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\begin{document}

\begin{centerline}
{\bf Solutions to Linear Algebra Practice Problems 1}
\end{centerline}


\vskip 6pt

\begin{enumerate}


\item{ (a.)

\begin{eqnarray*} \begin{split}
 & \left[ \begin{array}{cccc|c} 2&-1&0&0&1
 \\ -1&2&-1&0 & 0\\0&-1&2&-1 & 0\\0&0&-1&2 &6 \end{array} \right]
& \widetilde{r_2 + \frac{1}{2}r_1}
  & \left[ \begin{array}{cccc|c}
  2&-1&0&0&1
 \\ 0&\frac{3}{2}&-1&0 &\frac{1}{2}
 \\0&-1&2&-1 & 0
 \\0&0&-1&2 &6 \end{array} \right]\\
 \widetilde{ \frac{2}{3}r_2}
 & \left[ \begin{array}{cccc|c}
  2&-1&0&0&1
 \\ 0&1&- \frac{2}{3}&0 &\frac{1}{3}
 \\0&-1&2&-1 & 0
 \\0&0&-1&2 &6  \end{array} \right]
& \widetilde{r_3 + r_2} &
  \left[ \begin{array}{cccc|c}
  2&-1&0&0&1
 \\ 0&1&- \frac{2}{3}&0 &\frac{1}{2}
 \\0&0&\frac{4}{3}&-1 & \frac{1}{3}
 \\0&0&-1&2 &6\end{array} \right]\\
 \widetilde{\frac{3}{4}r_3} &
  \left[ \begin{array}{cccc|c}
   2&-1&0&0&1
 \\ 0&1&- \frac{2}{3}&0 &\frac{1}{2}
 \\0&0&1&-\frac{3}{4}& \frac{1}{4}
 \\0&0&-1&2 &6\end{array} \right]
& \widetilde{r_4 + r_3}
 & \left[ \begin{array}{cccc|c}
  2&-1&0&0&1
 \\ 0&1&- \frac{2}{3}&0 &\frac{1}{2}
 \\0&0&1&-\frac{3}{4}& \frac{1}{4}
 \\0&0&0&\frac{5}{4} &\frac{25}{4} \end{array} \right]\\
\widetilde{\frac{5}{4} r_4}
 & \left[ \begin{array}{cccc|c}
  2&-1&0&0&1
 \\ 0&1&- \frac{2}{3}&0 &\frac{1}{2}
 \\0&0&1&-\frac{3}{4}& \frac{1}{4}
 \\0&0&0&1 &5 \end{array} \right]
 & \widetilde{r_3+\frac{3}{4} r_4}
 & \left[ \begin{array}{cccc|c}
  2&-1&0&0&1
 \\ 0&1&- \frac{2}{3}&0 &\frac{1}{2}
 \\0&0&1&0& 4
 \\0&0&0&1 &5 \end{array} \right]\\
\widetilde{r_2+\frac{3}{2} r_3}
 & \left[ \begin{array}{cccc|c}
  2&-1&0&0&1
 \\ 0&1&0&0 &3
 \\0&0&1&0& 4
 \\0&0&0&1 &5 \end{array} \right]
 & \widetilde{r_1+ r_2}
 & \left[ \begin{array}{cccc|c}
  2&0&0&0&4
 \\ 0&1&0&0 &3
 \\0&0&1&0& 4
 \\0&0&0&1 &5 \end{array} \right]\\
 \widetilde{\frac{1}{2}r_1}
 & \left[ \begin{array}{cccc|c}
  1&0&0&0&2
 \\ 0&1&0&0 &3
 \\0&0&1&0& 4
 \\0&0&0&1 &5 \end{array} \right]
 \end{split}
 \end{eqnarray*}




Thus $(x_1, x_2, x_3, x_4)=(2, 3, 4, 5)$.

(b)  The solution in part (a) implies that

$2  \left[ \begin{matrix} 2
 \\ -1 \\0 \\0 \end{matrix} \right]
+3  \left[ \begin{matrix} -1
 \\ 2 \\-1 \\0 \end{matrix} \right]
+ 4  \left[ \begin{matrix}0
 \\ -1 \\2 \\-1 \end{matrix} \right]
+ 5  \left[ \begin{matrix} 0
 \\0 \\-1 \\2 \end{matrix} \right]=\left[ \begin{matrix}
1\\0 \\0 \\6 \end{matrix} \right]$.}

\item {

$M=\left[ \begin{matrix} 1&3&-4&7 \\ 2&6&5&1 \\3&9&4&5
\end{matrix} \right]$.

\begin{eqnarray*} \begin{split}
 & \left[ \begin{matrix} 1&3&-4&7 \\ 2&6&5&1 \\3&9&4&5
\end{matrix} \right]
& \widetilde{r_2 -2r_1, r_3 +(-3)r_1}
  & \left[ \begin{matrix}
   1&3&-4&7 \\ 0&0&13&-13 \\0&0&16&-16
\end{matrix} \right]\\
  \widetilde{\frac{1}{13}r_2 , \frac{1}{16}r_3,
r_3-r_2}
  & \left[ \begin{matrix}
   1&3&-4&7 \\ 0&0&1&-1 \\0&0&0&0
\end{matrix} \right]
& \widetilde{r_1 + 4r_2} &
  \left[ \begin{matrix}
 1&3&0&3 \\ 0&0&1&-1 \\0&0&0&0 \end{matrix} \right]
 \end{split}
 \end{eqnarray*}








So the reduced row echelon form of $M$ is is $\left[ \begin{matrix} 1&3&0&3 \\
0&0&1&-1 \\0&0&0&0\end{matrix} \right]$. The solution is $x_1+3x_2+
3x_4=0$ and $x_3-x_4=0$. Thus $x_1= -3x_2-3x_4$, $x_3=x_4$.


 Any solution is of the form\[
\vec{x}=\left[ \begin{matrix} x_1\\x_2\\x_3\\x_4\end{matrix}
\right]=\left[ \begin{matrix} -3x_2-3x_4\\x_2\\x_4\\x_4\end{matrix}
\right]=  x_2\left[ \begin{matrix} -3\\1\\0\\0\end{matrix} \right]
+x_4\left[
\begin{matrix}-3\\0\\1\\1\end{matrix} \right]\] where $x_2$ and $x_4$ are any numbers.

}

\item



\begin{enumerate}
\item
 \begin{eqnarray*} \begin{split}& \left[\begin{array}{ccc|c}1&1&0&2\\1 &
2&0&1\\3&5&a&b\end{array}\right] & \widetilde{r_2 + (-1)r_1, r_3+
(-3)r_1} &
  \left[ \begin{array}{ccc|c}
  1&1&0&2\\0 &1&0&-1\\0&2&a&b-6\end{array} \right]\\
  \widetilde{r_3 + (-2)r_2} &
  \left[ \begin{array}{ccc|c}
  1&1&0&2\\0 &1&0&-1\\0&0&a&b-4\end{array} \right]
  & \widetilde{r_1 + (-1)r_2} &
  \left[ \begin{array}{ccc|c}
  1&0&0&3\\0 &1&0&-1\\0&0&a&b-4\end{array} \right]
  \end{split}
\end{eqnarray*}

 The system will have a unique solution when
$a\not= 0$. The solution is $(3, -1, \frac{b-4}{a})$.
\item The system will have
infinitely many solutions if $a=0$ and $b=4$.

\item The system will be
inconsistent if $a=0$ and $b\not=4$
\end{enumerate}

\item
{

\[ A=
 \left[ \begin{matrix} 1&2&3&4
 \\ 2&3&4&1 \\3&4&1&2 \\4&1&2&3 \end{matrix} \right]
\ \text{and} \  B=
 \left[ \begin{matrix} 1&1&-1
 \\ 1&-1&1 \\1&1&1\\ 1&1&1\end{matrix} \right].
 \]  \[AB =  \left[ \begin{matrix} 1&2&3&4
 \\ 2&3&4&1 \\3&4&1&2 \\4&1&2&3 \end{matrix} \right]
 \left[ \begin{matrix} 1&1&-1
 \\ 1&-1&1 \\1&1&1\\ 1&1&1\end{matrix} \right]=\left[ \begin{matrix} 10 &6& 10
 \\ 10&4&10 \\10&2&10\\ 10&8&10\end{matrix} \right]
.\]

\[B^T=\left[ \begin{matrix} 1&1&1 & 1
 \\ 1&-1&1 & 1 \\ -1 &1&1&1\end{matrix} \right].\]
 \[B^TA^T
 =(AB)^T = \left(\left[ \begin{matrix} 10 &6& 10
 \\ 10&4&10 \\10&2&10\\ 10&8&10\end{matrix} \right]\right)^T
= \left[ \begin{matrix}
  10&10&10 & 10 \\ 6 &4&2&8 \\ 10&10&10 & 10 \end{matrix} \right]
.\]

}

\item {
  \[\left[ \begin{matrix} 0&1&3&0
 \\ 0&0&1&4 \\0&0&0&1 \\2&0&0&0 \end{matrix} \right]
 \widetilde{\ move\ the \ last \ row \ to \ the \ first \ row}
\left[ \begin{matrix} 2&0&0&0 \\0&1&3&0
 \\ 0&0&1&4 \\0&0&0&1 \end{matrix} \right]
 \] This matrix has four pivot vectors. So the columns of the matrix form a linearly
independent set.

  \begin{eqnarray*} \begin{split}&
 \left[ \begin{matrix}
 -4&-3&0 \\0&-1&4 \\1&0&3\\ 5&4&6 \end{matrix} \right]
&\widetilde{interchange\ first \ and \ third \ row}& \left[
\begin{matrix}
 1&0&3 \\0&-1&4 \\ -4&-3&0\\ 5&4&6 \end{matrix} \right]\\
\widetilde{r_3 + 4r_1 , r_4 + (-5) r_1}& \left[
\begin{matrix}
 1&0&3 \\0&-1&4 \\ 0&-3&12\\ 0&4&-9 \end{matrix} \right]&
 \widetilde{ (-1) r_2}& \left[
\begin{matrix}
 1&0&3 \\0&1&-4 \\ 0&-3&12\\ 0&4&-9 \end{matrix} \right]\\
 \widetilde{r_3 + 3r_2 , r_4 + (-4) r_2}& \left[
\begin{matrix}
 1&0&3 \\0&1&-4 \\ 0&0&0\\ 0&0&7 \end{matrix} \right]&
 \widetilde{ interchange \ 3rd \ and \ 4th \ row, \frac{1}{7} r_4}& \left[
\begin{matrix}
 1&0&3 \\0&1&-4 \\ 0&0&1\\ 0&0&0 \end{matrix} \right]
  \end{split}
\end{eqnarray*}
This matrix has three pivot vectors. So the columns of the matrix
form a linearly independent set.




The column vectors of \[\left[ \begin{matrix}
 -4&-3&1 & 5 &1 \\ 2&-1&4 & -1 & 2\\1&2&3 & 6 & -3 \\ 5&4&6 & -3 & 2  \end{matrix}
 \right] \]form a dependent set since we have five column vectors in
 $R^4$.

}

\item{

 \begin{eqnarray*} \begin{split}
 \left[ \begin{matrix} 0&1&0&t
 \\-1& 0& t&0 \\0&- t&0&1 \\- t&0&-1&0\end{matrix} \right]
 \widetilde{interchange \ 1st \ and \ 2nd \ row, (-1) r_1}
 \left[ \begin{matrix}1& 0& -t&0 \\ 0&1&0&t
 \\0&- t&0&1 \\- t&0&-1&0\end{matrix} \right]\\
 \widetilde{ r_4 + t \cdot r_1}
 \left[ \begin{matrix}1& 0& -t&0 \\ 0&1&0&t
 \\0&- t&0&1 \\0&0&-1-t^2&0\end{matrix} \right]
 \widetilde{ r_3 + t \cdot r_2}
 \left[ \begin{matrix}1& 0& -t&0 \\ 0&1&0&t
 \\0&0&0&1+t^2 \\0&0&-1-t^2&0\end{matrix} \right]\\
 \widetilde{\frac{1}{1+t^2} r_3 ,\frac{-1}{1+t^2} r_4, r_3 \leftrightarrow r_4} \left[ \begin{matrix}1& 0& -t&0 \\ 0&1&0&t
 \\0&0&1&0 \\0&0&0&1\end{matrix}\right]
\end{split}
\end{eqnarray*}
$M$ has four pivot vectors so $M$ is invertible. Note that we have
used the fact that $1+t^2 \neq 0$ in the computation.

, \begin{eqnarray*} \begin{split}A= \left[ \begin{matrix}1&1&0&0
 \\1& 0& 1&0 \\0&1&-1&0 \\1&1&1&1\end{matrix} \right]
 \widetilde{r_2+(-1) r_1, r_4 + (-1)r_1}
\left[ \begin{matrix}1&1&0&0
 \\0& -1& 0&0 \\0&1&-1&0 \\0&0&1&1\end{matrix} \right]\\
 \widetilde{-(1)r_2, r_3 + (-1)r_2}
\left[ \begin{matrix}1&1&0&0
 \\0& 1& 0&0 \\0&0&-1&0 \\0&0&1&1\end{matrix} \right]
 \widetilde{-(1)r_3, r_4 + (-1)r_3}
\left[ \begin{matrix}1&1&0&0
 \\0& 1& 0&0 \\0&0&1&0 \\0&0&0&1\end{matrix} \right]
 \end{split}
\end{eqnarray*}.
$A$ has four pivot vectors so $A$ is invertible.



 $B= \left[ \begin{matrix}1&1&0&0
 \\0& 0& 1&0 \\0&0&0&1 \\0&1&0&0\end{matrix} \right]\widetilde{}
\left[ \begin{matrix}1&1&0&0\\0&1&0&0
 \\0& 0& 1&0 \\0&0&0&1 \end{matrix} \right]
 $. $B$ has four pivot vectors so $B$ is invertible.
}

\item{

(a) Let $A=\left[ \begin{matrix} 3&7&-3 \\ 3&8&-1 \\-6&-14&7
\end{matrix} \right]$.

 \begin{eqnarray*} \begin{split} \left[ \begin{array}{ccc|ccc}
   3&7&-3 & 1 & 0 & 0\\ 3&8&-1 & 0 & 1 & 0\\-6&-14&7 & 0 & 0 & 1\end{array} \right]
   \widetilde{r_2 + (-1) r_1, r_3 + 2r_1}
   \left[ \begin{array}{ccc|ccc}
   3&7&-3 & 1 & 0 & 0\\ 0&1&2 & -1 & 1 & 0\\0&0&1 & 2 & 0 & 1\end{array}
   \right]\\
   \widetilde{r_2 + (-2) r_3, r_1 + 3r_3}
   \left[ \begin{array}{ccc|ccc}
   3&7&0 & 7 & 0 &3\\ 0&1&0 & -5 & 1 & -2\\0&0&1 & 2 & 0 & 1\end{array}
   \right]\widetilde{r_1 + (-7) r_2}
   \left[ \begin{array}{ccc|ccc}
   3&0&0 & 42 & -7 &17\\ 0&1&0 & -5 & 1 & -2\\0&0&1 & 2 & 0 &
   1\end{array}  \right]\\ \widetilde{\frac{1}{3}r_1 }
   \left[ \begin{array}{ccc|ccc}
   1&0&0 & 14 & \frac{-7}{3} & \frac{17}{3}\\ 0&1&0 & -5 & 1 & -2\\0&0&1 & 2 & 0 &
   1\end{array}
   \right]
\end{split}
\end{eqnarray*}
 So $A^{-1} =\left[ \begin{array}{ccc}
   14 & \frac{-7}{3} & \frac{17}{3}\\  -5 & 1 & -2\\ 2 & 0 &
   1\end{array}
   \right].$



(b)Since $A^{-1}$ exists, So the vector $b=\left[ \begin{matrix} 1
\\2\\3 \end{matrix} \right]$ lies in the span of the column vectors
of $A$.


}




%\item{Let $P$ be the set of all polynomials $f(x)$ and let $Q$ be the
%subset of $P$ consisting of all polynomials $f(x)$ so
%$f(0)=f(1)=0$. Show that $Q$ is a subspace of $P$.

%Answer: If $f,g\in Q$, then$(f+g)(0)=f(0)+g(0)=0$ and $(f+g)
%(1)=f(1)+g(1)=0$. So $f+g\in Q$. If $f\in Q $ and $r\in {\Bbb R}
%$}, then $(rf)(0)=r f(0)=0$ and $(rf)(1)=r f(1)=0$. So $rf\in Q$.
%Therefore $Q$ is a subspace of $P$.



\item{ (a) \begin{eqnarray*} \begin{split}M=\left[ \begin{matrix} 1&1&2 \\ 1&a+1&3 \\1&a&a+1
\end{matrix} \right]\widetilde{r_2+ (-1)r_1, r_3 + (-1)r_1}
\left[ \begin{matrix} 1&1&2 \\ 0&a&1 \\0&a-1&a-1
\end{matrix} \right]\\
\widetilde{\frac{1}{a-1} r_3 \ if  a-1 \neq 0, r_2 \leftrightarrow
r_3} \left[
\begin{matrix} 1&1&2
\\ 0&1&1 \\0&a&1
\end{matrix} \right]\widetilde{ r_3 + (-a) r_2 \leftrightarrow
r_3} \left[
\begin{matrix} 1&1&2
\\ 0&1&1 \\0&0&1-a
\end{matrix} \right]
\end{split}
\end{eqnarray*}

Thus $M$ is invertible if $a \neq 1$.

(b)  The column vectors are independent if $a \neq 1$.  The column
vectors are dependent if $a = 1$.

}



\item{
First, note that \begin{eqnarray*} \begin{split}T (\left[
\begin{matrix} 1
\\2\\3 \end{matrix} \right])=  & T(\left[ \begin{matrix} 1
\\0\\0 \end{matrix} \right]
+2\left[ \begin{matrix} 0
\\1\\0 \end{matrix} \right]+
3\left[ \begin{matrix} 0
\\0\\1 \end{matrix} \right]= T(\left[ \begin{matrix} 1
\\0\\0 \end{matrix} \right])
+T(2\left[ \begin{matrix} 0
\\1\\0 \end{matrix} \right])+T(
3\left[ \begin{matrix} 0
\\0\\1 \end{matrix} \right]
)\\
= & T(\left[ \begin{matrix} 1
\\0\\0 \end{matrix} \right])
+2T(\left[ \begin{matrix} 0
\\1\\0 \end{matrix} \right])+3T(
\left[ \begin{matrix} 0
\\0\\1 \end{matrix}\right] = T(e_1) + 2 T(e_2) + 3 T(e_3).\\
\end{split}
\end{eqnarray*}

Adding $T(e_1 + e_2) = \left[ \begin{matrix} 1
\\-1
\end{matrix} \right]$ and $T(e_1 - e_2) = \left[ \begin{matrix} 2
\\3
\end{matrix} \right]$, we get $T(2e_1)=\left[ \begin{matrix} 3
\\2
\end{matrix} \right]$ and $T(e_1) = \left[ \begin{matrix} \frac{3}{2}
\\1
\end{matrix} \right]$. Similarly, $T(e_1 + e_2)- T(e_1-e_2)=\left[ \begin{matrix}
-1
\\-4
\end{matrix} \right]$. So $2T(e_2)=\left[ \begin{matrix}
-1
\\-4
\end{matrix} \right]$ and $T(e_2) = \left[ \begin{matrix}
-\frac{1}{2}
\\-2
\end{matrix} \right]$. From $T(e_1 + e_2+ e_3)-T(e_1+e_2)=\left[ \begin{matrix} 1
\\-2
\end{matrix} \right]-\left[ \begin{matrix} 1
\\-1
\end{matrix} \right]$, we get $T(e_3)=\left[ \begin{matrix} 0
\\-1
\end{matrix} \right]$. Hence
$T (\left[
\begin{matrix} 1
\\2\\3 \end{matrix} \right])
=T(e_1) + 2 T(e_2) + 3 T(e_3)=\left[ \begin{matrix} \frac{3}{2}
\\1
\end{matrix} \right]+ 2\cdot \left[ \begin{matrix}
- \frac{1}{2}
\\-2
\end{matrix} \right] + 3 \cdot \left[ \begin{matrix} 0
\\-1
\end{matrix} \right]= \left[ \begin{matrix} \frac{1}{2}
\\-6
\end{matrix} \right].$
 }


\end{enumerate}



\end{document}
