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{\bf MATH 221 Solution to Linear Algebra Quiz V}
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(I) (1) T, (2) T, (3) T, (4) T, (5) T, (6) T, (7) T, (8) T,
(9) T, (10) F, (11) T, (12) F, (13) F, (14) F, (15) T, (16)
T, (17) F, (18) F, (19) T, (20) F.

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(II) (2), (4), (5), (7)

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(III) Reduce $A$ to echelon form, resulting in
$$B = \begin{pmatrix}
3&2&-4&1&5\\
0&0&1&1&-9\\
0&0&0&0&25\\
0&0&0&0&0
\end{pmatrix}$$
It is clear that
$$\dim RS(A) = \dim CS(A) = rank(A) = 3.$$By Rank Theorem
$\dim CS(A) + \dim NS(A) = n = 5$ hence $\dim NS(A) = 2$.

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The non-zero rows of $B$ form a basis for $RS(B) = RS(A)$ so
a basis of $RS(A)$ is given by
$$(3, 2, -4, 1, 5), \;(0, 0, 1, 1, - 9),\;(0, 0, 0, 0, 25).$$
The pivot columns are the first, the third
and the 5th hence the following vectors form
a basis of $CS(A)$:
$$\begin{pmatrix} 3\\
6\\-3\\9
\end{pmatrix},\;\begin{pmatrix} -4\\
-7\\6\\-11
\end{pmatrix},\;\begin{pmatrix} 5\\
1\\2\\6
\end{pmatrix}.$$

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To find the null space we further reduce the matrix $B$
to reduced echelon form
$$C = \begin{pmatrix}
1&2/3&0&5/3&0\\
0&0&1&1&0\\
0&0&0&0&1\\
0&0&0&0&0
\end{pmatrix}$$
Since
$NS(A) = NS(B) = NS(C)$ the null space consists of
sloutions of the equation
$C {\bf x} = 0$. The free variables are $x_2$ and $x_4$
hence the solutions are given by
$${\bf x} = \begin{pmatrix} x_1\\
x_2\\x_3\\x_4\\x_5
\end{pmatrix} = x_2 \begin{pmatrix} - \frac{3}{2}\\
1\\0\\0\\0\end{pmatrix} + x_4 \begin{pmatrix} - \frac{5}{2}\\
0\\-1\\1\\0\end{pmatrix}$$

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