\input{decls.tex} \title{Linear Algebra} \begin{document} \maketitle \section*{Allowable Operations on a Linear System} Solutions invariant. \begin{itemize} \item Multiply an equation by a non-zero scalar \item Swap two equations \item Add a multiple of one equation to another \end{itemize} \subsection*{Example} \begin{align*} &\systeme{ x - 2y + 2z = 6, -x + 3y + 4z = 2, 2x + y - 2z = -2 }\\\\ E_2 & \implies E_2 + E_1 \\ E_3 & \implies E_3 + E_1 \\ &\systeme{ x - 2y + 2z = 6, y + 6z = 8, 5y - 6z = -14 }\\\\ E_3 & \implies E_3 - 5E_2 \\ &\systeme{ x - 2y + 2z = 6, y + 6z = 8, z = \frac{3}{2} }\\\\ E_1 & \implies E_1 - 2E_3 \\ E_2 & \implies E_2 - 6E_3 \\ &\systeme{ x - 2y = 3, y = -1, z = \frac{3}{2} }\\\\ E_1 & \implies E_1 + 2E_2 \\ &\systeme{ x = 1, y = -1, z = \frac{3}{2} }\\\\ \end{align*} \section*{As Matrices} \begin{align*} \systeme{ x + 2y = 1, 2x - y = 3 } \quad=\quad \begin{pmatrix}[cc|c] 1 & 2 & 1 \\ 2 & -1 & 3 \end{pmatrix} & \systeme{ x - y + z = -2, 2x + 3y + z = 7, x - 2y - z = -2 } \quad=\quad \begin{pmatrix}[ccc|c] 1 & -1 & 1 & -2 \\ 2 & 3 & 1 & 7 \\ 1 & -2 & -1 & -2 \end{pmatrix} \\ \grstep[R_3 - R_1]{R_2 - 2R_1} & \begin{pmatrix}[ccc|c] 1 & -1 & 1 & -2 \\ 0 & 5 & -1 & 11 \\ 0 & -1 & -2 & 0 \end{pmatrix} \\ \grstep{5R_3 + R_2} & \begin{pmatrix}[ccc|c] 1 & -1 & 1 & -2 \\ 0 & 5 & -1 & 11 \\ 0 & 0 & -11 & 11 \\ \end{pmatrix} \\ \grstep{-11^{-1}R_3} & \begin{pmatrix}[ccc|c] 1 & -1 & 1 & -2 \\ 0 & 5 & -1 & 11 \\ 0 & 0 & 1 & -1 \end{pmatrix} \\ \grstep[R_1 - R_3]{R_2 + R_3} & \begin{pmatrix}[ccc|c] 1 & -1 & 0 & -1 \\ 0 & 5 & 0 & 10 \\ 0 & 0 & 1 & -1 \end{pmatrix} \\& \grstep{5^{-1}R_2} & \begin{pmatrix}[ccc|c] 1 & -1 & 0 & -1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \\ \end{pmatrix} \\ \grstep{R_1 + R_2} & \begin{pmatrix}[ccc|c] 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \end{pmatrix} \\ = & \quad \left\{ \subalign{ x & ~= ~1 \\ y & ~= ~2 \\ z & ~= ~-1 } \right. \end{align*} \section*{Row-Echelon Form} \begin{description} \item[Row-Echelon Form] The leading entry in each row is 1 and is further to the right than the previous row's leading entry, all 0 rows are at the end \item[Reduced Row-Echelon Form] every other entry in a column containing a leading 1 is 0 \item[Theorem:] A matrix can be transformed to reduced row-echelon form using a finite number of allowable row operations \end{description} \subsection*{Example} \begin{align*} & \systeme{3x_1 + 2x_2 = 1, x_1 - x_2 = 4, 2x_1 + x_2 = 5} = \begin{pmatrix}[cc|c] 3 & 2 & 1 \\ 1 & -1 & 4 \\ 2 & 1 & 5 \end{pmatrix} \\ \grstep{R_1\swap R_2} & \begin{pmatrix}[cc|c] 1 & -1 & 4 \\ 3 & 2 & 1 \\ 2 & 1 & 5 \end{pmatrix} \\ \grstep[R_2 - 3R_1]{R_3 - 2R_1} & \begin{pmatrix}[cc|c] 1 & -1 & 4 \\ 0 & 5 & -11 \\ 0 & 3 & -3 \end{pmatrix} \\ \grstep{5^{-1}R_2} & \begin{pmatrix}[cc|c] 1 & -1 & 4 \\ 0 & 1 & \frac{-11}{5} \\ 0 & 3 & -3 \end{pmatrix} \\ \grstep{R_3 - 2R_2} & \begin{pmatrix}[cc|c] 1 & -1 & 4 \\ 0 & 1 & \frac{-11}{5} \\ 0 & 0 & \frac{18}{5} \end{pmatrix} \\ = & \systeme{ x_1 - x_2 = 4, x_2 = \frac{-11}{5}, 0x_1 + 0x_2 = \frac{18}{5} } \end{align*} \begin{align*} & \begin{pmatrix}[cccc|c] 1 & -1 & 1 & 1 & 6 \\ -1 & 1 & -2 & 1 & 3 \\ 2 & 0 & 1 & 4 & 1 \\ \end{pmatrix} \\ \grstep[R_2 + R_1]{R_3 - 2R_1} & \begin{pmatrix}[cccc|c] 1 & -1 & 1 & 1 & 6 \\ 0 & 0 & -1 & 2 & 9 \\ 0 & 2 & -1 & 2 & -11 \end{pmatrix} \\ \grstep[R_2\swap R_3]{2^{-1}R_3} & \begin{pmatrix}[cccc|c] 1 & -1 & 1 & 1 & 6 \\ 0 & 1 & \frac{1}{2} & 1 & \frac{-11}{2} \\ 0 & 0 & -1 & 2 & 9 \\ \end{pmatrix} \\ \grstep[R_1 + R_3]{R_2 - 2^{-1}R_3} & \begin{pmatrix}[cccc|c] 1 & -1 & 0 & 3 & 15 \\ 0 & 1 & 0 & 0 & -10 \\ 0 & 0 & -1 & 2 & 9 \\ \end{pmatrix} \\ \grstep[-R_3]{R_1 + R_2} & \begin{pmatrix}[cccc|c] 1 & 0 & 0 & 3 & 15 \\ 0 & 1 & 0 & 0 & -10 \\ 0 & 0 & 1 & -2 & -9 \\ \end{pmatrix} \\ = & \systeme{ x_1 + 3x_4 = 5, x_2 = -10, x_3 - 2x_4 = -9 } \\ = & \left\{\substack{ x_1 = 5 - 3t \\ x_2 = -10 \\ x_3 = -9 + 2t }\right. \end{align*} \section*{Determinants} The determinant of a matrix is defined only for square matrices. \[\det{A} \neq 0 \iff \exists \text{ a unique solution to the linear system represented by } A\] Let \[A = \begin{pmatrix} a_{11} & a_{12} & a_{1n} \\ a_{21} & \ddots & \vdots \\ a_{31} & \ldots & a_{3n} \\ \end{pmatrix} \] \begin{description} \item[$i, j$ minor of $A$] an $n$x$n$ matrix constructed by removing the $i^\text{th}$ row and $j^\text{th}$ column of $A$ \\ Denoted by $A_{ij}$ \end{description} \begin{align*} & \det{A} \text{ where } n = 1. = a_{11} \\ & \det{A} = a_{11}\det{A_{11}} - a_{12}\det{A_{12}} + ... + (-1)^{n+1}a_{1n} \tag{Laplace expansion of the first row} \\ & \qquad \text{or laplace expansion along other row or column} \text{For } n = 2:& \\ & \det{A} = a_{11}\cdot a_{22} - a_{12}\cdot a_{21} \end{align*} \begin{description} \item[Upper Triangular] lower left triangle is 0 - $d_{ij} = 0 \quad \forall{i > j}$ \item[Lower Triangular] upper right triangle is 0 - $d_{ij} = 0 \quad \forall{i < j}$ \item[Diagonal] only values on the diagonal - $d_{ij} = 0 \quad \forall{i \neq j}$ \\ $\det{A} = \prod^{N}_{i=0}~a_{ij} \forall~\text{ row-echelon }A$ \end{description} \begin{itemize} \item Multiplying a row of a square matrix $A$ by $r$ multiplies $\det{A}$ by $r$ \item Swapping two rows of a square matrix $A$ multiplies $\det{A}$ by $-1$ \item Adding a multiple of a row does not effect the determinant \end{itemize} \section*{Transposition} \begin{description} \item[$A^T$] $a^T_{ij} = a_{ji}~ \forall~i,j$ \end{description} Note: $\det{A} = \det{A^T}~\forall~A$ \section*{Matrix Multiplication} LHS has columns $=$ rows of RHS It's the cartesian product \[A\times B = (a_{i1}b_{j1} + a_{i2}b_{2j} + \ldots + a_{im}b_{mj})_{ij}\] \begin{align*} \begin{pmatrix}[c|c|c] 2 & 1 + 1 & 3 + 6 \\ 4(2) & 4 + 1 & 3(4) + 6 \\ 0 & 2 & 2(6) \\ \end{pmatrix} = \begin{pmatrix} 2 & 2 & 9 \\ 8 & 5 & 18 \\ 0 & 2 & 12 \end{pmatrix} \end{align*} \begin{align*} \begin{pmatrix}1 \\ 2 \\ 3 \end{pmatrix}\begin{pmatrix}1 & 2 & 3 & 4\end{pmatrix} + \begin{pmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ \end{pmatrix} \end{align*} \end{document}