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Problems #1,11 on pg.236 are graded.
#1, pg.236: Assume that the matrix $A$ is row equivalent to the matrix $B$. Determine the rank of $A$ and the dimension of $\mathrm{Nul} \hspace{2pt} A$ without calculations. Find the bases for $\mathrm{Col} \hspace{2pt} A$, $\mathrm{Row} \hspace{2pt} A$ and $\mathrm{Nul} \hspace{2pt} A$:
$$A = \begin{bmatrix} 1 &-4 & 9 & -7 \\ -1& 2 & -4 & 1 \\ 5 & -6 & 10 & 7 \end{bmatrix},$$
$$B = \begin{bmatrix} 1 & 0 & -1 & 5 \\ 0&-2&5&-6 \\ 0&0&0&0 \end{bmatrix}.$$
Solution: By observation we know that $\mathrm{rank}(A)=2$ because there are two nonzero rows (and hence a basis of Row A containing two elements --- this can also be seen by noting that there are only two pivot columns). The rank-nullity theorem now shows that
$$\mathrm{dim \hspace{2pt} Nul \hspace{2pt}} A = 4 - \mathrm{rank} \hspace{2pt} A = 4 - 2 = 2.$$
Also by observation we can see that a basis for the row space of $A$ is
$$\mathscr{B}_{\mathrm{row}} = \left\{ \begin{bmatrix}1 \\ 0 \\ -1 \\ 5 \end{bmatrix}^T, \begin{bmatrix} 0 \\ -2 \\ 5 \\ -6 \end{bmatrix}^T \right\}$$
and a basis for the column space
$$\mathscr{B}_{\mathrm{col}} = \left\{ \begin{bmatrix} 1 \\ -1 \\ 5 \end{bmatrix}, \begin{bmatrix} -4 \\ 2 \\ -6 \end{bmatrix} \right\}.$$
To find a basis for the nullspace, we will first compute the reduced echelon form of $A$ (start computing this from $B$!) and we see
$$A \sim \begin{bmatrix} 1 & 0 & -1 & 5 \\ 0&1 & -\frac{5}{2} & 3 \\ 0&0&0&0 \end{bmatrix}.$$
To find a basis for the nullspace we need to solve the equation $A \vec{x}=\vec{0}$ and our row reduction shows that solving this equation yields the augmented matrix
$$\begin{bmatrix} 1&0&-1&5&0 \\ 0&1&-\frac{5}{2}&3&0 \\ 0&0&0&0&0 \end{bmatrix}.$$
From this we write the associated system of equations
$$\left\{ \begin{array}{ll}
x_1 -x_3+5x_4 &= 0 \\
x_2 - \dfrac{5}{2}x_3 + 3x_4 &= 0 \\
0&=0
\end{array} \right.$$
and so we see that all solutions $\vec{x}$ of $A\vec{x}=\vec{0}$ can be expressed as
$$\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} x_3 - 5x_4 \\ \frac{5}{2}x_3 - 3x_4 \\ x_3 \\ x_4 \end{bmatrix} = x_3 \begin{bmatrix} 1 \\ \frac{5}{2} \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -5 \\ -3 \\ 0 \\ 1 \end{bmatrix}.$$
Therefore we see that a basis for the nullspace of $A$ is given by
$$\mathscr{B}_{\mathrm{Nul}} = \left\{ \begin{bmatrix} 1 \\ \frac{5}{2} \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -5 \\ -3 \\ 0 \\ 1 \end{bmatrix} \right\}.$$
#11,pg.237: If the nullspace of an $8 \times 5$ matrix $A$ is $3$-dimensional, what is the dimension of the row space of $A$?
Solution: The rank-nullity theorem tells us that
$$\mathrm{rank} \hspace{2pt} A = 5 - \mathrm{dim \hspace{2pt} Nul \hspace{2pt}} A= 5 - 3 = 2.$$
The other part of the rank-nullity theorem tells us that $\mathrm{rank} \hspace{2pt} A = \mathrm{dim \hspace{2pt} Row \hspace{2pt}} A$ and so we see
$$\mathrm{dim \hspace{2pt} Row \hspace{2pt}}A = 2.$$
