\begin{theorem} Main Theorem for Invertible Matrices \\ Let $A$ be a square matrix. Then the following are equivalent \begin{enumerate} \item $A$ is an invertible matrix. \item Linear System $Ax = b$ has a unique solution \item Linear System $Ax = 0$ has only the trivial solution \item RREF of $A$ is $I$ \item A is the product of elementary matrices \end{enumerate} \end{theorem} \begin{theorem} Find Inverse \begin{itemize} \item Let $A$ be an invertible Matrix. \item RREF of $(A | I)$ is $(I | A^{-1})$ \end{itemize} How to identify if Square Matrix is invertible? \begin{itemize} \item Square matrix is invertible \subitem $\iff$ RREF is $I$ \subitem $\iff$ All columns in its REF are pivot \subitem $\iff$ All rows in REF are nonzero \item Square matrix is singular \subitem $\iff$ RREF is \textbf{NOT} $I$ \subitem $\iff$ Some columns in its REF are non-pivot \subitem $\iff$ Some rows in REF are zero. \item $A$ and $B$ are square matrices such that $AB = I$ \subitem then $A$ and $B$ are invertible \end{itemize} \end{theorem} \begin{defn}[LU Decomposition with Type 3 Operations]\ \\ \begin{itemize} \item Type 3 Operations: $(R_i + cR_j, i > j)$ \item Let $A$ be a $m \times n$ matrix. Consider Gaussian Elimination $A \dashrightarrow R$ \item Let $R \dashrightarrow A$ be the operations in reverse \item Apply the same operations to $I_m \dashrightarrow L$. Then $A = LR$ \item $L$ is a \hyperref[def:ltm]{lower triangular matrix} with 1 along diagonal \item If $A$ is square matrix, $R = U$ \end{itemize} Application: \begin{itemize} \item $A$ has LU decomposition $A = LU$, $Ax = b$ i.e., $LUx = b$ \item Let $y = Ux$, then it is reduced to $Ly = b$ \item $Ly = b$ can be solved with forward substitution. \item $Ux = y$ is the REF of A. \item $Ux = y$ can be solved using backward substitution. \end{itemize} \end{defn} \begin{defn}[LU Decomposition with Type II Operations]\ \\ \begin{itemize} \item Type 2 Operations: $(R_i \leftrightarrow R_j)$, where 2 rows are swapped \item $A \xrightarrow[]{E_1} \bullet \xrightarrow[]{E_2}\bullet \xrightarrow[E_3]{R_i \iff R_j}\bullet \xrightarrow[]{E_4}\bullet \xrightarrow[]{E_5} R$ \item $A = E^{-1}_1E^{-1}_2E^{}_3E^{-1}_4E^{-1}_5R$ \item $E_3A = (E_3E^{-1}_1E^{-1}_2E_3)E^{-1}_4E^{-1}_5R$ \item $P = E_3, L = (E_3E^{-1}_1E^{-1}_2E_3)E^{-1}_4E^{-1}_5, R = U$, $PA = LU$ \end{itemize} \end{defn} \begin{defn}[Column Operations]\ \\ \begin{itemize} \item Pre-multiplication of Elementary matrix $\iff$ Elementary row operation \subitem $A \to B \iff B = E_1E_2...E_kA$ \item Post-Multiplication of Elementary matrix $\iff$ Elementary Column Operation \subitem $A \to B \iff B = AE_1E_2...E_k$ \item If $E$ is obtained from $I_n$ by single elementary column operation, then \subitem $I \xrightarrow[]{kC_i}E \iff I \xrightarrow[]{kR_i}E$ \subitem $I \xrightarrow[]{C_i \leftrightarrow C_j}E \iff I \xrightarrow[]{R_i \leftrightarrow R_j}E$ \subitem $I \xrightarrow[]{C_i + kC_j}E \iff I \xrightarrow[]{R_j + kR_i}E$ \end{itemize} \end{defn} \subsection{Determinants} \begin{defn}[Determinants of $2 \times 2$ Matrix]\ \\ \begin{itemize} \item Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ \item $\det(A) = |A| = ad - bc$ \item $\det(I_2) = 1$ \item $A \xrightarrow{cR_i} B \implies \det(B) = c\det(A)$ \item $A \xrightarrow{R_1 \leftrightarrow R_2} B \implies \det(B) = -\det(A)$ \item $A \xrightarrow{R_i + cR_j} B \implies \det(B) = \det(A), i \neq j$ \end{itemize} Solving Linear equations with determinants for $2 \times 2$ \begin{itemize} \item $x_1 = \dfrac{\begin{vmatrix} b_1 & a_{12} \\ b_2 & a_{22} \end{vmatrix}} {\begin{vmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}}$, $x_2 = \dfrac{\begin{vmatrix} a_{11} & b_1 \\ a_{21} & b_2 \end{vmatrix}} {\begin{vmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}}$ \end{itemize} \end{defn} \begin{defn}[Determinants of $3 \times 3$ Matrix]\ \\ \begin{itemize} \item Suppose $A$ is invertible, then there exists EROs such that \item $A \xrightarrow{ero_1} A_1 \rightarrow ... \rightarrow A_{k-1} \xrightarrow{ero_k}A_k = I$ \item Then $\det(A)$ can be evaluated backwards. \subitem E.g. $A \xrightarrow{R_1 \leftrightarrow R_3} \bullet \xrightarrow{3R_2} \bullet \xrightarrow{R_2 + 2R_4} I \implies det(A) = 1 \to 1 \to \frac{1}{3} \to -\frac{1}{3}$ \end{itemize} \end{defn}