Show that the matrix $$ \mathrm{A}=\begin{array}{cll} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 14 & 7 & 10 \end{array} $$ is equivalent to \(\mathrm{D}^{33}{ }_{2}\) where $\mathrm{D}^{\mathrm{m}, \mathrm{n}} \mathrm{r}$ denotes the canonical form under equivalence of A. D \(^{\mathrm{m}, \mathrm{n}}_{\mathrm{r}}\) is the echelon form that has one's along the diagonal and zeros elsewhere, and where all the zero rows are consigned to the depths of the matrix.

Let the homogeneous linear system \(\mathrm{AX}=\mathrm{B}\) be given by \(\begin{array}{llll}1 & 2 & 0 \mid & \left|x_{1}\right| & |0|\end{array}\) \(|0 \quad 1 \quad 3| \quad\left|\mathrm{x}_{2}\right|=|0|\) $\mid \begin{array}{llll}2 & 1 & 3 \mid & \left|\mathrm{x}_{3}\right| & |0| \text { . }\end{array}$ Show that \(\mathrm{A}\) has only the trivial solution, $\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\right)=(0,0,0)$

Let \(\mathrm{V}\) be the subspace of \(\mathrm{R}^{4}\) spanned by $\mathrm{S}=\left\\{\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}\right\\}\( where \)\alpha_{1}=[1,2,1,2], \alpha_{2}=[2,1,2,1]$ \(\alpha_{3}=[3,2,3,2], \alpha_{4}=[3,3,3,3], \alpha_{5}=[5,3,5,3]\). Find a basis for \(\mathrm{V}\).

Find the rank of matrix A where $$ \mathrm{A}=\begin{array}{cccc} 1 & 0 & 2 & 3 \mid \\ 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 \end{array} $$

Let \(\mathrm{P}_{4}\), denote the vector space of all polynomials of degree at most equal to four. Let \(\mathrm{V}\) be the subspace of \(\mathrm{P}_{4}\). spanned by $\mathrm{S}=\left\\{\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right\\}\( where \)\alpha_{1}=\mathrm{t}^{4}+\mathrm{t}^{2}+2 \mathrm{t}+1$ \(\alpha_{2}=t^{4}+t^{2}+2 t+2, \alpha_{3}=2 t^{4}+t^{3}+t+2\) and \(\alpha_{4}=t^{4}+t^{3}-t^{2}\) \(-\mathrm{t}\). Find a basis for \(\mathrm{V}\).

Find the rank of matrix A where: $\begin{array}{rrrrr}\text { (i) } \mathrm{A}= & \mid 1 & 3 & 1 & -2 & -3 \mid \\\ & 11 & 4 & 3 & -1 & -4 \\ & 2 & 3 & -4 & -7 & -3 \\ & 13 & 8 & 1 & -7 & -8\end{array}$ (ii) $\begin{array}{rrr}\mathrm{A}= & \mid \begin{array}{rrr}1 & 2 & -3\end{array} \\ & \mid 2 & 1 & 0 \\ & \mid-2 & -1 & 3 \mid \\ & -1 & 4 & -2\end{array}$ (iii) $\begin{array}{rr}\mathrm{A}= & \mid 1 & 3 \mid \\ & \mid 0 & -2 \mid \\\ & \mid 5 & -1 \\ & \mid-2 & 3\end{array} \mid$