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% Document Info
\author{gustavhelms}
\pdfinfo{
  /Title (linear-algebra.pdf)
  /Creator (Cheatography)
  /Author (gustavhelms)
  /Subject (Linear Algebra Cheat Sheet)
}

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\noindent
\begin{multicols}{3}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Linear Algebra Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{gustavhelms} via \textcolor{DarkBackground}{\uline{cheatography.com/146840/cs/31828/}}}
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\noindent
\begin{multicols}{3}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}gustavhelms \\
  \uline{cheatography.com/gustavhelms} \\
  \end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 11th June, 2022.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{x{1.64241 cm} x{3.33459 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Linear Systems}}  \tn
% Row 0
\SetRowColor{LightBackground}
Solution & Can have {\bf{no solution}}, {\bf{one solution}} or {\bf{infinitly many}}. The solution is the {\bf{interserction}} \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
Solution set & The set of all possible solutions \tn 
% Row Count 6 (+ 2)
% Row 2
\SetRowColor{LightBackground}
Consistency & A system is {\bf{consistent}} if there is at least one solution otherwise it is {\bf{inconsistent}} \tn 
% Row Count 10 (+ 4)
% Row 3
\SetRowColor{white}
Equivalent & Linear systems are {\bf{equivalent}} if they have the same {\bf{solution set}} \tn 
% Row Count 13 (+ 3)
% Row 4
\SetRowColor{LightBackground}
Row operations & {\bf{Replacement}}, {\bf{interchange}} and {\bf{scaling}} \tn 
% Row Count 15 (+ 2)
% Row 5
\SetRowColor{white}
Row equivalent & If there is a sequence of {\bf{row operations}} between two linear systems then the systems are {\bf{row equivalent}}. Systems that are row equivalent has the same {\bf{solution set}}. \tn 
% Row Count 22 (+ 7)
% Row 6
\SetRowColor{LightBackground}
Existence & If a system has a solution (i.e. {\bf{consistent}}) \tn 
% Row Count 24 (+ 2)
% Row 7
\SetRowColor{white}
Uniqueness & Is the solution unique \tn 
% Row Count 25 (+ 1)
% Row 8
\SetRowColor{LightBackground}
Homogenous & A system is {\bf{homogenous}} if it can be written in the form A{\bf{x}} = 0 \tn 
% Row Count 28 (+ 3)
% Row 9
\SetRowColor{white}
Trivial solution & If a system only has a the solution {\bf{x}} = 0. A system with no {\bf{free variable}} only have the trivial solution. \tn 
% Row Count 33 (+ 5)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{x{1.64241 cm} x{3.33459 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Linear Systems (cont)}}  \tn
% Row 10
\SetRowColor{LightBackground}
Non-trivial solution & A {\bf{nonzero}} vector that satisfies A{\bf{x}} = 0. Has free variable. \tn 
% Row Count 3 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Inverser of a Matrix}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{C}} is invertible if {\bf{CA}} = I\textasciicircum{}n\textasciicircum{} and {\bf{AC}} = I\textasciicircum{}n\textasciicircum{}} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{If {\bf{A}} is (2x2) then,  A\textasciicircum{}-1\textasciicircum{} =} \tn 
% Row Count 3 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{(A\textasciicircum{}-1\textasciicircum{})\textasciicircum{}-1\textasciicircum{}}} = {\bf{A}}} \tn 
% Row Count 4 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{(AB)\textasciicircum{}-1\textasciicircum{}}}= {\bf{B\textasciicircum{}-1\textasciicircum{}A\textasciicircum{}-1\textasciicircum{}}}} \tn 
% Row Count 5 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{(A\textasciicircum{}T\textasciicircum{})\textasciicircum{}-1\textasciicircum{}}}= {\bf{(A\textasciicircum{}-1\textasciicircum{})\textasciicircum{}T\textasciicircum{}}}} \tn 
% Row Count 6 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Linear Transformations}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Tranformation/mapping} & T(x) from R\textasciicircum{}n\textasciicircum{} to R\textasciicircum{}m\textasciicircum{} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
Image & For x in R\textasciicircum{}n\textasciicircum{}the vector T(x) in R\textasciicircum{}m\textasciicircum{}is called the image \tn 
% Row Count 5 (+ 3)
% Row 2
\SetRowColor{LightBackground}
Range & The set of all {\bf{images}} of the vectors in the {\bf{domain}} of T(x) \tn 
% Row Count 9 (+ 4)
% Row 3
\SetRowColor{white}
Criterion for a transformation to be linear & 1. T(u + v) = T(u) + T(v)\{\{nl\}\} 2. T(cU) = cT(U) \tn 
% Row Count 12 (+ 3)
% Row 4
\SetRowColor{LightBackground}
Standard Matrix & The matrix A for a linear transformation T, that satisfies T(x) = Ax for all x in R\textasciicircum{}n\textasciicircum{} \tn 
% Row Count 17 (+ 5)
% Row 5
\SetRowColor{white}
Onto & A mapping T is said to be {\bf{onto}} if each b in the codomain is the image of at least one x in the domain. Range = Codomain. Solution existance. ColA must match codomain. \tn 
% Row Count 26 (+ 9)
% Row 6
\SetRowColor{LightBackground}
One-to-one & If each b in the {\bf{codomain}} is only the image {\bf{at most}} one x in the {\bf{domain}}. Solution Uniqueness. \tn 
% Row Count 32 (+ 6)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Linear Transformations (cont)}}  \tn
% Row 7
\SetRowColor{LightBackground}
 & T is one-to-one if and only if the cols of A are {\bf{linearly independent}} \tn 
% Row Count 4 (+ 4)
% Row 8
\SetRowColor{white}
Free variable? & If the system has a free variable, then the system is {\bf{not}} one-to-one. I.e. the homogenous system only has the trivial solution \tn 
% Row Count 11 (+ 7)
% Row 9
\SetRowColor{LightBackground}
Pivot in every row? & Then T is {\bf{onto}} \tn 
% Row Count 12 (+ 1)
% Row 10
\SetRowColor{white}
Pivot in every column? & Then T is {\bf{one-to-one}} \tn 
% Row Count 14 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{To determine whether a vector {\bf{c}} is in the range of a {\bf{T}}. {\bf{Solution}}: Let T(x) = Ax. Solve the matrix equation Ax = c. If the system is {\bf{consistent}}, then c is in the range of T.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{The Invertible Matrix Theorem}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{ The following statements are equivalent i.e. either they are all true or all false. Let A be a ({\emph{n}}x{\emph{n}}) matrix}}} \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{𝐴 is an invertible matrix.} \tn 
% Row Count 4 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{𝐴 is row equivalent to the 𝑛×𝑛 identity matrix} \tn 
% Row Count 6 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{𝐴 has 𝑛 pivot positions.} \tn 
% Row Count 7 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{The equation 𝐴𝐱 = 𝟎 has only the trivial solution} \tn 
% Row Count 9 (+ 2)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{The columns of 𝐴 form a linearly independent set} \tn 
% Row Count 11 (+ 2)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{The linear transformation 𝐱 ↦ 𝐴𝐱 is one-to-one.} \tn 
% Row Count 13 (+ 2)
% Row 7
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{The equation 𝐴𝐱 = 𝐛 has at least one solution for each 𝐛 in R\textasciicircum{}n\textasciicircum{}} \tn 
% Row Count 15 (+ 2)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{The columns of 𝐴 span R\textasciicircum{}n\textasciicircum{}} \tn 
% Row Count 16 (+ 1)
% Row 9
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{The linear transformation 𝐱 ↦ 𝐴𝐱 maps R\textasciicircum{}n\textasciicircum{} onto R\textasciicircum{}n\textasciicircum{}} \tn 
% Row Count 18 (+ 2)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{There is an 𝑛×𝑛 matrix 𝐶 such that 𝐶𝐴 = 𝐼.} \tn 
% Row Count 20 (+ 2)
% Row 11
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{There is an 𝑛×𝑛 matrix 𝐷 such that 𝐴𝐷 = 𝐼.} \tn 
% Row Count 22 (+ 2)
% Row 12
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{𝐴\textasciicircum{}T\textasciicircum{} is an invertible matrix.} \tn 
% Row Count 23 (+ 1)
% Row 13
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{The columns of A form a basis of R\textasciicircum{}n} \tn 
% Row Count 24 (+ 1)
% Row 14
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Col A = R\textasciicircum{}n} \tn 
% Row Count 25 (+ 1)
% Row 15
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Dim Col A = n} \tn 
% Row Count 26 (+ 1)
% Row 16
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Rank A = n} \tn 
% Row Count 27 (+ 1)
% Row 17
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Nul A = \{0\}} \tn 
% Row Count 28 (+ 1)
% Row 18
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Dim Nul A = 0} \tn 
% Row Count 29 (+ 1)
% Row 19
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{The number 0 is not an eigenvalue of A} \tn 
% Row Count 30 (+ 1)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{The Invertible Matrix Theorem (cont)}}  \tn
% Row 20
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{The determinant of A is not 0} \tn 
% Row Count 1 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Elementary Matrices}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Elementary Matrix}} Is obtained by performing a single elementary row operation on an {\bf{identity matrix}}} \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Each elementary matrix {\bf{E}} is {\bf{invertible}}} \tn 
% Row Count 4 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{A nxn matrix A is invertible if and only if A is row equivalent to I\textasciicircum{}n\textasciicircum{}.} \tn 
% Row Count 6 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{A = E\textasciicircum{}-1\textasciicircum{}I\textasciicircum{}n\textasciicircum{} and A\textasciicircum{}-1\textasciicircum{} = EI\textasciicircum{}n\textasciicircum{}= I\textasciicircum{}n\textasciicircum{}} \tn 
% Row Count 7 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Row reduce the augmented matrix {[} A I {]} to {[} I A\textasciicircum{}-1\textasciicircum{}{]} \{\{nl\}\} {\bf{NOTE}} If A is not row equivalent to I then A is not invertible} \tn 
% Row Count 10 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Linear Independence}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{A set of vectors are {\bf{linearly independent}} if they cannot be created by any linear combinations of earlier vectors in the set.} \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{If a set of vectors are {\bf{linear independent}}, then the solution is {\bf{unique}}} \tn 
% Row Count 5 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{If the vector equeation c1{\bf{{\emph{v}}1 + c2{\bf{}}v}}2 + ... + cp{\bf{*v}}p = 0 only has a {\bf{trivial solution}} the set of vectors are {\bf{linearly independent}}} \tn 
% Row Count 8 (+ 3)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Theorem}}: If a set contains more vectors than there are entries in each vector, then the set is {\bf{linearly dependent}}} \tn 
% Row Count 11 (+ 3)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Theorem}}: If a set of vectors containt the zero vector, then the set is {\bf{linearly dependent}}} \tn 
% Row Count 13 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.89126 cm} x{3.08574 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Algebraic properties of a matrix}}  \tn
% Row 0
\SetRowColor{LightBackground}
Matrix and vector sum & A({\bf{u}} + {\bf{v}}) = A{\bf{u}} + A{\bf{v}} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
Matrix, vector and scalar & A({\emph{c{\bf{}}u}}) = {\emph{c}}(A{\bf{u}}) \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
Associative law & A(BC) = (AB)C \tn 
% Row Count 5 (+ 1)
% Row 3
\SetRowColor{white}
Left distributive law & A (B + C) = AB + AC \tn 
% Row Count 7 (+ 2)
% Row 4
\SetRowColor{LightBackground}
Right distributive law & (B + C) = BA + BC \tn 
% Row Count 9 (+ 2)
% Row 5
\SetRowColor{white}
Scalar multiplication & r(AB) = (rA)B = A(rB) \tn 
% Row Count 11 (+ 2)
% Row 6
\SetRowColor{LightBackground}
Identity matrix multi & I\textasciicircum{}m\textasciicircum{}A = A = AI\textasciicircum{}n\textasciicircum{} \tn 
% Row Count 13 (+ 2)
% Row 7
\SetRowColor{white}
Commute & If AB = BA then we say that A and B  {\bf{commute}} with each others \tn 
% Row Count 16 (+ 3)
% Row 8
\SetRowColor{LightBackground}
 & (A\textasciicircum{}T\textasciicircum{})\textasciicircum{}T\textasciicircum{} = A \tn 
% Row Count 17 (+ 1)
% Row 9
\SetRowColor{white}
 & (A + B)\textasciicircum{}T\textasciicircum{} = A\textasciicircum{}T\textasciicircum{} + B\textasciicircum{}T\textasciicircum{} \tn 
% Row Count 18 (+ 1)
% Row 10
\SetRowColor{LightBackground}
 & (AB)\textasciicircum{}T\textasciicircum{} = B\textasciicircum{}T\textasciicircum{} A\textasciicircum{}T\textasciicircum{} \tn 
% Row Count 19 (+ 1)
% Row 11
\SetRowColor{white}
 & For any scalar r, (rA)\textasciicircum{}T\textasciicircum{} = rA\textasciicircum{}T\textasciicircum{} \tn 
% Row Count 21 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{LU Factorization}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Factorization}} of a matrix A is an equation that expresses A as a {\bf{product}} of two or more matrices: \{\{nl\}\} {\bf{Synthesis}}: BC = A \{\{nl\}\} {\bf{Analysis}}: A= BC} \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Assumption:}} A is a {\emph{m}}x{\emph{n}} matrix that can be row reduced without {\bf{interchanges}}} \tn 
% Row Count 6 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{L}}: is a {\emph{m}}x{\emph{m}} unit lower triangular with 1's on the diagonal} \tn 
% Row Count 8 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{U}}: is a {\emph{m}}x{\emph{n}} echelon form of A} \tn 
% Row Count 9 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{U}} is equal to {\bf{E*A = U}}, why {\bf{A = E\textasciicircum{}-1\textasciicircum{}U}} = LU where {\bf{L}} = {\bf{E\textasciicircum{}-1\textasciicircum{}}}} \tn 
% Row Count 11 (+ 2)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{See figure ** for how to find L and U} \tn 
% Row Count 12 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Find {\bf{x}} by first solving {\bf{Ly = b}} and then solving {\bf{Ux = y}}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.64241 cm} x{3.33459 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Row Reduction and Echelon forms}}  \tn
% Row 0
\SetRowColor{LightBackground}
Leading entry & A leading entry refers to the leftmost {\bf{non-zero}} entry in a row \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
Echelon form & Row equivalent systems can be reduced into {\bf{several different}} echelon forms \tn 
% Row Count 6 (+ 3)
% Row 2
\SetRowColor{LightBackground}
Reduced echelon form & A system is only row equivalent to {\bf{one}} REF \tn 
% Row Count 8 (+ 2)
% Row 3
\SetRowColor{white}
Forward phase & Reducing an augmented matrix A into an {\bf{echelon form}} \tn 
% Row Count 11 (+ 3)
% Row 4
\SetRowColor{LightBackground}
Backward phase & Reducing an augmented matrix A into a {\bf{reduced echelon form}} \tn 
% Row Count 14 (+ 3)
% Row 5
\SetRowColor{white}
Basic variables & Variables in {\bf{pivot columns}}. \tn 
% Row Count 16 (+ 2)
% Row 6
\SetRowColor{LightBackground}
Free variables & Variables that are not in {\bf{pivot columns}}. When a system has a free variable the system is {\bf{consistent}} but not {\bf{unique}} \tn 
% Row Count 21 (+ 5)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Subspaces of R\textasciicircum{}n}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{A {\bf{subspace}} of R\textasciicircum{}n\textasciicircum{} is any set H in R\textasciicircum{}n\textasciicircum{} that has three properties:} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{- The zero vector is in H} \tn 
% Row Count 3 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{- For each {\bf{u}} and {\bf{v}} in H, the sum {\bf{u + v}} is in H} \tn 
% Row Count 5 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{- For each {\bf{u}} in {\emph{H}} and each scalar {\emph{c}}, the vector {\emph{c{\bf{}}u}} is in {\emph{H}}} \tn 
% Row Count 7 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Zero subspace}} is the set containing only the {\bf{zero vector}} in R\textasciicircum{}n\textasciicircum{}} \tn 
% Row Count 9 (+ 2)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Column space}} is the set of all linear combinations of the columns of A.} \tn 
% Row Count 11 (+ 2)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Null space (Nul A)}} is the set of all solutions of the equation {\bf{Ax = 0}}} \tn 
% Row Count 13 (+ 2)
% Row 7
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Basis}} for a subspace H is the set of {\bf{linearly independent}} vectors that span H} \tn 
% Row Count 15 (+ 2)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{In general, the {\bf{pivot columns}} of A form a basis for col A} \tn 
% Row Count 17 (+ 2)
% Row 9
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{The number of vectors in any basis is {\bf{unique}}. We call this number {\bf{dimension}}} \tn 
% Row Count 19 (+ 2)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{The {\bf{rank}} of a matrix 𝐴, denoted by {\bf{rank 𝐴}}, is the {\bf{dimension}} of the {\bf{column space}} of 𝐴} \tn 
% Row Count 22 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Determine whether b is in the {\bf{col A}}. {\bf{Solution}}: b is only in col A if the equation {\bf{Ax = b}} has a solution}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Page break}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{} \tn 
% Row Count 0 (+ 0)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Algebraic properties of a vector}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{u}} + {\bf{v}} = {\bf{v}} + {\bf{u}}} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{({\bf{u}} + {\bf{v}}) + {\bf{w}} = {\bf{u}} + ({\bf{v}} + {\bf{w}})} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{u}} + (-{\bf{u}}) = -{\bf{u}} + {\bf{u}}} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{c}}({\bf{u}} + {\bf{v}}) = {\emph{c{\bf{}}u}} + {\emph{c{\bf{}}v}}} \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{c}}({\emph{d{\bf{}}u}}) = ({\emph{cd}}){\bf{u}}} \tn 
% Row Count 5 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}