\subsection{Introduction} \begin{defn}[Matrix]\ \\ \begin{itemize} \item $\begin{pmatrix} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots \\ a_{m1} & a_{m2} & ... & a_{mn} \end{pmatrix}$ \item $m$ is no of rows, $n$ is no of columns \item size is $m \times n$ \item $A = (a_{ij})_{m \times n}$ \end{itemize} \end{defn} \subsubsection{Special Matrix} \begin{note}[Special Matrices]\ \\ \begin{itemize} \item Row Matrix : $\begin{pmatrix} 2 & 1 & 0 \end{pmatrix}$ \item Column Matrix \subitem $\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ \item \textbf{Square Matrix}, $n \times n$ matrix / matrix of order $n$. \subitem Let $A = (a_{ij})$ be a square matrix of order $n$ \item Diagonal of $A$ is $a_{11}, a_{22}, ..., a_{nn}$. \item \textbf{Diagonal Matrix} if Square Matrix and non-diagonal entries are zero \subitem Diagonals can be zero \subitem \textbf{Identity Matrix} is a special case of this \item \textbf{Square Matrix} if Diagonal Matrix and diagonal entries are all the same. \item \textbf{Identity Matrix} if Scalar Matrix and diagonal = 1 \subitem $I_n$ is the identity matrix of order $n$. \item \textbf{Zero Matrix} if all entries are 0. \subitem Can denote by either $\overrightarrow{0}, 0$ \item Square matrix is \textbf{symmetric} if symmetric wrt diagonal \subitem $A = (a_{ij})_{n \times n}$ is symmetric $\iff a_{ij} = a_{ji},\ \forall i, j$ \item \textbf{Upper Triangular} if all entries \textbf{below} diagonal are zero. \subitem $A = (a_{ij})_{n \times n}$ is upper triangular $\iff a_{ij} = 0 \text{ if } i > j$ \item \textbf{Lower Triangular} if all entries \textbf{above} diagonal are zero. \label{def:ltm} \subitem $A = (a_{ij})_{n \times n}$ is lower triangular $\iff a_{ij} = 0 \text{ if } i < j$ \subitem if Matrix is both Lower and Upper triangular, its a Diagonal Matrix. \end{itemize} \end{note} \subsection{Matrix Operations} \begin{defn}[Matrix Operations]\ \\ Let $A = (a_{ij})_{m \times n}, B = (b_{ij})_{m \times n}$ \begin{itemize} \item Equality: $B = (b_{ij})_{p \times q}$, $A = B \iff m = p \ \& \ n = q \ \& \ a_{ij} = b_{ij} \forall i,j$ \item Addition: $A + B = (a_{ij} + b_{ij})_{m \times n}$ \item Subtraction: $A - B = (a_{ij} - b_{ij})_{m \times n}$ \item Scalar Mult: $cA = (ca_{ij})_{m \times n}$ \end{itemize} \end{defn} \begin{defn}[Matrix Multiplication] \ \\ Let $A = (a_{ij})_{m \times p}, B = (b_{ij})_{p \times n}$ \begin{itemize} \item $AB$ is the $m \times n$ matrix s.t. $(i,j)$ entry is $$a_{i1}b_{1j} + a_{i2}b_{2j} + ... + a_{ip}b_{pj} = \sum^p_{k=1}a_{ik}b_{kj}$$ \item No of columns in $A$ = No of rows in $B$. \item Matrix multiplication is \textbf{NOT commutative} \end{itemize} \end{defn} \begin{theorem}[Matrix Properties]\ \\ Let $A, B, C$ be $m \times p, p \times q, q \times n$ matrices \begin{itemize} \item Associative Law: $A(BC) = (AB)C$ \item Distributive Law: $A(B_1 + B_2) = AB_1 + AB_2$ \item Distributive Law: $(B_1 + B_2)A = B_1A + B_2A$ \item $c(AB) = (cA)B = A(cB)$ \item $A\textbf{0}_{p \times n} = \textbf{0}_{m \times n}$ \item $A\textbf{I}_{n} = \textbf{I}_{n}A = A$ \end{itemize} \end{theorem} \begin{defn}[Powers of Square Matricss]\ \\ Let $A$ be a $m \times n$. $AA$ is well defined $\iff m = n \iff A$ is square. \textbf{Definition.} Let $A$ be square matrix of order $n$. Then Powers of a are $$ A^k = \begin{cases} I_n & \text{if } k = 0 \\ AA...A & \text{if } k \geq 1. \end{cases} $$ \textbf{Properties.} \begin{itemize} \item $A^mA^n = A^{m+n}, (A^m)^n = A^{mn}$ \item $(AB)^2 = (AB)(AB) \neq A^2B^2 = (AA)(BB)$ \end{itemize} \end{defn} Matrix Multiplication Example: \begin{itemize} \item Let $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 2 & 3 \\ -1 & -2 \end{pmatrix}$ \item Let $a_1 = \begin{pmatrix}1 & 2 & 3 \end{pmatrix}, a_2 = \begin{pmatrix}4 & 5 & 6 \end{pmatrix}$ \item $AB = \begin{pmatrix} a_1 & a_2 \end{pmatrix}B = \begin{pmatrix} a_1B \\ a_2B \end{pmatrix}$. \item $\begin{pmatrix} \begin{pmatrix}1 & 2 & 3 \end{pmatrix} & \begin{pmatrix} 1 & 1 \\ 2 & 3 \\ -1 & -2 \end{pmatrix} \\ \begin{pmatrix}4 & 5 & 6 \end{pmatrix} & \begin{pmatrix} 1 & 1 \\ 2 & 3 \\ -1 & -2 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} \begin{pmatrix}2 & 1\end{pmatrix} \\ \begin{pmatrix}8 & 7\end{pmatrix} \\ \end{pmatrix} $ \end{itemize} \begin{note}[Representation of Linear System] \ \\ \begin{itemize} \item $\begin{cases} a_{11}x_1 + a_{12}x_2 + ... + a_{1n}x_n & = b_1 \\ a_{21}x_1 + a_{22}x_2 + ... + a_{2n}x_n & = b_2 \\ \vdots & \vdots \\ a_{m1}x_1 + a_{m2}x_2 + ... + a_{mn}x_n & = b_m \\ \end{cases}$ \item A = $\begin{pmatrix} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & ... & a_{mn} \\ \end{pmatrix}$, Coefficient Matrix, $A_{m\times n}$ \item $x = \begin{pmatrix} x_{1} \\ \vdots \\ x_{n} \\ \end{pmatrix}$, Variable Matrix, $x_{n \times 1}$ \item $b = \begin{pmatrix} b_{1} \\ \vdots \\ b_{m} \\ \end{pmatrix}$, Constant Matrix, $b_{m \times 1}$. Then $Ax = b$ \item $A = (a_{ij})_{m\times n} $ \item $m$ linear equations in $n$ variables, $x_1, ..., x_n$ \item $a_{ij}$ are coefficients, $b_i$ are the constants \item Let $u = \begin{pmatrix} u_1 \\ \vdots \\ u_n \end{pmatrix}$. \subitem $x_1 = u_1, \hdots, x_n = u_n$ is a solution to the system \subitem $\iff Au = b \iff u$ is a solution to $Ax = b$ \item Let $a_j$ denote the $j$th column of $A$. Then \subitem $b = Ax = x_1a_1 + ... + x_na_n = \sum^n_{j=1}x_ja_j$ \end{itemize} \end{note} \begin{defn}[Transpose]\ \\ \begin{itemize} \item Let $A = (a_{ij})_{m\times n}$ \item The transpose of $A$ is $A^T = (a_{ji})_{n \times m}$ \item $(A^T)^T = A$ \item A is symmetric $\iff A = A^T$ \item Let $B$ be $m \times n$, $(A+B)^T = A^T + B^T$ \item Let $B$ be $n \times p$, $(AB)^T = B^TA^T$ \end{itemize} \end{defn} \begin{defn}[Inverse]\ \\ \begin{itemize} \item Let $A, B$ be matrices of same size \subitem $A + X = B \implies X = B - A = B + (-A)$ \subitem $-A$ is the \textit{additive inverse} of $A$ \item Let $A_{m\times n}, B_{m\times p}$ matrix. \subitem $AX = B \implies X = A^{-1}B$. \end{itemize} Let A be a \textbf{square matrix} of order $n$. \begin{itemize} \item If there exists a square matrix $B$ of order $N$ s.t. $AB = I_{n}$ and $BA = I_{n}$, then $A$ is \textbf{invertible} matrix and $B$ is inverse of $A$. \item If $A$ is not invertible, A is called singular. \item suppose $A$ is invertible with inverse $B$ \item Let $C$ be any matrix having the same number of rows as $A$. $$\begin{aligned} AX = C &\implies B(AX) = BC \\ &\implies (BA)X = BC \\ &\implies X = BC. \end{aligned}$$ \end{itemize} \end{defn} \begin{theorem}[Properties of Inversion]\ \\ Let $A$ be a square matrix. \begin{itemize} \item Let $A$ be an invertible matrix, then its inverse is unique. \item Cancellation Law: Let $A$ be an invertible matrix \subitem $AB_1 = AB_2 \implies B_1 = B_2$ \subitem $C_1A = C_2A \implies C_1 = C_2$ \subitem $AB = 0 \implies B = 0, CA = 0 \implies C = 0$ ($A$ is invertible, A cannot be 0) \subitem This fails if $A$ is singular \item Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ \subitem $A$ is invertible $\iff ad - bc \neq 0$ \subitem $A$ is invertible $A^{-1} = \dfrac{1}{ad - bc} \begin{pmatrix}d & -b \\ -c & a \end{pmatrix}$ \end{itemize} Let $A$ and $B$ be invertible matrices of same order \begin{itemize} \item Let $c \neq 0$. Then $cA$ is invertible, $(cA^{-1} = \frac{1}{c}A^{-1}$ \item $A^T$ is invertible, $(A^T)^{-1} = (A^{-1})^T$ \item $AB$ is invertible, $(AB)^{-1} = (B^{-1}A^{-1})$ \end{itemize} Let $A$ be an invertible matrix. \begin{itemize} \item $A^{-k} = (A^{-1})^k$ \item $A^{m+n} = A^mA^n$ \item $(A^m)^n = A^{mn}$ \end{itemize} \end{theorem} \begin{defn}[Elementary Matrices] If it can be obtained from $I$ by performing single elementary row operation \begin{itemize} \item $cRi, c \neq 0: \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}(cR_3)$ \item $R_i \leftrightarrow R_j, i \neq j,: \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}(R_2 \leftrightarrow R_4)$ \item $R_i + cR_j, i \neq j,: \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & c \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}(R_2 + cR_4)$ \item Every elementary Matrix is invertible \end{itemize} \end{defn} $A = \begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ a_{41}&a_{42}&a_{43} \end{pmatrix}$, $E = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&c&0\\ 0&0&0&1 \end{pmatrix}(cR_3)$, $EA = \begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ ca_{31}&ca_{32}&ca_{33}\\ a_{41}&a_{42}&a_{43} \end{pmatrix}$ \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 $\det(A) = |A| = ad - bc$ \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]\ \\ \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. $A \xrightarrow{R_1 \leftrightarrow R_3} \bullet \xrightarrow{3R_2} \bullet \xrightarrow{R_2 + 2R_4} I \implies det(A) = 1 \to 1 \to \frac{1}{3} \to -\frac{1}{3}$ \item Let $M_{ij}$ be submatrix where the $i$th row and $j$th column are deleted \item Let $A_{ij} = (-1)^{i+j}\det(M_{ij})$, which is the $(i, j)$-cofactor \item $\det(A) = a_{11}A_{11} + a_{12}A_{12} + ... + a_{1n}A_{1n}$ \item $\det(I) = 1$ \item $A \xrightarrow{cR_i} B \implies \det(B) = c\det(A)$ \subitem $I \xrightarrow{cR_i} E \implies \det(E) = c$ \item $A \xrightarrow{R_1 \leftrightarrow R_2} B \implies \det(B) = -\det(A)$ \subitem $I \xrightarrow{R_1 \leftrightarrow R_2} E \implies \det(E) = -1$ \item $A \xrightarrow{R_i + cR_j} B \implies \det(B) = \det(A), i \neq j$ \subitem $I \xrightarrow{R_i + cR_j} E \implies \det(E) = 1$ \item $\det(EA) = \det(E)\det(A)$ \end{itemize} Calculating determinants easier \begin{itemize} \item Let $A$ be square matrix. Apply Gaussian Elimination to get REF $R$ \item $A \xrightarrow{E_1} \bullet \xrightarrow{E_2} \bullet ... \bullet \xrightarrow{E_k} R$ \item $A \xleftarrow{E^{-1}_1} \bullet \xleftarrow{E^{-1}_2} \bullet ... \bullet \xleftarrow{E^{-1}_k} R$ \item Since $E_i$ and $E^{-1}_k$ is type $II$ or $III$, $\det(E_i) = -1 / 1$ \subitem $\det(A) = (-1)^t\det(R)$, where $t$ is no of type $II$ or $III$ operations \item If $A$ is singluar, then $R$ has a zero row, and then $det(A) = 0$ \item If A is invertible, then all rows of $R$ are nonzero \subitem $\det(R) = a_{11}a_{22}...{a_nn}$, the product of diagonal entries. \end{itemize} \end{defn} \subsection{Recap} \begin{itemize} \item If A has a REF \subitem If there is a zero row => Singular matrix \subitem All rows are nonzero => invertible Matrix \item If A is invertible, Using Gauss Jordan Elim $(A | I) \to (I | A^{-1})$ \item \end{itemize} \subsection{More about Determinants} \begin{defn}[Determinant Properties]\ \\ $A$ is a Square Matrix \begin{itemize} \item $\det(A) = 0 \implies A$ is singular \item $\det(A) \neq 0 \implies A$ is invertible \item $\det(A) = \det(A^T)$ \item $\det(cA) = c^n\det(A)$, where $n$ is the order of the matrix \item If $A$ is triangular, $\det(A)$ product of diagonal entries \item $\det(AB) = \det(A)\det(B)$ \item $\det(A^{-1}) = [\det(A)]^{-1}$ \end{itemize} Cofactor Expansion: \begin{itemize} \item To eavluate determinant using cofactor expansion, expand row/column with most no of zeros. \end{itemize} \end{defn} \subsection{Finding Determinants TLDR} \begin{defn}[Finding Determinants]\ \\ \begin{itemize} \item If $A$ has zero row / column, $\det(A) = 0$ \item If $A$ is triangular, $det(A) = a_{11}a_{22}...a_{nn}$ \item If Order $n = 2 \to \det(A) = a_{11}a_{22} - a_{12}a_{21}$ \item If row/column has many 0, use cofactor expansion \item Use Gaussian Elimination to get REF \subitem $\det(A) = (-1)^t\det(R), t$ is no of type $II$ operations \end{itemize} \end{defn} \begin{defn}[Finding Inverse with Adjoint Matrix]\ \\ \begin{itemize} \item $\text{adj}(A) = (A_{ji})_{n\times n} = (A_{ij})^T_{n\times n}$ \item $A^{-1} = [\det(A)]^{-1}\text{adj}(A)$ \end{itemize} \end{defn} \begin{defn}[Cramer's Rule] Suppose $A$ is an invertible matrix of order $n$ \begin{itemize} \item Liner system $Ax = b$ has unique solution \item $x = \dfrac{1}{\det(A)}\begin{pmatrix}\det(A_1) \\ \vdots \\ det(A_n) \end{pmatrix}$, \item $A_j$ is obtained by replacing the $j$th column in $A$ with $b$. \end{itemize} \end{defn}