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We are given two vectors:

\[
a = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}, \quad b = \begin{bmatrix} -1 \\ 4 \\ 2 \end{bmatrix}
\]

And we are asked to find scalars \(s\) and \(t\) such that:

\[
s \cdot a + t \cdot b = \begin{bmatrix} -8 \\ 24 \\ 12 \end{bmatrix}
\]

This results in the following system of equations, where we solve for \(s\) and \(t\):

\[
s \cdot 1 + t \cdot (-1) = -8
\]
\[
s \cdot 2 + t \cdot 4 = 24
\]
\[
s \cdot (-1) + t \cdot 2 = 12
\]

### Solving the system:

#### Equation 1:
\[
s - t = -8 \quad \text{(1)}
\]

#### Equation 2:
\[
2s + 4t = 24 \quad \text{(2)}
\]

#### Equation 3:
\[
-s + 2t = 12 \quad \text{(3)}
\]

#### Step 1: Solve equation (1) for \(s\):
\[
s = t - 8 \quad \text{(4)}
\]

#### Step 2: Substitute \(s = t - 8\) into equations (2) and (3):

For equation (2):
\[
2(t - 8) + 4t = 24
\]
\[
2t - 16 + 4t = 24
\]
\[
6t - 16 = 24
\]
\[
6t = 40
\]
\[
t = \frac{40}{6} = \frac{20}{3}
\]

For equation (3):
\[
-(t - 8) + 2t = 12
\]
\[
- t + 8 + 2t = 12
\]
\[
t + 8 = 12
\]
\[
t = 4
\]

There seems to be a contradiction in these steps, let me investigate this further. Please wait while I correct the solution.



Let a = [1, 2, -1] and b = [-1, 4, 2]. Show that there are scalars s and t such that s * a + t * b = [-8, 24, 12].

Learn how to solve for scalars s and t such that the linear combination of vectors sa + tb equals a given vector. The problem involves solving a system of linear equations and is suitable for advanced high school or college students.

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