2. Define null space and column space of a matrix. $$ \left.\qquad \begin{array}{rrrr}2 & 4 & -2 & 1 \ -2 & -5 & 7 & 3 \ 3 & 7 & -8 & 6\end{array} ight], \vec{u}=\left[\begin{array}{r}3 \ -2 \ -1 \ 0\end{array} ight], \vec{v}\left[\begin{array}{r}3 \ -1 \ 3\end{array} ight] $$ a. Determine if $$ \vec{u} $$ is in NulA. Could $$ \vec{u} $$ be in ColA? b. Determine if $$ \vec{v} $$ is in ColA. Could $$ \vec{v} $$ be in NulA?

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2. Define null space and column space of a matrix. $$ \left.\qquad \begin{array}{rrrr}2 & 4 & -2 & 1 \ -2 & -5 & 7 & 3 \ 3 & 7 & -8 & 6\end{array}ight], \vec{u}=\left[\begin{array}{r}3 \ -2 \ -1 \ 0\end{array}ight], \vec{v}\left[\begin{array}{r}3 \ -1 \ 3\end{array}ight] $$ a. Determine if $$ \vec{u} $$ is in NulA. Could $$ \vec{u} $$ be in ColA? b. Determine if $$ \vec{v} $$ is in ColA. Could $$ \vec{v} $$ be in NulA?

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a. To determine if $$ \vec{u} $$ is in the null space (NulA) of the matrix $$ A $$, we need to check if $$ A\vec{u} = \vec{0} $$. If it is, then $$ \vec{u} $$ is in the null space.

Let's calculate $$ A\vec{u} $$:

$$
A\vec{u} = \left.\qquad \begin{array}{rrrr}
2 & 4 & -2 & 1 \
-2 & -5 & 7 & 3 \
3 & 7 & -8 & 6
\end{array}\right] \cdot \left[\begin{array}{r}
3 \
-2 \
-1 \
0
\end{array}\right]
$$

$$
= \left[\begin{array}{r}
2(3) + 4(-2) - 2(-1) + 1(0) \
-2(3) - 5(-2) + 7(-1) + 3(0) \
3(3) + 7(-2) - 8(-1) + 6(0)
\end{array}\right]
$$

$$
= \left[\begin{array}{r}
6 - 8 + 2 + 0 \
-6 + 10 - 7 + 0 \
9 - 14 + 8 + 0
\end{array}\right]
$$

$$
= \left[\begin{array}{r}
0 \
-3 \
3
\end{array}\right]
$$

Since $$ A\vec{u} \neq \vec{0} $$, $$ \vec{u} $$ is not in the null space of $$ A $$.

To determine if $$ \vec{u} $$ could be in the column space (ColA) of $$ A $$, we need to check if $$ \vec{u} $$ can be written as a linear combination of the columns of $$ A $$. Since $$ \vec{u} $$ is not the zero vector and $$ A\vec{u} \neq \vec{0} $$, $$ \vec{u} $$ cannot be in the column space of $$ A $$.

b. To determine if $$ \vec{v} $$ is in the column space (ColA) of the matrix $$ A $$, we need to check if $$ \vec{v} $$ can be written as a linear combination of the columns of $$ A $$. If it is, then $$ \vec{v} $$ is in the column space of $$ A $$.

Let's calculate $$ A\vec{v} $$:

$$
A\vec{v} = \left.\qquad \begin{array}{rrrr}
2 & 4 & -2 & 1 \
-2 & -5 & 7 & 3 \
3 & 7 & -8 & 6
\end{array}\right] \cdot \left[\begin{array}{r}
3 \
-1 \
3 \
0
\end{array}\right]
$$

$$
= \left[\begin{array}{r}
2(3) + 4(-1) - 2(3) + 1(0) \
-2(3) - 5(-1) + 7(3) + 3(0) \
3(3) + 7(-1) - 8(3) + 6(0)
\end{array}\right]
$$

$$
= \left[\begin{array}{r}
6 - 4 - 6 + 0 \
-6 + 5 + 21 + 0 \
9 - 7 - 24 + 0
\end{array}\right]
$$

$$
= \left[\begin{array}{r}
-4 \
20 \
-16
\end{array}\right]
$$

Since $$ A\vec{v} \neq \vec{0} $$, $$ \vec{v} $$ is not in the column space of $$ A $$.

To determine if $$ \vec{v} $$ could be in the null space (NulA) of $$ A $$, we need to check if $$ A\vec{v} = \vec{0} $$. Since $$ A\vec{v} \neq \vec{0} $$, $$ \vec{v} $$ is not in the null space of $$ A $$.
a. $$ \vec{u} $$ is not in NulA and cannot be in ColA. b. $$ \vec{v} $$ is not in ColA and cannot be in NulA.

Define null space and column space of a matrix.

a. Determine if is in NulA. Could be in ColA?
b. Determine if is in ColA. Could be in NulA?

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Define null space and column space of a matrix. a. Determine if is in NulA. Could be in ColA? b. Determine if is in ColA. Could be in NulA?