Express the column matrix b as a linear combination of the columns of A. (Use A1, A2, and A3 respectively for the columns of A.)
A = 

1	−3	−1	

2	3	1
,    b = 

−7	

7


We are tasked with expressing the vector \( \mathbf{b} \) as a linear combination of the columns of matrix \( A \). Let's begin by setting up the equation:

\[
A = \begin{pmatrix}
1 & -3 & -1 \\
2 & 3 & 1
\end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix}
-7 \\
7
\end{pmatrix}
\]

Let the columns of \( A \) be denoted as \( A_1, A_2, \) and \( A_3 \):
\[
A_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, \quad A_2 = \begin{pmatrix} -3 \\ 3 \end{pmatrix}, \quad A_3 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}
\]

We want to express \( \mathbf{b} \) as a linear combination of \( A_1, A_2, \) and \( A_3 \). This means we want to find scalars \( x_1, x_2, x_3 \) such that:

\[
x_1 A_1 + x_2 A_2 + x_3 A_3 = \mathbf{b}
\]

Substitute the column vectors:

\[
x_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} + x_2 \begin{pmatrix} -3 \\ 3 \end{pmatrix} + x_3 \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -7 \\ 7 \end{pmatrix}
\]

This gives the following system of equations:

\[
x_1 - 3x_2 - x_3 = -7 \quad \text{(equation 1)}
\]
\[
2x_1 + 3x_2 + x_3 = 7 \quad \text{(equation 2)}
\]

Now, let's solve this system of equations.

### Step 1: Solve equation 1 for \( x_1 \)
\[
x_1 = 3x_2 + x_3 - 7
\]

### Step 2: Substitute \( x_1 \) into equation 2
\[
2(3x_2 + x_3 - 7) + 3x_2 + x_3 = 7
\]
Simplify:
\[
6x_2 + 2x_3 - 14 + 3x_2 + x_3 = 7
\]
\[
9x_2 + 3x_3 - 14 = 7
\]
\[
9x_2 + 3x_3 = 21
\]
\[
3x_2 + x_3 = 7 \quad \text{(equation 3)}
\]

### Step 3: Solve equation 3 for \( x_3 \)
\[
x_3 = 7 - 3x_2
\]

### Step 4: Substitute \( x_3 \) into the expression for \( x_1 \)
\[
x_1 = 3x_2 + (7 - 3x_2) - 7
\]
\[
x_1 = 0
\]

### Step 5: Substitute \( x_1 = 0 \) into equation 1
\[
0 - 3x_2 - (7 - 3x_2) = -7
\]
Simplify:
\[
-3x_2 - 7 + 3x_2 = -7
\]
\[
-7 = -7
\]

Thus, \( x_2 \) can be any real number, meaning we have infinitely many solutions. Therefore, we express the vector \( \mathbf{b} \) as a linear combination with one free parameter. For example, taking \( x_2 = 1 \), we get:

\[
x_3 = 7 - 3(1) = 4
\]
\[
x_1 = 0
\]

Thus, one possible solution is:

\[
\mathbf{b} = 0A_1 + 1A_2 + 4A_3
\]

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. What does it mean for a vector to be a linear combination of others?
2. How do we interpret the solution space of a system with infinite solutions geometrically?
3. Can we find other sets of solutions for this problem by changing the free parameter?
4. What is the role of matrix rank in determining if a vector can be expressed as a linear combination?
5. How do we express the general solution for the linear combination in this case?

**Tip:** When dealing with systems of linear equations, always check for consistency to determine the solution type (unique, none, or infinite).

Express the column matrix b as a linear combination of the columns of A. (Use A1, A2, and A3 respectively for the columns of A.) A = 
1 −3 −1 
2 3 1, b = −7 7

Learn how to express matrix b as a linear combination of the columns of matrix A. This problem demonstrates key linear algebra concepts, including solving a system of linear equations, understanding linear combinations, and finding solutions with free parameters. Suitable for undergraduate students studying linear algebra.

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