Math

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\begin{document}

\textbf{Math 1180 Final Review}\\
\begin{enumerate}
\item $\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\cdot\|\vec{v}\|}=\frac{-1+2+2}{\sqrt{6}\cdot\sqrt{6}}=\frac{3}{6}=\frac{1}{2}$, $\theta=\frac{\pi}{3}$\\

\item $\vec{p}_o=\begin{bmatrix}1 \\ -1 \\ 2\end{bmatrix}$, $\vec{d}=\begin{bmatrix}1\\-1\\2\end{bmatrix}-\begin{bmatrix}-1\\-2\\1\end{bmatrix}=\begin{bmatrix}2\\1\\1\end{bmatrix}$\\

    vector form: $\vec{x}=\vec{p}_o+t\vec{d}=\begin{bmatrix}1 \\ -1 \\ 2\end{bmatrix}+t\begin{bmatrix}2\\1\\1\end{bmatrix}$, parametric $\begin{cases}x=1+2t\\y=-1+t\\z=2+t\end{cases}$\\

\item \textit{Proof.} Consider $c_1(\vec{u})+c_2(\vec{u}-\vec{v})+c_3(\vec{u}-\vec{v}-\vec{w})=\0$ $\circledast$\\

$(c_1+c_2)\vec{u}-(c_2+c_3)\vec{v}-c_3\vec{w}=\0$\\

    Since $\vec{u}$, $\vec{v}$, $\vec{w}$ are lin. indep. $\implies$ $\begin{cases}c_1+c_2\qquad=0 \\ \qquad c_2+c_3=0 \\ \qquad\qquad\mspace{2mu} c_3=0 \end{cases}$ $\implies$ $\begin{cases}c_1=0\\c_2=0\\c_3=0\end{cases}$\\

    So $\circledast$ is true iff $c_1=c_2=c_3=0$.\\

    Therefore, $\vec{u}$, $\vec{u}-\vec{v}$, $\vec{u}-\vec{v}-\vec{w}$ are lin. indep.\\

\item \textit{Proof.} Let $X=I-A-A^2$.

\begin{flalign*}(I+A)X &=(I+A)(I-A-A^2)=(I-\cancel{A}-\cancel{A^2})+(\cancel{A}-\cancel{A^2}-A^3)\\
&=I-A^3=I-0=1\text{ since }A^3=0&\end{flalign*} \\
By Thm 3.13, $I+A$ is invertible and $(I+A)^{-1}=X=I-A-A^2$

\item \begin{enumerate}
\item[(1)]$\det A=1\cdot1\cdot(-1)$\\$\det B=\begin{vmatrix}1 & -2 & 1\\1 & -1 & 0\\1 & -1 & 1\\\end{vmatrix}=\begin{vmatrix}1 & -2 & 1\\0 & 1 & -1\\0 & 1 & 0\end{vmatrix}=\begin{vmatrix}1 & -2 & 1\\0 & 1 & -1\\0 & 0 & 0\end{vmatrix}=1\cdot1\cdot1=1$\\
$\det{(AB)}=\det A\cdot\det B=(-1)\cdot1=1$
\item[(2)]Since $\det A \neq 0$, $\det B \neq 0$, $\det AB \neq 0$, $A$, $B$ \& $AB$ are invertible.\\

\begin{flalign*}
[A\vert I_n]&=
\left[\begin{array}{ccc|ccc}
1 & -1 & 1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & -1 & 0 & 0 & 1 \\
\end{array}\right]\\
&\rightarrow \left[\begin{array}{ccc|ccc}
1 & 0 & 2 & 1 & 1 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 \\
\end{array}\right]\\
&\rightarrow\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & 1 & 2 \\
0 & 1 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & -1 \\
\end{array}\right]\\
A^{-1}&=\begin{bmatrix}
1&1&2\\0&1&1\\0&0&-1
\end{bmatrix}&
\end{flalign*}

\begin{flalign*}
[B\vert I_n]&=
\left[\begin{array}{ccc|ccc}
1 & -2 & 1 & 1 & 0 & 0 \\
1 & -1 & 0 & 0 & 1 & 0 \\
1 & -1 & 1 & 0 & 0 & 1 \\
\end{array}\right]\\
&\rightarrow \left[\begin{array}{ccc|ccc}
1 & -2 & 1 & 1 & 0 & 0 \\
0 & 1 & -1 & -1 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 \\
\end{array}\right]\\
&\rightarrow\left[\begin{array}{ccc|ccc}
1 & 0 & -1 & -1 & 2 & 0 \\
0 & 1 & -1 & -1 & 1 & 0 \\
0 & 0 & 1 & 0 & -1 & 1 \\
\end{array}\right]\\
&\rightarrow\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & -1 & 1 & 1 \\
0 & 1 & 0 & -1 & 0 & 1 \\
0 & 0 & 1 & 0 & -1 & 1 \\
\end{array}\right]\\
B^{-1}&=\begin{bmatrix}
-1&-1&1\\1&0&-1\\0&1&-1
\end{bmatrix}&
\end{flalign*}

$(AB)^{-1}=B^{-1}A^{-1}=\begin{bmatrix}-1&-1&1\\1&0&-1\\0&1&-1\end{bmatrix}\begin{bmatrix}1&1&2\\0&1&1\\0&0&-1\end{bmatrix}=\begin{bmatrix}-1&0&-2\\-1&-1&-3\\0&-1&-2\end{bmatrix}$
\end{enumerate}

\item
$A\rightarrow\begin{bmatrix}1 & 0 & 2 & 4 \\0 & 1 & -3 & -1 \\0 & 4 & -12 & 4 \\0 & -1 & 3 & 1 \end{bmatrix}\rightarrow\begin{blockarray}{*5{c}}\begin{block}{[ccccc]}1 & 0 & 2 & 4 \\0 & 1 & -3 & -1 \\0 & 4 & -12 & 4 \\0 & -1 & 3 & 1\\\end{block}\uparrow & \uparrow & s & t\end{blockarray}$\\

$x_3=s, \, x_4=t \\x_2-3x_3-x_4=0 \\\implies x_2=3s+t \\x_1+2x_3+4x_4=0 \\\implies x_1=-2s-4t$\\

$\vec{x}=\begin{bmatrix}
-2s-4t\\
\:\: 3s+\; t\\
s\quad \; \\
\; \, \; \qquad t
\end{bmatrix}=s\begin{bmatrix}-2\\3\\1\\0\end{bmatrix}+t\begin{bmatrix}-4\\1\\0\\1\end{bmatrix}$\\

$\text{A basis for }\mathnormal{row}(A)=\left \{\begin{bmatrix}1&0&2&4\end{bmatrix},\,\begin{bmatrix}0&1&-3&-1\end{bmatrix}\right\}$\\

$\text{A basis for }\mathnormal{col}(A)=\left\{\begin{bmatrix}1\\0\\3\\0\end{bmatrix},\,\begin{bmatrix}0\\1\\4\\-1\end{bmatrix}\right\}$\\

$\text{A basis for }\mathnormal{null}(A)=\left\{\begin{bmatrix}-2\\3\\1\\0\end{bmatrix},\,\begin{bmatrix}-4\\1\\0\\1\end{bmatrix}\right\}$\\

$\mathnormal{rank}(A)=2,\,\mathnormal{nullity}(A)=2$

\item
\begin{enumerate}
\item [(1)]
\begin{flalign*}
\det(A-\lambda I)
&=\begin{vmatrix}
1-\lambda & 2 & 0\\
-1 & -1-\lambda & 1\\
0 & 1 & 1-\lambda
\end{vmatrix}=
(1-\lambda)\begin{vmatrix}
-1-\lambda&1\\1&1-\lambda\end{vmatrix}-(-1)\begin{vmatrix}2&0\\1&1-\lambda\end{vmatrix}\\
&=(1-\lambda)[(-1-\lambda)(1-\lambda)-1]+2(1-\lambda)=(1-\lambda)[\lambda^2-1-1+2]=-\lambda^2(1-\lambda)\\
&=0,\,(A-\lambda I)\vec{x}=\0&
\end{flalign*}


\item [(2)]
\end{enumerate}
\end{enumerate}
\end{document}