1. **State the problem**: Given matrix $$A=\begin{bmatrix}1 & 2 \\ 1 & -2\end{bmatrix}$$ and vector $$b=\begin{bmatrix}4 \\ 0\end{bmatrix},$$ we want to express $b$ as a linear combination of $A$'s columns, then find solutions to $Ax=b$, and analyze solution uniqueness. 2. **Part (a)**: Write $b$ as a linear combination of the columns $a_1$ and $a_2$ of $A$. Columns are $$a_1 = \begin{bmatrix}1 \\ 1\end{bmatrix}, \quad a_2 = \begin{bmatrix}2 \\ -2\end{bmatrix}.$$ We look for scalars $\alpha, \beta$ such that $$b = \alpha a_1 + \beta a_2.$$ The problem states $b = 2a_1 + 1a_2$, check: $$2a_1 + 1a_2 = 2\begin{bmatrix}1 \\ 1\end{bmatrix} + \begin{bmatrix}2 \\ -2\end{bmatrix} = \begin{bmatrix}2 \\ 2\end{bmatrix} + \begin{bmatrix}2 \\ -2\end{bmatrix} = \begin{bmatrix}4 \\ 0\end{bmatrix} = b.$$ This verifies the linear combination. 3. **Part (b)**: Use this to solve $Ax=b$. We want $x = \begin{bmatrix}x_1 \\ x_2\end{bmatrix}$ such that $$Ax = x_1 a_1 + x_2 a_2 = b.$$ From part (a), $b = 2a_1 + 1a_2$, so $$x_1=2, \quad x_2=1.$$ Hence, a solution is $$x = \begin{bmatrix}2 \\ 1\end{bmatrix}.$$ 4. **Part (c)**: Does the system have other solutions? Since $A$ is square and its columns are linearly independent (determinant $$\det(A) = 1\times(-2) - 2\times 1 = -2 - 2 = -4 \neq 0$$), the system has a unique solution. So, no other solutions. 5. **Part (d)**: Changing the right-hand side to $c$, how many solutions? Because $A$ is invertible (determinant nonzero), for any vector $c$ in $\mathbb{R}^2$, the system $Ax=c$ has exactly one solution. **Final answers:** (a) $b = 2a_1 + 1a_2$ (b) $x = \begin{bmatrix}2 \\ 1\end{bmatrix}$ (c) No other solutions (d) Always one solution for any vector $c$