MATH 332 — Fall 2026

Inverses and Invertibility
Anton, Rorres, & Kaul §§1.4–1.6, 1.11
Quiz 1 · Sep 10

Fred J. Hickernell

October 9, 2026

Course Map

Course decks

In this deck

\[ \mat{A}\vct{x}=\vct{b} \;\implies\; \vct{x}=\mat{A}^{-1}\vct{b} \qquad\text{if }\mat{A}^{-1}\text{ exists} \]

From Row Operations to Inverses

Undoing a row operation leads directly to the idea of an inverse matrix

Why square matrices?

Systems and Matrices: rectangular allowed

\[ \mat{A}\in\reals^{m\times n} \]

  • \(m\) equations
  • \(n\) unknowns
  • \(m\) and \(n\) need not agree

Inverses: square throughout

\[ \mat{A}\in\reals^{n\times n} \]

  • \(n\) equations
  • \(n\) unknowns
  • a two-sided inverse may exist

Two-sided inverse — undo \(\mat{A}\) in both orders

\[ \mat{A}^{-1}\mat{A}=\mat{I}_n, \qquad \mat{A}\mat{A}^{-1}=\mat{I}_n \]

  • This deck — square matrices
  • Rectangular matrices — no two-sided inverse
  • Pseudoinverses — later

Reversible means invertible

Systems and Matrices — reversible means undoable

For square \(\mat{A}\)

  • action \(\vct{x}\mapsto\mat{A}\vct{x}\)
  • reversible exactly when \(\mat{A}^{-1}\) undoes both directions

\[ \vct{x} \xrightarrow{\ \mat{A}\ } \mat{A}\vct{x} \xrightarrow{\ \mat{A}^{-1}\ } \vct{x}, \qquad \vct{x} \xrightarrow{\ \mat{A}^{-1}\ } \mat{A}^{-1}\vct{x} \xrightarrow{\ \mat{A}\ } \vct{x} \]

Equivalently,

\[ \mat{A}^{-1}\mat{A}=\mat{I}, \qquad \mat{A}\mat{A}^{-1}=\mat{I} \]

  • \(\mat{A}\) — invertible or nonsingular
  • \(\mat{A}^{-1}\) — inverse of \(\mat{A}\)

Reversible matrix action \(=\) multiplication by an invertible matrix

Elimination matrices are invertible

From Systems and Matrices

  • \(\mat{M}_1\) — zeros the first column below its pivot
  • \(\mat{N}_1\) — reverses that elimination stage

\[ \mat{M}_1= \begin{bmatrix} 1&0&0\\-2&1&0\\-3&0&1 \end{bmatrix}, \qquad \mat{N}_1= \begin{bmatrix} 1&0&0\\2&1&0\\3&0&1 \end{bmatrix} \]

A direct check gives

\[ \mat{N}_1\mat{M}_1=\mat{I} =\mat{M}_1\mat{N}_1 \]

Now we can write \(\mat{N}_1=\mat{M}_1^{-1}\)

  • \(\mat{M}_j\) — an invertible product of elementary matrices
  • \(\mat{M}_j^{-1}\) — reverses the whole column-elimination stage

Invertible or singular

For a square matrix \(\mat{A}\):

Invertible = nonsingular

\(\mat{A}^{-1}\) exists

  • for every \(\vct{b}\), \(\mat{A}\vct{x}=\vct{b}\) has exactly one solution
  • distinct inputs produce distinct outputs—no information is lost
  • in echelon form, every row and column has a pivot

Singular

\(\mat{A}^{-1}\) does not exist

  • some \(\vct{b}\) cannot be reached as \(\mat{A}\vct{x}\)
  • distinct inputs can produce the same output—information is lost
  • in echelon form, at least one row and one column have no pivot
  • Pivot test — row-reduce \(\mat{A}\) and inspect its pivots
  • Determinants — deferred; elimination shows pivots directly
  • Near singularity — computed determinants may be unreliable

Inverse Algebra

Unique inverse, essential order

Left and right inverses must agree

Suppose \(\mat{B}\) is a left inverse and \(\mat{C}\) is a right inverse of \(\mat{A}\):

\[ \mat{B}\mat{A}=\mat{I}, \qquad \mat{A}\mat{C}=\mat{I} \]

Associativity connects them:

\[ \mat{B} =\mat{B}\mat{I} =\mat{B}(\mat{A}\mat{C}) =(\mat{B}\mat{A})\mat{C} =\mat{I}\mat{C} =\mat{C} \]

Thus any left inverse and right inverse of the same matrix must agree

  • Two-sided inverse — unique and denoted \(\mat{A}^{-1}\)
  • Special commuting pair — \(\mat{A}^{-1}\mat{A}=\mat{I}=\mat{A}\mat{A}^{-1}\)

Undo products in reverse order

If \(\mat{A}\) and \(\mat{B}\) are invertible, then

\[ (\mat{A}\mat{B})^{-1}=\mat{B}^{-1}\mat{A}^{-1} \]

Associativity lets us regroup without changing the order:

\[\begin{align*} (\mat{A}\mat{B})(\mat{B}^{-1}\mat{A}^{-1}) &=\mat{A}(\mat{B}\mat{B}^{-1})\mat{A}^{-1} =\mat{A}\mat{A}^{-1}=\mat{I},\\ (\mat{B}^{-1}\mat{A}^{-1})(\mat{A}\mat{B}) &=\mat{B}^{-1}(\mat{A}^{-1}\mat{A})\mat{B} =\mat{B}^{-1}\mat{B}=\mat{I} \end{align*}\]

  • No standard matrix quotient \(\mat{B}/\mat{A}\)
  • \(\mat{A}^{-1}\mat{B}\) and \(\mat{B}\mat{A}^{-1}\) generally different
  • Rightmost factor acts first
  • Inverse reverses factor order

Compute an Inverse

Gauss–Jordan reduction records the row operations that turn \(\mat{A}\) into \(\mat{I}\)

Reduce beside the identity

\(\mat{G}\) — product of the Gauss–Jordan elementary matrices

Same row operations on both blocks

\[ \mat{G}[\,\mat{A}\mid\mat{I}\,] = [\,\mat{G}\mat{A}\mid\mat{G}\,] = [\,\mat{I}\mid\mat{G}\,] \]

From the left block

\[ \mat{G}\mat{A}=\mat{I} \implies \mat{A}=\mat{G}^{-1} \implies \mat{G}=\mat{A}^{-1} \]

  • Gaussian elimination — solve systems
  • Gauss–Jordan reduction — compute the inverse

Find an inverse

For \(\mat{A}=\begin{bmatrix}1&1\\2&3\end{bmatrix}\),

\[ \left[\begin{array}{cc|cc} 1&1&1&0\\2&3&0&1 \end{array}\right] \xrightarrow{R_2\leftarrow R_2-2R_1} \left[\begin{array}{cc|cc} 1&1&1&0\\0&1&-2&1 \end{array}\right] \]

\[ \xrightarrow{R_1\leftarrow R_1-R_2} \left[\begin{array}{cc|cc} 1&0&3&-1\\0&1&-2&1 \end{array}\right] \]

Thus \(\mat{A}^{-1}=\begin{bmatrix}3&-1\\-2&1\end{bmatrix}\)

Return to our three-equation matrix

Our three-equation system

  • Same forward-elimination steps
  • Both blocks of \([\,\mat{A}\mid\mat{I}\,]\)

\[\begin{align*} \left[ \begin{array}{rrr|rrr} 1&2&-1&1&0&0\\ 2&-1&1&0&1&0\\ 3&1&2&0&0&1 \end{array} \right] &\xrightarrow{\substack{R_2\leftarrow R_2-2R_1\\R_3\leftarrow R_3-3R_1}} \left[ \begin{array}{rrr|rrr} 1&2&-1&1&0&0\\ 0&-5&3&-2&1&0\\ 0&-5&5&-3&0&1 \end{array} \right] \\[0.8em] &\xrightarrow{R_3\leftarrow R_3-R_2} \left[ \begin{array}{rrr|rrr} 1&2&-1&1&0&0\\ 0&-5&3&-2&1&0\\ 0&0&2&-1&-1&1 \end{array} \right] \end{align*}\]

The left block is now in row echelon form

Finish the Gauss–Jordan reduction

Continue upward until the left block becomes \(\mat{I}\):

\[\begin{align*} \left[ \begin{array}{rrr|rrr} 1&2&-1&1&0&0\\ 0&-5&3&-2&1&0\\ 0&0&2&-1&-1&1 \end{array} \right] &\xrightarrow{\substack{R_3\leftarrow \tfrac12R_3\\R_1\leftarrow R_1+R_3\\R_2\leftarrow R_2-3R_3}} \left[ \begin{array}{rrr|rrr} 1&2&0&\tfrac12&-\tfrac12&\tfrac12\\ 0&-5&0&-\tfrac12&\tfrac52&-\tfrac32\\ 0&0&1&-\tfrac12&-\tfrac12&\tfrac12 \end{array} \right] \\[0.8em] &\xrightarrow{\substack{R_2\leftarrow -\tfrac15R_2\\R_1\leftarrow R_1-2R_2}} \left[ \begin{array}{rrr|rrr} 1&0&0&\tfrac3{10}&\tfrac12&-\tfrac1{10}\\ 0&1&0&\tfrac1{10}&-\tfrac12&\tfrac3{10}\\ 0&0&1&-\tfrac12&-\tfrac12&\tfrac12 \end{array} \right] \end{align*}\]

The left block becomes \(\mat{I}\)

The right block becomes \(\mat{A}^{-1}\)

A missing pivot proves singularity

For \(\mat{A}=\begin{bmatrix}1&2\\2&4\end{bmatrix}\),

\[ \left[\begin{array}{cc|cc} 1&2&1&0\\2&4&0&1 \end{array}\right] \xrightarrow{R_2\leftarrow R_2-2R_1} \left[\begin{array}{cc|cc} 1&2&1&0\\0&0&-2&1 \end{array}\right] \]

  • Second pivot missing
  • Left block cannot become \(\mat{I}\)
  • Missing pivot — \(\mat{A}\) singular
  • No inverse

Invertibility and Linear Systems

One structural property controls every system \(\mat{A}\vct{x}=\vct{b}\)

One theorem, many viewpoints

For a square \(n\times n\) matrix \(\mat{A}\), these statements are equivalent:

  • \(\mat{A}\) is invertible
  • \(\mat{A}\) row-reduces to \(\mat{I}\)
  • an echelon form of \(\mat{A}\) has \(n\) pivots
  • \(\mat{A}\vct{x}=\vct{0}\) has only \(\vct{x}=\vct{0}\)
  • \(\mat{A}\vct{x}=\vct{b}\) has exactly one solution for every \(\vct{b}\)
  • \(\mat{A}\) is a product of elementary matrices

Algebraic, computational, and geometric views of reversibility

Reversible versus information-losing

\(\mat{A}\) invertible

No pivot is missing, so

\[ \mat{A}\vct{x}=\vct{0} \quad\Longrightarrow\quad \vct{x}=\vct{0} \]

For every \(\vct{b}\),

\[ \vct{x}=\mat{A}^{-1}\vct{b} \]

exists and is unique

\(\mat{A}\) singular

  • row reduction has a missing pivot
  • the homogeneous system is always consistent
  • a nonpivot column gives a free variable

Choose the free variable(s) to obtain \(\vct{z}\ne\vct{0}\) with

\[ \mat{A}\vct{z}=\vct{0} \]

  • Singular action — information loss
  • \(\mat{A}(\vct{x}+t\vct{z})=\mat{A}\vct{x}\) for every \(t\)

The right-hand side decides

For singular square \(\mat{A}\), the outcome of \(\mat{A}\vct{x}=\vct{b}\) depends on \(\vct{b}\)

Inconsistent

If elimination produces a contradiction,

\[ [\,0\ \cdots\ 0\mid c\,], \qquad c\ne0, \]

then there is no solution

Consistent

If \(\vct{x}_0\) is one solution and \(\mat{A}\vct{z}=\vct{0}\) with \(\vct{z}\ne\vct{0}\), then

\[\begin{align*} \mat{A}(\vct{x}_0+t\vct{z}) &=\mat{A}\vct{x}_0+t\mat{A}\vct{z}\\ &=\vct{b}+t\vct{0}=\vct{b} \end{align*}\]

For every \(t\) — infinitely many solutions

  • Singular square \(\mat{A}\) — never exactly one solution
  • Depending on \(\vct{b}\) — none or infinitely many

Choose the least computation

One or a few right-hand sides

Use Gaussian elimination and back substitution:

\[ [\,\mat{A}\mid\vct{b}_1\ \cdots\ \vct{b}_k\,] \]

Do not compute the entire inverse

Many systems, same \(\mat{A}\)

Reuse the elimination structure or a factorization

Compute \(\mat{A}^{-1}\) when the inverse itself is the object of interest

  • Unnecessary computation — more time
  • More opportunities for round-off error

Elimination leaves reusable factors

For \(\mat{A}\in\reals^{m\times n}\) and \(q=\min(m-1,n)\), suppose no row exchanges are needed:

\[ \mat{U}=\mat{M}_q\cdots\mat{M}_1\mat{A} \]

Undoing the stages in reverse order gives

\[ \mat{A}=\mat{L}\mat{U}, \qquad \mat{L}=\mat{M}_1^{-1}\cdots\mat{M}_q^{-1}, \]

Full-form factors

  • \(\mat{L}\in\reals^{m\times m}\) — unit lower triangular
  • \(\mat{U}\in\reals^{m\times n}\) — zeros below its diagonal
  • Square \(\mat{U}\) — upper triangular

Next in Matrix Transformations — no-exchange LU and permutation matrices for row exchanges

Leontief Input–Output Models

An economy must meet external demand while supporting its own production

Production creates internal demand

For sectors \(i\) and \(j\), define

\[\begin{align*} x_i & = \alerttext{gross production}\text{ of sector }i, \\ d_i & = \alerttext{external demand}\text{ for sector }i, \\ c_{ij} & = \text{output from sector }i\text{ needed per unit produced by sector }j \end{align*}\]

For all sectors at once,

\[ \underbrace{\vct{x}}_{\text{total}} =\underbrace{\mat{C}\vct{x}}_{\text{internal}} +\underbrace{\vct{d}}_{\text{external}} \]

Therefore

\[ (\mat{I}-\mat{C})\vct{x}=\vct{d} \]

  • \(\vct{x}\) — gross production plan meeting \(\vct{d}\)
  • Static balance model — neither an optimization nor a dynamic steady-state model

One forecast or a response map?

Consider a two-sector economy:

\[ \mat{C}=\begin{bmatrix}.20&.10\\.30&.25\end{bmatrix}, \qquad \vct{d}=\begin{bmatrix}72\\30\end{bmatrix} \]

One demand forecast

Solve directly:

\[ \vct{x}=\begin{bmatrix}100\\80\end{bmatrix} \]

Gross production for the specified \(\vct{d}\)

Many possible demands

The response map is

\[ (\mat{I}-\mat{C})^{-1} \approx \begin{bmatrix}1.316&.175\\.526&1.404\end{bmatrix} \]

Check the model, not only the arithmetic

For \(\vct{x}=(100,80)^{\mathsf T}\),

\[ \mat{C}\vct{x}+\vct{d} =\begin{bmatrix}28\\50\end{bmatrix} +\begin{bmatrix}72\\30\end{bmatrix} =\begin{bmatrix}100\\80\end{bmatrix} =\vct{x} \]

  • Inverse columns — production responses to added demand
  • Nonnegative gross production — meaningful plan
  • Singular \(\mat{I}-\mat{C}\) — a nonzero \(\vct{z}\) satisfies \(\vct{z}=\mat{C}\vct{z}\)
  • Along \(\vct{z}\), gross-output changes are absorbed internally
  • Compatible \(\vct{d}\) — \(\vct{x}+t\vct{z}\) is an algebraic solution for every \(t\in\reals\)
    Only nonnegative members are feasible plans
  • Some \(\vct{d}\) — no solution

Gaussian Covariance and Precision

Preview for statistics, AI, and data science — no background required

Covariance describes joint variation

Gaussian-process values at selected inputs \(t_1,\ldots,t_n\)

\[ \vf= \begin{bmatrix} f(t_1)&\cdots&f(t_n) \end{bmatrix}^{\mathsf T} \sim\Norm(\vmu,\mSigma), \qquad \mSigma_{ij}=K(t_i,t_j) \]

Diagonal entries

\[ \mSigma_{ii}=\var(f(t_i)) \]

describe uncertainty at individual inputs

Off-diagonal entries

\[ \mSigma_{ij}=\cov(f(t_i),f(t_j)) \]

describe how values at two inputs vary together

\(\mSigma\) — uncertainty and joint variation

Precision describes conditional structure

When \(\mSigma\) is invertible, its precision matrix is

\[ \mQ=\mSigma^{-1} \]

Statistical meaning

  • \(\vr^{\mathsf T}\mQ\vr\) — departure \(\vr=\vf-\vmu\) weighted against expected joint variation
  • For \(i\ne j\): \(\mQ_{ij}=0\) exactly when \(f_i\) and \(f_j\) are conditionally independent given the rest

Computational meaning

If a calculation needs

\[ \valpha=\mQ\vr, \]

solve

\[ \mSigma\valpha=\vr \]

rather than forming \(\mQ\)

Preview only — covariance: joint variation; precision: conditional structure

Compute a precision matrix

A Gaussian vector has covariance matrix

\[ \mSigma= \begin{bmatrix} 1&1&0\\ 1&2&1\\ 0&1&2 \end{bmatrix} \]

\(\exstar\) Compute the precision matrix \(\mQ=\mSigma^{-1}\) by Gauss–Jordan reduction of \([\,\mSigma\mid\mat{I}\,]\) Verify that \(\mSigma\mQ=\mat{I}\) Compare \(\mSigma_{13}\) with \(\mQ_{13}\) and interpret the different zero patterns

Big Ideas

  • Two-sided inverse — \(\mat{A}^{-1}\mat{A}=\mat{I}=\mat{A}\mat{A}^{-1}\)
  • Invertible — pivot in every row and column; one solution for every \(\vct{b}\)
  • Inverse of a product — reverse factor order
  • Gauss–Jordan on \([\,\mat{A}\mid\mat{I}\,]\) computes \(\mat{A}^{-1}\)
  • Least computation — solve a few right-hand sides directly
    Reuse factors or response maps for many
  • Applications — Leontief maps demand to production
    Gaussian precision captures conditional structure

How Far We Have Come

From equations to reversible matrix actions

  • Systems and Matrices — \(\mat{A}\vct{x}=\vct{b}\) becomes the augmented matrix \([\,\mat{A}\mid\vct{b}\,]\): coefficients and right-hand side together
  • Elimination — row operations preserve all solutions; reduced row echelon form reveals inconsistencies (contradictions), pivots, and free variables
  • Inverses — a two-sided inverse undoes a square matrix action: \(\mat{A}^{-1}\mat{A}=\mat{I}=\mat{A}\mat{A}^{-1}\)
  • Gauss–Jordan — reduce \([\,\mat{A}\mid\mat{I}\,]\) to \([\,\mat{I}\mid\mat{A}^{-1}\,]\); a missing pivot in \(\mat{A}\) means no inverse
  • Inverse of a product — for invertible \(\mat{A},\mat{B}\), undo the factors in reverse order: \((\mat{A}\mat{B})^{-1}=\mat{B}^{-1}\mat{A}^{-1}\)
  • Choose the computation — solve a few right-hand sides directly; a computed inverse supplies the reusable response map \(\vct{b}\mapsto\mat{A}^{-1}\vct{b}\)

For square \(\mat{A}\): a pivot in every column \(\iff\) \(\mat{A}^{-1}\) exists \(\iff\) every \(\vct{b}\) has exactly one solution

What Comes Next

Inverses — undoing a matrix action

\[ \vct{x}\xrightarrow{\mat{A}}\mat{A}\vct{x} \xrightarrow{\mat{A}^{-1}}\vct{x} \]

Matrix Transformations — structure reveals the action

  • diagonal, triangular, and symmetric matrices
  • reflections, rotations, projections, and composition

Next — Matrix Transformations: structure, transformations, and geometry

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