Vector Span Cb90C7

1. **State the problem:** We want to determine if $\vec{w}$ can be written as a linear combination of $\vec{v}_1$, $\vec{v}_2$, and $\vec{v}_3$. That means finding scalars $a$, $b$, and $c$ such that: $$a\vec{v}_1 + b\vec{v}_2 + c\vec{v}_3 = \vec{w}$$ 2. **Write the equation in component form:** $$a \begin{bmatrix}1 \\ 1 \\ 2 \\ 2\end{bmatrix} + b \begin{bmatrix}1 \\ 2 \\ 1 \\ 2\end{bmatrix} + c \begin{bmatrix}1 \\ 0 \\ 0 \\ 1\end{bmatrix} = \begin{bmatrix}-1 \\ 0 \\ 2 \\ -3\end{bmatrix}$$ This gives the system of equations: 1. $a + b + c = -1$ 2. $a + 2b + 0c = 0$ 3. $2a + b + 0c = 2$ 4. $2a + 2b + c = -3$ 3. **Solve the system:** From equation (2): $$a + 2b = 0 \implies a = -2b$$ Substitute $a = -2b$ into equation (1): $$-2b + b + c = -1 \implies -b + c = -1 \implies c = -1 + b$$ Substitute $a = -2b$ into equation (3): $$2(-2b) + b = 2 \implies -4b + b = 2 \implies -3b = 2 \implies b = -\frac{2}{3}$$ Then: $$a = -2b = -2 \times \left(-\frac{2}{3}\right) = \frac{4}{3}$$ $$c = -1 + b = -1 - \frac{2}{3} = -\frac{5}{3}$$ Check equation (4) with these values: $$2a + 2b + c = 2 \times \frac{4}{3} + 2 \times \left(-\frac{2}{3}\right) + \left(-\frac{5}{3}\right) = \frac{8}{3} - \frac{4}{3} - \frac{5}{3} = \frac{8 - 4 - 5}{3} = -\frac{1}{3} \neq -3$$ 4. **Interpretation:** The last equation is not satisfied, so there is no solution for $a$, $b$, and $c$ that satisfies all four equations. 5. **Conclusion:** The vector $\vec{w}$ is **not** in the span of $\vec{v}_1$, $\vec{v}_2$, and $\vec{v}_3$. It cannot be expressed as a linear combination of these vectors.