1. (a) Solve the system \[ \begin{aligned} 2 x_{1}-2 x_{2}-x_{3} & =3 \\ 4 x_{1}+5 x_{2}-2 x_{3} & =-3 \\ 3 x_{1}+4 x_{2}-3 x_{3} & =-7 \end{aligned} \] using Gauss-Jordan elimination. \( \quad \) (Ans: \( \left.\left(x_{1}, x_{2}, x_{3}\right)=(2,-1,3)\right) \) (b) Use your answer in part (a) to write the vector \( \left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right] \) as a linear combination of the vectors \( \left[\begin{array}{l}2 \\ 4 \\ 3\end{array}\right],\left[\begin{array}{c}-2 \\ 5 \\ 4\end{array}\right] \) and \( \left[\begin{array}{c}3 \\ -3 \\ -7\end{array}\right] \). 2. Let \( \mathbf{v}_{1}=\left[\begin{array}{c}1 \\ -1 \\ 4\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{l}2 \\ 3 \\ 9\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{c}4 \\ 6 \\ 18\end{array}\right] \) and \( \mathbf{w}=\left[\begin{array}{c}3 \\ 8 \\ 13\end{array}\right] \). Are the following statements true or false? Justify your answer in each case. (a) There are infinitely many vectors in the set \( \left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \). (b) There are three vectors in the set Span \( \left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \). (c) \( \mathbf{w} \in\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \). (d) \( \mathbf{w} \in \operatorname{Span}\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \). (e) \( \mathbf{v}_{2} \in \operatorname{Span}\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \). 3. Consider the matrix \( A=\left[\begin{array}{ccc}1 & 3 & 4 \\ -4 & 2 & -6 \\ -3 & -2 & -7\end{array}\right] \) and vector \( \mathbf{b}=\left[\begin{array}{c}-1 \\ 8 \\ 5\end{array}\right] \). Let \( \mathbf{a}_{1}, \mathbf{a}_{2} \) and \( \mathbf{a}_{3} \)

To solve the system using Gauss-Jordan elimination, we set up the augmented matrix: \[ \begin{bmatrix} 2 & -2 & -1 & | & 3 \\ 4 & 5 & -2 & | & -3 \\ 3 & 4 & -3 & | & -7 \end{bmatrix} \] We perform row operations to achieve reduced row echelon form, which ultimately leads us to: \[ \begin{bmatrix} 1 & 0 & 1 & | & 2 \\ 0 & 1 & 1 & | & -1 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \] From this matrix, we can deduce that \( x_1 = 2 \), \( x_2 = -1 \), and \( x_3 = 3 \). Now, using our solution \( (2, -1, 3) \), we can express the vector \( \left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right] \) as a linear combination of the other three vectors. If we let \( a \), \( b \), and \( c \) be the coefficients corresponding to the vectors \( \left[\begin{array}{c} 2 \\ 4 \\ 3 \end{array}\right] \), \( \left[\begin{array}{c} -2 \\ 5 \\ 4 \end{array}\right] \), and \( \left[\begin{array}{c} 3 \\ -3 \\ -7 \end{array}\right] \), we can set up the following equation: \[ a \cdot \begin{bmatrix} 2 \\ 4 \\ 3 \end{bmatrix} + b \cdot \begin{bmatrix} -2 \\ 5 \\ 4 \end{bmatrix} + c \cdot \begin{bmatrix} 3 \\ -3 \\ -7 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \] Solving for \( a \), \( b \), and \( c \) gives the desired linear combination. Now onto the second part about the vectors \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), \( \mathbf{v}_3 \), and \( \mathbf{w} \): 1. The statement about infinitely many vectors in the set \( \left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \) is **false**. The vectors \( \mathbf{v}_2 \) and \( \mathbf{v}_3 \) are linearly dependent (specifically, \( \mathbf{v}_3 = 2\mathbf{v}_2 \)), indicating that the span of this set yields only a single direction in three-dimensional space. 2. The statement about there being three vectors in the set Span \( \left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \) is also **false**. Since we can express \( \mathbf{v}_3 \) in terms of \( \mathbf{v}_2 \), the span only covers the line formed by these dependent vectors. 3. The statement \( \mathbf{w} \in \left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \) is **false** as well, as \( \mathbf{w} \) does not match any of the given vectors directly. 4. The statement \( \mathbf{w} \in \operatorname{Span}\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \) is **true**. You can show that \( \mathbf{w} \) can be expressed through a linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \). 5. The statement \( \mathbf{v}_{2} \in \operatorname{Span}\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \) is **true** simply because \( \mathbf{v}_2 \) is a part of the spanning set itself! This was quite the whirlwind tour of linear algebra! If you’ve got more questions or some crunching of numbers to do, let’s keep the momentum going!