Invertible Matrices and Determinants: Principles

 # Relationship Between Matrix Operations and Inverses

  • Contextual Recap of Chapter Two     - In previous sessions, multiplication and addition were defined, alongside various properties like the transpose.     - Multiplication was defined specifically in terms of the dot product, necessitating a detour into Chapter 4 (vectors in Rn\mathbb{R}^n) and Section 5.1 (the dot product itself) to deepen the understanding of rows and columns.

  • Arithmetic Analogies in Addition     - Addition involves three key components: $A + B$, the additive identity (OO, the zero matrix), and the additive inverse (A-A).     - Fundamental relationships: A+O=O+A=AA + O = O + A = A and A+(A)=A+A=OA + (-A) = -A + A = O.     - Note on Universality: The additive inverse (A-A) exists for all matrices AA, regardless of whether they are square or rectangular.     - Note on Uniqueness: The zero matrix (OO) is not unique across the board; its size must match the matrix AA it is paired with.

  • Extending Analogies to Multiplication     - Multiplicative Identity (II): Multiplication is size-sensitive. If AA is an n×mn \times m matrix, the identity depends on the side of multiplication:         - Right side: A×Im=AA \times I_m = A (Requires an m×mm \times m identity).         - Left side: In×A=AI_n \times A = A (Requires an n×nn \times n identity).     - Multiplicative Inverse (A1A^{-1}): We seek a matrix such that combining it with AA yields the identity matrix II.

Defining and Validating Inverse Matrices

  • Theoretical Questions Regarding Inverses     - For a given matrix AA, are there matrices BB and CC such that BA=IBA = I and AC=IAC = I?     - Non-commutativity in multiplication suggests no immediate reason that BB must equal CC or that the resulting identities must be the same size.

  • Existence of Inverses     - Inverses for all AA: False. The zero matrix (OO) cannot have an inverse because any matrix multiplied by OO results in OO, never II.     - Inverses for some AA: True. For example, if A=IA = I, then I×I=II \times I = I.     - Inverses for none: False (proven by the existence of II).

  • Proof of Uniqueness (B=CB = C)     - Assume BA=IBA = I and AC=IAC = I.     - Start with BA=IBA = I and multiply both sides by CC on the right: (BA)C=IC(BA)C = IC.     - Using the associative property: B(AC)=CB(AC) = C.     - Since AC=IAC = I, then BI=CBI = C.     - Conclusion: B=CB = C. This means for any invertible matrix, the left inverse and right inverse are identical, denoted as A1A^{-1}.

  • Notation and Requirements     - The notation is A1A^{-1}, read as "A inverse."     - In order for AA1=A1A=IA A^{-1} = A^{-1} A = I to hold, the matrix AA must be square (n×nn \times n).

Methods for Finding Inverses

  • System of Equations Method (Inefficient)     - Example: Given A=(3amp;1 2amp;1)A = \begin{pmatrix} 3 & -1 \ 2 & 1 \end{pmatrix}, find A1=(aamp;b camp;d)A^{-1} = \begin{pmatrix} a & b \ c & d \end{pmatrix}.     - AA1=(3acamp;3bd 2a+camp;2b+d)=(1amp;0 0amp;1)A A^{-1} = \begin{pmatrix} 3a-c & 3b-d \ 2a+c & 2b+d \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}.     - This results in two separate $2 \times 2$ systems of equations:         1. 3ac=13a - c = 1 and 2a+c=02a + c = 0.         2. 3bd=03b - d = 0 and 2b+d=12b + d = 1.     - This can be solved via a $4 \times 4$ checkboard augmented matrix, but it is inefficient due to numerous zeros.

  • The Augmented Matrix Method ([AI][A | I])     - Based on the system Ax=bAx = b. If A1A^{-1} exists, multiplying both sides on the left yields: A1Ax=A1bA^{-1}Ax = A^{-1}b, resulting in Ix=A1bIx = A^{-1}b or x=A1bx = A^{-1}b.     - Process: Augment matrix AA with the identity II: [AI][A | I].     - Perform row operations until the left side is transformed into II.     - The right side will simultaneously transform into A1A^{-1}.     - Final form: [IA1][I | A^{-1}].

  • Elementary Matrices     - Every row operation can be represented as multiplication by an "Elementary Matrix" (EE).     - If kk operations are required to transform AA into II, then EkE2E1A=IE_k \dots E_2 E_1 A = I.     - Consequently, EkE2E1I=A1E_k \dots E_2 E_1 I = A^{-1}.

  • Example: Finding the inverse of a $3 \times 3$ matrix     - Matrix A=(1amp;2amp;0 1amp;1amp;1 0amp;0amp;1)A = \begin{pmatrix} 1 & 2 & 0 \ -1 & 1 & 1 \ 0 & 0 & 1 \end{pmatrix}.     - Set up augmented matrix: (1amp;2amp;0amp;amp;1amp;0amp;0 1amp;1amp;1amp;amp;0amp;1amp;0 0amp;0amp;1amp;amp;0amp;0amp;1)\begin{pmatrix} 1 & 2 & 0 & | & 1 & 0 & 0 \ -1 & 1 & 1 & | & 0 & 1 & 0 \ 0 & 0 & 1 & | & 0 & 0 & 1 \end{pmatrix}.     - Operation: R1+R2R2R_1 + R_2 \rightarrow R_2 yields (1amp;2amp;0amp;amp;1amp;0amp;0 0amp;3amp;1amp;amp;1amp;1amp;0 0amp;0amp;1amp;amp;0amp;0amp;1)\begin{pmatrix} 1 & 2 & 0 & | & 1 & 0 & 0 \ 0 & 3 & 1 & | & 1 & 1 & 0 \ 0 & 0 & 1 & | & 0 & 0 & 1 \end{pmatrix}.     - Operation: R2R3R2R_2 - R_3 \rightarrow R_2 yields (1amp;2amp;0amp;amp;1amp;0amp;0 0amp;3amp;0amp;amp;1amp;1amp;1 0amp;0amp;1amp;amp;0amp;0amp;1)\begin{pmatrix} 1 & 2 & 0 & | & 1 & 0 & 0 \ 0 & 3 & 0 & | & 1 & 1 & -1 \ 0 & 0 & 1 & | & 0 & 0 & 1 \end{pmatrix}.     - Operation: 3R12R2R13R_1 - 2R_2 \rightarrow R_1 yields (3amp;0amp;0amp;amp;1amp;2amp;2 0amp;3amp;0amp;amp;1amp;1amp;1 0amp;0amp;1amp;amp;0amp;0amp;1)\begin{pmatrix} 3 & 0 & 0 & | & 1 & -2 & 2 \ 0 & 3 & 0 & | & 1 & 1 & -1 \ 0 & 0 & 1 & | & 0 & 0 & 1 \end{pmatrix}.     - Final Scaling: 13R1\frac{1}{3}R_1 and 13R2\frac{1}{3}R_2 yields (1amp;0amp;0amp;amp;1/3amp;2/3amp;2/3 0amp;1amp;0amp;amp;1/3amp;1/3amp;1/3 0amp;0amp;1amp;amp;0amp;0amp;1)\begin{pmatrix} 1 & 0 & 0 & | & 1/3 & -2/3 & 2/3 \ 0 & 1 & 0 & | & 1/3 & 1/3 & -1/3 \ 0 & 0 & 1 & | & 0 & 0 & 1 \end{pmatrix}.     - Resulting A1=(1/3amp;2/3amp;2/3 1/3amp;1/3amp;1/3 0amp;0amp;1)A^{-1} = \begin{pmatrix} 1/3 & -2/3 & 2/3 \ 1/3 & 1/3 & -1/3 \ 0 & 0 & 1 \end{pmatrix}.

Theorems and Properties of Inverses

  • Invertible Matrix Facts     - If AA and BB are invertible matrices, kk is a positive integer, and cc is a nonzero scalar:         - A1A^{-1}, AkA^k, cAcA, and ATA^T (transpose) are all invertible.         - (A1)1=A(A^{-1})^{-1} = A         - (cA)1=1cA1(cA)^{-1} = \frac{1}{c} A^{-1}         - (Ak)1=(A1)k=Ak(A^k)^{-1} = (A^{-1})^k = A^{-k}         - (AB)1=B1A1(AB)^{-1} = B^{-1} A^{-1} (Note the reversed order).         - (AT)1=(A1)T(A^T)^{-1} = (A^{-1})^T

  • Elementary Matrices     - All elementary matrices (EE) are invertible because row operations are reversible.

The Fundamental Theorem of Invertible Matrices (Equivalent Conditions)

  • Core Concept     - Let AA be an n×nn \times n matrix. The following statements are equivalent (if one is true, all are; if one is false, all are):         1. AA is an invertible matrix.         2. Ax=bAx = b has a unique solution for all vectors bb.         3. The homogeneous system Ax=0Ax = 0 has only the trivial solution (x=0x = 0).         4. AA is row equivalent to the identity matrix II.         5. AA can be written as a product of elementary matrices.         6. The determinant of AA is non-zero (det(A)0\det(A) \neq 0).

Chapter Three: Determinants

  • Definition and Domain     - Determinant is a function det(A):Set of all n×n matricesR\det(A): \text{Set of all } n \times n \text{ matrices} \rightarrow \mathbb{R}.     - Result of a determinant is always a real number.     - Notation: det(A)\det(A) or vertical bars A|A|. Brackets [][ ] denote a matrix, while vertical bars | | denote a determinant scalar.

  • Specific Cases for Computation     - 2x2 Determinant:         - det(aamp;b camp;d)=adbc\det \begin{pmatrix} a & b \ c & d \end{pmatrix} = ad - bc.         - Example: det(2amp;5 3amp;7)=(2×7)(5×3)=14+15=29\det \begin{pmatrix} 2 & 5 \ -3 & 7 \end{pmatrix} = (2 \times 7) - (5 \times -3) = 14 + 15 = 29.     - 3x3 Determinant (The Basket Technique/Basket Weaving):         - Copy the first two columns to the right of the matrix.         - Sum the products of the diagonals going down from left to right.         - Subtract the sum of the products of the diagonals going up from left to right.         - Warning: This trick only works for $3 \times 3$ matrices.

  • Computational Approaches     - Permutation Formula: Inefficient for manual calculation; used for theoretical motivation.     - Cofactor Expansion:         - Can expand about any row or column.         - Generally iterative (4×44 \times 4 matrix breaks into four 3×33 \times 3 matrices).         - Scaling: Grows factorially (O(n!)O(n!)). A $55 \times 55$ matrix would take longer than the age of the universe to compute via cofactor expansion on a supercomputer.         - Strategic Tip: Pick the row or column with the most zeros to minimize work.     - Row/Column Reduction: Efficient and scales well. Determinants allow both row and column operations (unlike standard matrices).

Properties of Determinant Operations

  • Cofactor Expansion Signs     - Follows a checkerboard pattern:         (+amp;amp;+ amp;+amp; +amp;amp;+)\begin{pmatrix} + & - & + \ - & + & - \ + & - & + \end{pmatrix}.

  • Triangular Matrices     - The determinant of an upper or lower triangular matrix is simply the product of its diagonal entries.

  • Row/Column Operations and Their Impact on Determinants     1. Swapping: Swapping any two rows or columns changes the sign of the determinant.     2. Scaling Matrix-wide: det(cA)=cndet(A)\det(cA) = c^n \det(A).     3. Scaling a Single Row/Column: If a row or column is multiplied by scalar cc, the determinant is multiplied by cc.     4. Row Addition: Replacing a row with the sum of itself and a multiple of another row (R2R2+cR1R_2 \rightarrow R_2 + cR_1) does not change the determinant value.         - Caution: If the row being changed is multiplied by a scalar (e.g., R22R2+R1R_2 \rightarrow 2R_2 + R_1), the determinant changes and must be adjusted (scaled by $1/c$).

  • Invertibility Connection     - A matrix is invertible if and only if det(A)0\det(A) \neq 0.     - If a matrix has a row of zeros in row echelon form, its determinant is zero, and it is therefore not invertible.

Questions & Discussion

  • Student Question: "Can we take a break, professor?"

  • Instructor Response: "Yeah. We'll take a break then. So why don't we come back at 07:45, and then we'll get back to work on chapter chapter three."