Solvability Condition 2D5Cc3

1. **State the problem:** We want to find the condition on $$\vec{b}$$ such that the system $$B\vec{v} = \vec{b}$$ is solvable, where $$B = \begin{bmatrix} 0 & 2 & 1 \\ 1 & 1 & 1 \\ -2 & 0 & 1 \end{bmatrix}$$ and $$\vec{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$$. 2. **Recall the solvability condition:** A system $$A\vec{x} = \vec{b}$$ has a solution if and only if $$\vec{b}$$ lies in the column space of $$A$$. Equivalently, $$\vec{b}$$ must satisfy the consistency condition given by the rank of $$A$$ and the augmented matrix $$[A|\vec{b}]$$. 3. **Calculate the determinant of $$B$$:** $$\det(B) = 0 \cdot \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ -2 & 0 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1 \times 1 - 1 \times 0 = 1$$ $$\begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} = 1 \times 1 - 1 \times (-2) = 1 + 2 = 3$$ $$\begin{vmatrix} 1 & 1 \\ -2 & 0 \end{vmatrix} = 1 \times 0 - 1 \times (-2) = 0 + 2 = 2$$ So, $$\det(B) = 0 \times 1 - 2 \times 3 + 1 \times 2 = 0 - 6 + 2 = -4$$ 4. **Interpretation:** Since $$\det(B) = -4 \neq 0$$, matrix $$B$$ is invertible. 5. **Conclusion for part (a):** Because $$B$$ is invertible, the system $$B\vec{v} = \vec{b}$$ has a unique solution for every $$\vec{b} \in \mathbb{R}^3$$. Therefore, there is **no condition** on $$\vec{b}$$ for solvability; the system is always solvable. --- 6. **Part (b):** We need an example of a $$3 \times 3$$ matrix $$C$$ such that the system $$C\vec{v} = \vec{b}$$ has no solution or infinitely many solutions depending on $$\vec{b}$$. 7. **Key idea:** Such a matrix $$C$$ must be singular (non-invertible), i.e., $$\det(C) = 0$$, so that the system is not always solvable or has infinitely many solutions. 8. **Example matrix:** $$C = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix}$$ Note that the second row is twice the first row, so $$\det(C) = 0$$. 9. **Explanation:** - For some $$\vec{b}$$, the system $$C\vec{v} = \vec{b}$$ is consistent and has infinitely many solutions (if $$\vec{b}$$ lies in the column space of $$C$$). - For other $$\vec{b}$$, the system has no solution (if $$\vec{b}$$ is not in the column space). 10. **Summary:** - (a) Since $$\det(B) \neq 0$$, the system $$B\vec{v} = \vec{b}$$ is solvable for all $$\vec{b}$$. - (b) The matrix $$C = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix}$$ is an example where solvability depends on $$\vec{b}$$. **Final answers:** - (a) No condition on $$\vec{b}$$; system always solvable. - (b) Example matrix $$C$$ given above.