We are given the solution set for an inhomogeneous linear system, where:

$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -6 \\ 8 \\ 7 \end{bmatrix} + t \cdot \begin{bmatrix} 1 \\ 5 \\ -13 \end{bmatrix} + s \cdot \begin{bmatrix} 2 \\ 3 \\ -5 \end{bmatrix}$ with $t, s \in \mathbb{R}$.

We are asked to determine a value $c$ so that: $\begin{bmatrix} 9 \\ 4 \\ c \end{bmatrix}$ becomes a solution to the associated homogeneous system.

Solution:

To find $c$, we should assume that the vector $\begin{bmatrix} 9 \\ 4 \\ c \end{bmatrix}$ can be expressed as a linear combination of the vectors $\begin{bmatrix} 1 \\ 5 \\ -13 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 3 \\ -5 \end{bmatrix}$, since these vectors span the solution space of the homogeneous system. Thus, we need to find $t$ and $s$ such that:

$\begin{bmatrix} 9 \\ 4 \\ c \end{bmatrix} = t \cdot \begin{bmatrix} 1 \\ 5 \\ -13 \end{bmatrix} + s \cdot \begin{bmatrix} 2 \\ 3 \\ -5 \end{bmatrix}$

Expanding this equation gives:

1. $9 = t + 2s$
2. $4 = 5t + 3s$
3. $c = -13t - 5s$

We can solve these equations for $t$, $s$, and ultimately $c$.

1. From the first equation, $t = 9 - 2s$.

2. Substitute $t = 9 - 2s$ into the second equation:

   $4 = 5(9 - 2s) + 3s$ $4 = 45 - 10s + 3s$ $4 = 45 - 7s$ $-41 = -7s$ $s = \frac{41}{7}$

3. Substitute $s = \frac{41}{7}$ back into $t = 9 - 2s$:

   $t = 9 - 2 \cdot \frac{41}{7}$ $t = 9 - \frac{82}{7}$ $t = \frac{63 - 82}{7}$ $t = -\frac{19}{7}$

4. Finally, substitute $t = -\frac{19}{7}$ and $s = \frac{41}{7}$ into the third equation to find $c$:

   $c = -13 \left(-\frac{19}{7}\right) - 5 \cdot \frac{41}{7}$ $c = \frac{247}{7} - \frac{205}{7}$ $c = \frac{42}{7}$ $c = 6$

Answer:

$c = 6$

Would you like more details, or do you have any questions?

Related Questions:

1. How would the solution change if the inhomogeneous term were different?
2. What is the geometric interpretation of the solution set for this system?
3. How can we check if a vector belongs to the solution space of a homogeneous system?
4. What are the properties of a homogeneous system's solution set?
5. Can we apply the same method to a higher-dimensional system?

Tip:

In homogeneous systems, solutions can often be written as linear combinations of vectors that span the null space.