3=\begin{bmatrix}1\\-21\\4\end{bmatrix},\quad c_4=\begin{bmatrix}1\\5\\1\end{bmatrix}. \end{aligned} $$

Check that three columns are independent (rank is 3)

Take the $3\times 3$ matrix with columns $c_1, c_2, c_4$: $$ M=\begin{bmatrix}5&5&1\\-3&-2&5\\4&6&1\end{bmatrix}. $$ Compute its determinant: $$ \det(M)=5\bigl((-2)(1)-5\cdot 6\bigr)-5\bigl((-3)(1)-5\cdot 4\bigr)+1\bigl((-3)(6)-(-2)(4)\bigr)=-55\neq 0. $$ Since $\det(M)\neq 0$, the vectors $c_1, c_2, c_4$ are linearly independent. Because the columns live in $\mathbb{R}^3$, having 3 independent columns means the column space has dimension 3.

Show the remaining column is in their span

Solve for a linear combination of $c_1, c_2, c_4$ that equals $c_3$. One valid relation is: $$ c_3=-c_1+2c_2-4c_4. $$ So $c_3$ adds no new direction beyond $c_1, c_2, c_4$.

Match to the multiple-choice options

Option (1) is exactly $\{c_4, c_1, c_2\}$, which is a linearly independent set of columns that spans $\operatorname{Col}(A)$. Therefore, option (1) is a basis for the column space.