4. Given that the matrices $$ A $$ and $$ B $$ are row equivalent and vectors $$ \mathbf{u}, \mathbf{w} $$ as follows: $$ A:=\left[\begin{array}{lllll}1 & 0 & -3 & 0 & 0 \ 2 & 1 & 2 & 1 & 0 \ 1 & 1 & 5 & 1 & 0 \ 2 & 1 & 2 & 1 & 1 \ 0 & 0 & 0 & 0 & 1\end{array}\right] \text { and } B:=\left[\begin{array}{ccccc}1 & 0 & -3 & 0 & 0 \ 0 & 1 & 8 & 1 & 0 \ 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0\end{array}\right], \mathbf{u}:=\left[\begin{array}{lll}3 & 2 & 7\end{array} 21\right] \text {. } $$ Let $$ \mathbf{v}_{i} $$ be the $$ i $$ th row of $$ A, \mathbf{w}_{i} $$ be the $$ i $$ th column of $$ A $$ and $$ \mathbf{u}_{i} $$ be the $$ i $$ th column of $$ B $$ $$ i=1,2,3,4,5 $$. (a) Choose a set of vectors from $$ \left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}, \mathbf{v}_{4}, \mathbf{v}_{5}\right\} $$ that would form a basis for row $$ (A) $$. (b) Given that $$ \mathbf{u} \in \operatorname{row}(A) $$. Write $$ \mathbf{u} $$ as a linear combination of the basis found in (a). (c) Choose a set of vectors from $$ \left\{\mathbf{w}_{1}, \mathbf{w}_{2}, \mathbf{w}_{3}, \mathbf{w}_{4}, \mathbf{w}_{5}\right\} $$ that would form a basis for $$ \operatorname{col}(A) $$.

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y a Basis for Row Space of $$ A $$To find a basis for the row space of $$ A $$, we can look at the non-zero rows of the row echelon form of $$ A $$. The matrix $$ B $$ is already in row echelon form, and its non-zero rows will form a basis for the row space of $$ A $$.

The non-zero rows of $$ B $$ are:
1. $$ \mathbf{b}_1 = [1, 0, -3, 0, 0] $$
2. $$ \mathbf{b}_2 = [0, 1, 8, 1, 0] $$
3. $$ \mathbf{b}_3 = [0, 0, 0, 0, 1] $$

Thus, the set of vectors that form a basis for $$ \text{row}(A) $$ is:
$$
\{\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3\}
\end{step_content>
( \mathbf{u} $$ as a Linear Combination of the BasisWe need to express the vector $$ \mathbf{u} = [3, 2, 7, 21] $$ as a linear combination of the basis vectors found in Step 1.

Let:
\[
\mathbf{u} = c_1 \mathbf{b}_1 + c_2 \mathbf{b}_2 + c_3 \mathbf{b}_3
$$
This gives us the following system of equations:
$$
c_1(1, 0, -3, 0, 0) + c_2(0, 1, 8, 1, 0) + c_3(0, 0, 0, 0, 1) = (3, 2, 7, 21)
$$

This leads to the following equations:
1. $$ c_1 = 3 $$
2. $$ c_2 = 2 $$
3. $$ c_3 = 21 $$

Thus, we can express $$ \mathbf{u} $$ as:
\[
\mathbf{u} = 3 \mathbf{b}_1 + 2 \mathbf{b}_2 + 21 \mathbf{b}_3
\end{step_content>
y a Basis for Column Space of $$ A $$To find a basis for the column space of $$ A $$, we can look at the pivot columns of $$ B $$. The pivot columns correspond to the first, second, and fifth columns of $$ A $$.

Thus, the set of vectors that form a basis for $$ \text{col}(A) $$ is:
\[
\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_5\}
\end{step_content>
(a) A basis for $$ \text{row}(A) $$ is $$ \{\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3\} $$. (b) $$ \mathbf{u} = 3 \mathbf{b}_1 + 2 \mathbf{b}_2 + 21 \mathbf{b}_3 $$. (c) A basis for $$ \text{col}(A) $$ is $$ \{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_5\} $$.

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