Find a basis for the column space of the matrix.
-$$B = \left[ \begin{array} { r r r r }
1 & - 2 & 5 & - 3 \
2 & - 4 & 13 & - 2 \
- 3 & 6 & - 15 & 9
\end{array} \right]$$
A)
$$
\left\{\left[\begin{array}{r}
1 \
2 \
-3
\end{array}\right],\left[\begin{array}{r}
5 \
13 \
-15
\end{array}\right]\right\}
$$
B) $$
\left\{\left[\begin{array}{l}
1 \
0 \
0
\end{array}\right],\left[\begin{array}{l}
0 \
1 \
0
\end{array}\right]\right\}
$$
C)

$$
\left\{\left[\begin{array}{r}
1 \
2 \
-3
\end{array}\right],\left[\begin{array}{r}
-2 \
-4 \
6
\end{array}\right]\right\}
$$
D)
$$
\left\{\left[\begin{array}{l}
2 \
1 \
0 \
0
\end{array}\right],\left[\begin{array}{c}
\frac{29}{3} \
0 \
-\frac{4}{3} \
1
\end{array}\right]\right\}
$$

Question

Find a basis for the column space of the matrix.
-$$B = \left[ \begin{array} { r r r r } 
1 & - 2 & 5 & - 3 \
2 & - 4 & 13 & - 2 \
- 3 & 6 & - 15 & 9
\end{array} \right]$$
A)
$$
\left\{\left[\begin{array}{r}
1 \
2 \
-3
\end{array}\right],\left[\begin{array}{r}
5 \
13 \
-15
\end{array}\right]\right\}
$$
B) $$
\left\{\left[\begin{array}{l}
1 \
0 \
0
\end{array}\right],\left[\begin{array}{l}
0 \
1 \
0
\end{array}\right]\right\}
$$
C)

$$
\left\{\left[\begin{array}{r}
1 \
2 \
-3
\end{array}\right],\left[\begin{array}{r}
-2 \
-4 \
6
\end{array}\right]\right\}
$$
D)
$$
\left\{\left[\begin{array}{l}
2 \
1 \
0 \
0
\end{array}\right],\left[\begin{array}{c}
\frac{29}{3} \
0 \
-\frac{4}{3} \
1
\end{array}\right]\right\}
$$

Quizplus · Accepted Answer

The basis for the column space of a matrix consists of the linearly independent columns of that matrix. Observing the given matrix, we can see that the second column is a multiple of the first column (-2 times the first column), and the third column can be expressed as a linear combination of the first two columns. Therefore, the first and third columns are linearly independent and span the column space of the matrix. The fourth column is also a linear combination of the first two columns. Thus, the correct basis for the column space is the set containing the first and third columns.

Find a Basis for the Column Space of the Matrix B=[1−25−32−413−2−36−159]B = \left[ \begin{array} { r r r r } 1 & - 2 & 5 & - 3 \\2 & - 4 & 13 & - 2 \\- 3 & 6 & - 15 & 9\end{array} \right]B=​12−3​−2−46​513−15​−3−29​​

Find a Basis for the Column Space of the Matrix $B = \left[ \begin{array} { r r r r } 1 & - 2 & 5 & - 3 \\2 & - 4 & 13 & - 2 \\- 3 & 6 & - 15 & 9\end{array} \right]$