*Question 6* We can compute the Smith normal form of $A$ as follows $\begin{align} &\begin{pmatrix}1&-3&0&1\\0&0&1&2\\-2&0&1&0\end{pmatrix} \stackrel{A^{14}(-1) A^{12}(3) }{ \longrightarrow} \begin{pmatrix}1&0&0&0\\0&0&1&2\\-2&0&1&2\end{pmatrix} \stackrel{S^{24}}{ \longrightarrow} \begin{pmatrix}1&0&0&0\\0&2&1&0\\-2&2&1&0\end{pmatrix} \\ &\stackrel{ A^{23}(-1) M^{2}\left( \frac{1}{2} \right) }{ \longrightarrow} \begin{pmatrix}1&0&0&0\\0&1&0&0\\-2&1&0&0\end{pmatrix} \stackrel{A_{23}(-1) A_{13}(2) }{ \longrightarrow} \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\end{pmatrix} \end{align}$ *Question 7* (a) $A$ is not square so is not invertible. (b) Row reduce $(A \mid I_{2})$ as follows: $\begin{align} & \left( \begin{array}{cc|cc} 2& 4 & 1 & 0 \\ 6 & 8 &0 &1 \end{array} \right) \stackrel{M_{1}\left( \frac{1}{2} \right)}{\longrightarrow} \left( \begin{array}{cc|cc} 1& 2 & \frac{1}{2} & 0 \\ 6 & 8 &0 &1 \end{array} \right) \stackrel{A_{12}\left( -6 \right)}{\longrightarrow} \left( \begin{array}{cc|cc} 1& 2 & \frac{1}{2} & 0 \\ 0 & -4 &-3 &1 \end{array} \right) \\ & \stackrel{A_{21}(-2)M_{2}\left( - \frac{1}{4} \right)}{\longrightarrow} \left( \begin{array}{cc|cc} 1& 0 & -1 & \frac{1}{2} \\ 0 & 1 & \frac{3}{4} & -\frac{1}{4} \end{array} \right) \end{align}$hence $A^{-1} = \begin{pmatrix}1 & -2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\0 & -\frac{1}{4} \end{pmatrix}\begin{pmatrix} 1&0 \\ -6 & 1 \end{pmatrix}\begin{pmatrix} \frac{1}{2} &0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & \frac{1}{2} \\ \frac{3}{4} & -\frac{1}{4} \end{pmatrix}$ (c) Row reduce $(A \mid I_{3})$ as follows: $\begin{align} & \left( \begin{array}{ccc|ccc} 1 & -2 & 0 & 1 & 0 & 0 \\ 2 & -3 & 0 & 0 & 1& 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right) \stackrel{A_{12}(-2)}{\longrightarrow} \left( \begin{array}{ccc|ccc} 1 & -2 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & -2 & 1& 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right) \stackrel{A_{21}(2)}{\longrightarrow} \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & -3 & 2 & 0 \\ 0 & 1 & 0 & -2 & 1& 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right) \end{align}$hence $A^{-1}= \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} -3 & 2 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ *Question 8* Let $\underline{v}_{1},\underline{v}_{2},\underline{v}_{3}$ equal $(1,1,0)^{T},(0,1,1)^{T},(n,0,n(n-2))^{T}$ respectively. When $n=0$, $\underline{v}_{3}=(0,0,0)^{T}$. The following linear dependence relation $0\times \underline{v}_{1}+0\times \underline{v}_{2}+\underline{v}_{3}= \underline{0}$ shows that the vectors are linearly dependent. Suppose $\lambda_{1}\underline{v}_{1}+\lambda_{2}\underline{v}_{2}+\lambda \underline{v}_{3} = \underline{k}_{1}$ then considering $\underline{k}$ components gives $\lambda_{2}=1$, considering the $\underline{i}$ component gives $\lambda_{1}=1$ and considering the $\underline{j}$ component gives $\lambda_{2}=-\lambda_{1}=-1$ (which gives a contradiction) so the vectors do not span $\mathbb{R}^3$. When $n=1$, $\underline{v}_{3}= (1,0,-1)^{T}$. The following linear dependence relation $\underline{v}_{1} -\underline{v}_{2}-\underline{v}_{3}=\underline{0}$ shows that the vectors are linearly dependent. Since $\underline{k}$ cannot be written as a linear combination of $\underline{v}_{1}$ and $\underline{v}_{2}$ as shown above we have that it also cannot be written as a linear combination of the three vectors since $\underline{v}_{3}$ is a linear combination of the first two. The vectors therefore do not span $\mathbb{R}^{3}$. When $n=2$, $\underline{v}_{3}=(2,0,0)^{T}$ which is not in the span of the first two so the three vectors are linearly independent and span $\mathbb{R}^{3}$. Note that $-\underline{v}_{1}+\underline{v}_{2} + \frac{1}{2}\underline{v}_{3}=\underline{i}$, $\frac{1}{2}\underline{v}_{3}=\underline{k}$ and $\underline{v}_{1}-\frac{1}{2}\underline{v}_{3}=\underline{j}$ so the vectors span $\mathbb{R}^{3}$. *Question 9* Let $W = \begin{pmatrix}0 & 2 & 1\\ 1 & -1 & 1 \\ 1 & 2 & 2 \end{pmatrix}$ (i.e. its columns are $\underline{f}_{i}$ for $i=1,2,3$ ). Compute the RREF of $(W\mid I_{3})$ as follows $\begin{align} &\left(\begin{array}{ccc|ccc} 0 & 2 & 1 & 1 & 0 & 0 \\ 1 & -1 & 1 & 0 & 1 & 0 \\ 1 & 2 & 2 & 0 & 0 & 1 \end{array}\right) \stackrel{S_{13}}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1 & 2 & 2 & 0 & 0 & 1 \\ 1 & -1 & 1 & 0 & 1 & 0 \\ 0 & 2 & 1 & 1 & 0 & 0 \end{array}\right) \stackrel{A_{12}(-1)}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1 & 2 & 2 & 0 & 0 & 1 \\ 0 & -3 & -1 & 0 & 1 & -1 \\ 0 & 2 & 1 & 1 & 0 & 0 \end{array}\right) \\ &\stackrel{A_{23}(-2)A_{21}(-2)M_{2}\left( -\frac{1}{3} \right)}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1 & 0 & \frac{4}{3} & 0 & \frac{2}{3} & \frac{1}{3} \\ 0 & 1 & \frac{1}{3} & 0 & -\frac{1}{3} & \frac{1}{3} \\ 0 & 0 & \frac{1}{3} & 1 & \frac{2}{3} & -\frac{2}{3} \end{array}\right) \stackrel{A_{31}\left( -\frac{4}{3} \right)A_{31}\left( -\frac{1}{3} \right)M_{3}(3)}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & -4 & -2 & 3 \\ 0 & 1 & 0 & -1 & -1 & 1 \\ 0 & 0 & 1 & 3 & 2 & -2 \end{array}\right). \end{align}$So $W^{-1} = \begin{pmatrix}-4 & -2 & 3 \\ -1 & -1 & 1 \\ 3 &2 & -2 \end{pmatrix}.$ (a) The equation $\underline{w}=\lambda_{1}\underline{f}_{1} +\lambda_{2}\underline{f}_{2}+\lambda_{3}\underline{f}_{3}$ can be rewritten as $W \begin{pmatrix} \lambda_{1}\\ \lambda_{2} \\ \lambda_{3} \end{pmatrix} = \begin{pmatrix} 2\\ -5 \\ 1 \end{pmatrix}$so $\begin{pmatrix} \lambda_{1}\\ \lambda_{2} \\ \lambda_{3} \end{pmatrix} = W^{-1} \begin{pmatrix} 2\\ -5 \\ 1 \end{pmatrix} = \begin{pmatrix}-4 & -2 & 3 \\ -1 & -1 & 1 \\ 3 &2 & -2 \end{pmatrix} \begin{pmatrix} 2\\ -5 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \\ -6 \end{pmatrix}$ (b) The coordinates of $\underline{u}_{n}=(1,2,n)^{T}$ with respect to $\underline{f}_{1},\underline{f}_{2},\underline{f}_{3}$ are $\mu_{1},\mu_{2},\mu_{3}$ respectively where $\begin{pmatrix} \mu_{1}\\ \mu_{2} \\ \mu_{3} \end{pmatrix} = W^{-1} \begin{pmatrix} 1\\ 2 \\ n \end{pmatrix} = \begin{pmatrix}-4 & -2 & 3 \\ -1 & -1 & 1 \\ 3 &2 & -2 \end{pmatrix} \begin{pmatrix} 1\\ 2 \\ n \end{pmatrix} = \begin{pmatrix} -8+3n \\ -3+n \\ 7 - 2n \end{pmatrix}$