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denote the columns of the matrix a by v1, v2 and v3. if b is a linear conbination of the coumns of A, specify a linear conbination of the columns of A that equals b. If b is not a linear conbination of the columns of A, leave the spaces blank 

To solve the problem, we need to check if the vector **b** is a linear combination of the columns of matrix **A**. Let's analyze the image and extract the necessary dataTo solve the problem, we need to check if the vector **b** is a linear combination of the columns of matrix **A**. Let's analyze the image and extract the necessary 

### Given Matrix \( A \) and Vector \( b \):
Matrix \( A \):
\[
A = \begin{pmatrix}
1 & 2 & 1 \\
0 & 1 & -1 \\
1 & 3 & 1
\end{pmatrix}
\]
Vector \( b \):
\[
b = \begin{pmatrix}
2 \\
-3 \\
0
\end{pmatrix}
\]

### Step 1: Form the augmented matrix.
We need to set up the augmented matrix \([A \mid b]\) to solve the system \(Ax = b\):
\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 2 \\
0 & 1 & -1 & -3 \\
1 & 3 & 1 & 0
\end{array}\right]
\]

### Step 2: Perform row operations to check for consistency.
To see if the system is consistent (meaning \( b \) is a linear combination of the columns of \( A \)), we perform row reduction (Gaussian elimination) on the augmented matrix:

1. Subtract row 1 from row 3:
   \[
   R3 = R3 - R1 \Rightarrow \left[\begin{array}{ccc|c}
   1 & 2 & 1 & 2 \\
   0 & 1 & -1 & -3 \\
   0 & 1 & 0 & -2
   \end{array}\right]
   \]

2. Subtract row 2 from row 3:
   \[
   R3 = R3 - R2 \Rightarrow \left[\begin{array}{ccc|c}
   1 & 2 & 1 & 2 \\
   0 & 1 & -1 & -3 \\
   0 & 0 & 1 & 1
   \end{array}\right]
   \]

This matrix is in row-echelon form.

### Step 3: Interpret the results.
The system is consistent (no contradictory row like \([0 \ 0 \ 0 \mid k]\) where \(k\) is non-zero). Therefore, \( b \) **is** a linear combination of the columns of \( A \).

### Step 4: Back-substitution to find the linear combination.
Perform back-substitution to find the solution vector \( x \):

1. From the third row: \( x_3 = 1 \)
2. From the second row: \( x_2 - x_3 = -3 \Rightarrow x_2 = -2 \)
3. From the first row: \( x_1 + 2x_2 + x_3 = 2 \Rightarrow x_1 + 2(-2) + 1 = 2 \Rightarrow x_1 = 5 \)

Thus, \( x = \begin{pmatrix} 5 \\ -2 \\ 1 \end{pmatrix} \).

### Conclusion:
The vector \( b = 2v_1 - 2v_2 + v_3 \) is a linear combination of the columns of \( A \).

Would you like further details on any part of the solution? 

Here are some related questions:
1. What are the conditions for a vector to be a linear combination of other vectors?
2. How can the row-echelon form help in solving systems of linear equations?
3. What do the terms "consistent" and "inconsistent" mean in the context of linear equations?
4. How do we interpret free variables in the context of solutions to linear systems?
5. What is the geometric interpretation of a linear combination of vectors?

**Tip:** Always check the consistency of a system of linear equations before solving to determine if a solution exists.

Explore how to determine if vector b is a linear combination of matrix A columns using Gaussian elimination and row-echelon form. Step-by-step solution with detailed matrix operations and interpretations. Suitable for college-level students and beyond.

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