To determine whether the vector $\mathbf{b}$ can be written as a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$, we need to solve the equation:

$c_1 \mathbf{a}_1 + c_2 \mathbf{a}_2 = \mathbf{b}$

Where: $\mathbf{a}_1 = \begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix}, \quad \mathbf{a}_2 = \begin{pmatrix} -2 \\ -3 \\ 6 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} -5 \\ 6 \\ 3 \end{pmatrix}$

This gives the following system of equations:

$c_1(-1) + c_2(-2) = -5$ $c_1(3) + c_2(-3) = 6$ $c_1(-1) + c_2(6) = 3$

We now solve this system to determine if $c_1$ and $c_2$ exist.

Equation 1:

$-c_1 - 2c_2 = -5$

This simplifies to: $c_1 + 2c_2 = 5 \quad \text{(Eq 1)}$

Equation 2:

$3c_1 - 3c_2 = 6$

Dividing the entire equation by 3: $c_1 - c_2 = 2 \quad \text{(Eq 2)}$

Equation 3:

$-c_1 + 6c_2 = 3$

This simplifies to: $c_1 - 6c_2 = -3 \quad \text{(Eq 3)}$

Now, we can solve these equations step by step.

From Eq 2, we have $c_1 = c_2 + 2$.

Substitute $c_1 = c_2 + 2$ into Eq 1: $(c_2 + 2) + 2c_2 = 5$ $c_2 + 2 + 2c_2 = 5$ $3c_2 + 2 = 5$ $3c_2 = 3$ $c_2 = 1$

Now, substitute $c_2 = 1$ back into Eq 2: $c_1 - 1 = 2$ $c_1 = 3$

Check Solution with Eq 3:

Substitute $c_1 = 3$ and $c_2 = 1$ into Eq 3: $3 - 6(1) = -3$ $3 - 6 = -3$ This is true, so the solution is valid.

Thus, $\mathbf{b}$ can be written as a linear combination of $\mathbf{a}_1$ and $\mathbf{a}_2$: $\mathbf{b} = 3\mathbf{a}_1 + 1\mathbf{a}_2$

Final Answer:

$c_1 = 3, \quad c_2 = 1$

Do you need more details or have any questions?

Here are 5 relative questions to expand on this concept:

1. What is the geometric interpretation of linear combinations and vector spans?
2. How can you determine if three vectors are linearly dependent or independent?
3. What does it mean for a vector to be in the span of other vectors?
4. How does the row-reduction method help in solving systems of linear equations like this?
5. Can the cross product be used to find relationships between vectors in 3D space?

Tip: Always check your solution by substituting the found coefficients back into the original equations to ensure consistency.