Facts

If $f(x)=ax+b$, then increasing $x$ by one unit changes $f(x)$ by exactly $a$, no matter where you start: $$ \begin{aligned} &f(x+1)-f(x) \\ &\quad= \bigl(a(x+1)+b\bigr) - (ax+b) \\ &\quad= a\,. \end{aligned} $$ So $a>0$ gives a graph rising to the right, the steeper the larger $a$ is; $a<0$ one falling to the right, with steepness $|a|$; and $a=0$ the horizontal line $y=b$.

Let $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ be distinct points on a non-vertical straight line $\ell$. Then the slope of $\ell$ is $$ a = \frac{y_2-y_1}{x_2-x_1}\,, \tag{1.1} $$ independently of which two points of $\ell$ are chosen; and the line through $(x_1,y_1)$ with slope $a$ has equation $$ y - y_1 = a\,(x-x_1)\,. \tag{1.2} $$

A system of two linear equations in two unknowns has either

1. exactly one solution — the lines cross at a single point;
2. no solution — the lines are parallel and distinct;
3. infinitely many solutions — the two equations describe the same line.

In particular, it is impossible for such a system to have exactly two solutions.

For $a\ne 0$, $$ ax^2+bx+c = a\left(x+\frac{b}{2a}\right)^{\!2} - \frac{b^2-4ac}{4a}\,. \tag{1.4} $$

Let $f(x)=ax^2+bx+c$ with $a\ne0$, and set $x^{*} = -\dfrac{b}{2a}$.

• If $a>0$, then $f$ has a minimum at $x^{*}$, and no maximum.
• If $a<0$, then $f$ has a maximum at $x^{*}$, and no minimum.

In both cases the extreme value is $$ f(x^{*}) = c - \frac{b^2}{4a} = -\frac{b^2-4ac}{4a}\,, $$ and the parabola is symmetric about the vertical line $x=x^{*}$: for every $t$, $f(x^{*}+t) = f(x^{*}-t)$.

Let $P$ be a polynomial and $a$ a number. Then $$ x-a \text{ is a factor of } P(x) \;\Longleftrightarrow\; P(a)=0\,. $$

Suppose all the coefficients $a_n,\ldots,a_1,a_0$ in (1.5) are integers. Then every integer root of $P$ is a factor of the constant term $a_0$.

Let $P$ have degree $m$ and $Q$ have degree $n$, with $m\ge n$. Then there is exactly one pair of polynomials $q$ (of degree $m-n$) and $r$ (of degree less than $n$) with $$ \begin{aligned} &P(x) = q(x)\,Q(x) + r(x)\,, \\ &\text{equivalently} \\ &\frac{P(x)}{Q(x)} = q(x) + \frac{r(x)}{Q(x)} \end{aligned} \tag{1.6} $$ wherever $Q(x)\ne0$. We call $q$ the quotient and $r$ the remainder; if $r$ is the zero polynomial, $Q$ is a factor of $P$.

Every power function $y=x^{r}$ passes through $(1,1),$ since $1^{r}=1,$ and its shape on $x>0$ depends only on $r{:}$ for $r>1$ it increases and bends upwards $(y=x^2);$ for $r=1$ it is the line $y=x;$ for $0<r<1$ it increases and bends downwards $(y=\sqrt{x});$ for $r=0$ it is the constant $y=1;$ and for $r<0$ it decreases, with $y\to\infty$ as $x\to 0$ and $y\to0$ as $x\to\infty$ $(y=1/x).$

If $f(t)=Aa^{t},$ then for every $t,$ $$ f(t+1) = A a^{t+1} = a\cdot Aa^{t} = a\,f(t)\,. $$ So increasing $t$ by one unit multiplies the value by $a,$ no matter where you start, and $A=f(0).$ Hence $a>1$ gives growth, $0<a<1$ decay and $a=1$ a constant; writing $a = 1+p/100$ with $p>0,$ the function grows by $p\%$ per unit, and $a=1-p/100$ with $0<p<100$ makes it fall by $p\%$ per unit.

For all $x,y>0$ and all real $p,$ $$ \begin{aligned} \ln(xy) &= \ln x + \ln y\,,\\ \ln\frac{x}{y} &= \ln x - \ln y\,,\\ \ln x^{p} &= p\,\ln x\,, \end{aligned} $$ and $\ln$ and $\exp$ undo each other: $\ln e^{x}=x$ for all real $x,$ and $e^{\ln x}=x$ for all $x>0.$

Let $a>1$ and $A>0.$ The exponential function $f(t)=Aa^{t}$ has a unique doubling time $t^{*},$ the solution of $a^{t^{*}}=2,$ namely $$ t^{*} = \frac{\ln 2}{\ln a}\,. $$ It does not depend on $A,$ nor on the moment from which you start measuring.

The Newton quotient is the average rate of change of $f$ between $a$ and $a+h,$ so its limit $f'(a)$ is the instantaneous rate of change at $a;$ and for small $h,$ $f(a+h)-f(a)\approx f'(a)\cdot h.$

Let $f$ be differentiable on an interval $I.$ Then $$ \begin{aligned} &f'(x)\ge 0 \text{ for all } x\in I\\ &\qquad\iff f \text{ is increasing on } I\,,\\ &f'(x)\le 0 \text{ for all } x\in I\\ &\qquad\iff f \text{ is decreasing on } I\,,\\ &f'(x)= 0 \text{ for all } x\in I\\ &\qquad\iff f \text{ is constant on } I\,. \end{aligned} $$ Moreover $f'(x)>0$ for all $x\in I$ implies that $f$ is strictly increasing on $I,$ and similarly $f'(x)<0$ implies strictly decreasing.

Suppose $\lim_{x\to a}f(x)=A$ and $\lim_{x\to a}g(x)=B.$ Then $$ \begin{aligned} \lim_{x\to a}\bigl(f(x)\pm g(x)\bigr) &= A\pm B\,,\\ \lim_{x\to a} f(x)g(x) &= AB\,,\\ \lim_{x\to a}\frac{f(x)}{g(x)} &= \frac{A}{B}\ \ (B\ne0)\,, \end{aligned} $$ together with $\lim_{x\to a}c=c$ for a constant $c,$ and $\lim_{x\to a}x=a.$

Let $A$ be a constant and let $g$ and $h$ be differentiable. Then $$ \begin{aligned} f(x) = A &\Longrightarrow f'(x) = 0\\ f(x) = A + g(x) &\Longrightarrow f'(x) = g'(x)\\ f(x) = A\cdot g(x) &\Longrightarrow f'(x) = A\,g'(x)\\ f(x) = x^{a} &\Longrightarrow f'(x) = a\,x^{a-1}\\ f(x) = g(x)+h(x) &\Longrightarrow f'(x) = g'(x)+h'(x)\\ f(x) = g(x)-h(x) &\Longrightarrow f'(x) = g'(x)-h'(x) \end{aligned} $$ In words: additive constants disappear, multiplicative constants survive.

If $f$ and $g$ are differentiable, then $$ (fg)'(x) = f'(x)g(x) + f(x)g'(x)\,. \tag{3.1} $$

If $f$ and $g$ are differentiable and $g(x)\ne0,$ then $$ \left(\frac{f}{g}\right)'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}\,. \tag{3.2} $$

Wherever the quantities are nonzero, $$ \begin{aligned} \frac{(fg)'}{fg} &= \frac{f'}{f} + \frac{g'}{g}\,,\\ \frac{(f/g)'}{f/g} &= \frac{f'}{f} - \frac{g'}{g}\,. \end{aligned} $$ Relative rates add for products and subtract for quotients.

Let $F(x) = g\bigl(f(x)\bigr),$ with $f$ differentiable at $x$ and $g$ differentiable at $f(x).$ Then $$ F'(x) = g'\bigl(f(x)\bigr)\cdot f'(x)\,. \tag{3.3} $$ Writing $z=g(y)$ and $y=f(x),$ the same statement in Leibniz's notation is $$ \frac{dz}{dx} = \frac{dz}{dy}\cdot\frac{dy}{dx}\,. $$

On an interval where $f$ is twice differentiable,

• $f''>0{:}$ the slope is increasing, the curve bends upwards, and its tangents lie below it;
• $f''<0{:}$ the slope is decreasing, the curve bends downwards, and its tangents lie above it.

$\displaystyle\frac{d}{dx}e^{x} = e^{x}$, and by the chain rule (Fact 3.16), $\displaystyle\frac{d}{dx}e^{g(x)} = e^{g(x)}\,g'(x)$.

Since $a=e^{\ln a}$ we have $a^{x}=e^{x\ln a}$, and therefore $$ \frac{d}{dx}a^{x} = a^{x}\ln a\,. $$

For $x>0$, $$ \frac{d}{dx}\ln x = \frac{1}{x}\,, $$ and consequently, wherever $h(x)>0$, $$ \frac{d}{dx}\ln h(x) = \frac{h'(x)}{h(x)}\,. \tag{4.2} $$

For every real number $a$ and every $x>0$, $$ \frac{d}{dx}x^{a} = a\,x^{a-1}\,. $$

If $f$ is differentiable at $a$, then for $x$ near $a$, $$ f(x) \;\approx\; f(a) + f'(a)(x-a)\,. \tag{4.3} $$

$P_{n}$ is the unique polynomial of degree at most $n$ whose value and first $n$ derivatives at $a$ agree with those of $f$: $$ P_{n}^{(k)}(a) = f^{(k)}(a)\qquad\text{for } k=0,1,\ldots,n\,, $$ and $f(x)\approx P_{n}(x)$ for $x$ near $a$. The case $n=1$ is (4.3).

Let $f$ be $n+1$ times differentiable on an interval containing $a$ and $x$. Then $$ \begin{aligned} &f(x) = p_{n}(x) + R_{n+1}(x)\,, \\ &R_{n+1}(x) = \frac{f^{(n+1)}(z)}{(n+1)!}\,(x-a)^{n+1} \tag{5.1} \end{aligned} $$ for some number $z$ strictly between $a$ and $x$. In the case $a=0$, which is the one normally used, $R_{n+1}(x) = f^{(n+1)}(z)\,x^{n+1}/(n+1)!$ with $z$ between $0$ and $x$.

If $\bigl|f^{(n+1)}(t)\bigr| \le M$ for every $t$ between $a$ and $x$, then $$ \bigl|R_{n+1}(x)\bigr| \;\le\; \frac{M}{(n+1)!}\,|x-a|^{\,n+1}\,. \tag{5.2} $$

$\lim_{x\to a}f(x)$ exists if and only if both one-sided limits exist and are equal, in which case all three agree.

If $f$ is continuous at $a$ then $\lim_{x\to a}f(x)=f(a)$: you simply substitute. Moreover anything built from constants and $x$ using addition, subtraction, multiplication, division and composition is continuous wherever it is defined — so polynomials are continuous everywhere, rational functions off the zeros of the denominator, and $e^{x}$, $\ln x$ and powers on their domains.

Let $f$ and $g$ be differentiable near $a$ with $f(a)=g(a)=0$, and suppose $g'(a)\ne0$. Then $$ \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}\,. \tag{5.5} $$ More generally, and for $\tfrac{\infty}{\infty}$ as well, $\lim_{x\to a} f/g = \lim_{x\to a} f'/g'$ whenever the right-hand limit exists. If that limit is again indeterminate, apply the rule again.

Let $f$ be continuous on the closed interval $[a,b]$. Then $f$ takes every value between $f(a)$ and $f(b)$ somewhere on $[a,b]$. In particular, if $f(a)$ and $f(b)$ have opposite signs, there is at least one $c$ in $(a,b)$ with $f(c)=0$.

If $f$ is continuous on a closed and bounded interval $[a,b]$, then $f$ attains both a maximum and a minimum on $[a,b]$.

Let $c$ be an interior point of an interval on which $f$ is defined, let $f$ be differentiable at $c$, and let $c$ be a maximum or minimum point. Then $f'(c)=0$.

Let $f$ be differentiable on an interval $I$ and let $c\in I$. If $f'(x)\ge0$ for all $x<c$ in $I$ and $f'(x)\le0$ for all $x>c$ in $I$, then $c$ is a maximum point for $f$ over the whole of $I$. If the inequalities are reversed, $c$ is a minimum point.

Let $f$ be twice differentiable on an interval $I$.

1. $f$ is concave on $I$ if and only if $f''(x)\le0$ throughout $I$, and convex if and only if $f''(x)\ge0$ throughout. If $f''<0$ throughout then $f$ is strictly concave, though not conversely.
2. If $f$ is concave on $I$ and $c$ is an interior point with $f'(c)=0$, then $c$ is a maximum point for $f$ over the whole of $I$. If $f$ is convex, such a $c$ is a minimum point over the whole of $I$.

Let $c$ be an interior point with $f'(c)=0$, and let $f$ be twice differentiable near $c$. Then $$ \begin{aligned} &f''(c)<0 \;\Longrightarrow\;\\ &\qquad c \text{ is a local maximum point}\,,\\ &f''(c)>0 \;\Longrightarrow\;\\ &\qquad c \text{ is a local minimum point}\,. \end{aligned} $$ Alternatively, and always reliably, use the sign of $f'$: if $f'$ goes from $+$ to $-$ at $c$ then $c$ is a local maximum, from $-$ to $+$ a local minimum, and if the sign does not change, neither.

If $c$ is an inflection point of $f$ and $f''$ is continuous at $c$, then $f''(c)=0$.

If $F'=G'=f$ on an interval, then $F-G$ is constant there. So the $+C$ in Definition 7.12 describes all the freedom there is. The two antiderivatives to know at once are $$ \begin{aligned} &\int x^{a}\,dx = \frac{x^{a+1}}{a+1}+C \quad (a\ne-1)\,, \\ &\int \frac{1}{x}\,dx = \ln|x|+C\,, \end{aligned} $$ the second being the case the first cannot reach.

Let $f$ be continuous and non-negative on $[a,b]$, and let $A(t)$ be the area under the graph of $f$ between $a$ and $t$. Then $$ A'(t) = f(t)\,, $$ so $A$ is an antiderivative of $f$ with $A(a)=0$. Consequently the area under the graph between $a$ and $b$ is exactly $\int_{a}^{b}f(x)\,dx$.

For continuous $f$ and $g$ and constants $\alpha,\beta$, $$ \begin{aligned} &\int_{a}^{a}f = 0\,,\\ &\int_{b}^{a}f = -\int_{a}^{b}f\,,\\ &\int_{a}^{b}(\alpha f+\beta g) = \alpha\!\int_{a}^{b}\!f + \beta\!\int_{a}^{b}\!g\,, \end{aligned} $$ $$ \int_{a}^{b}f = \int_{a}^{c}f + \int_{c}^{b}f\,. $$ Each is immediate from (7.1).

For differentiable $u$ and $v$, $$ \begin{aligned} &\int u(x)\,v'(x)\,dx\\ &\quad= u(x)v(x) - \int u'(x)\,v(x)\,dx\,. \end{aligned} \tag{7.2} $$

If $F'=f$, then differentiating $F(g(x))$ by the chain rule gives $F'(g(x))g'(x)$, so $$ \int f\bigl(g(x)\bigr)g'(x)\,dx = F\bigl(g(x)\bigr) + C\,. \tag{8.1} $$ In practice one writes $u=g(x)$ and $du = g'(x)\,dx$, turning the left-hand side into $\int f(u)\,du$.

For $|k|<1$, $$ a + ak + ak^{2} + ak^{3} + \cdots = \frac{a}{1-k}\,, $$ and for $|k|\ge1$ (with $a\ne0$) the series diverges.

$$ \begin{aligned} &\int_{1}^{\infty} x^{-a}\,dx \text{ converges} \iff a>1\,, \\ &\sum_{n=1}^{\infty} n^{-p} \text{ converges} \iff p>1\,. \end{aligned} $$ Geometric decay — shrinking by a constant factor — always converges. Power decay converges only above the exponent $1$.

Write $r$ for the interest rate per period and $k = 1/(1+r)$ for the discount factor. A payment of $a$ at the end of each of the next $n$ periods has total present value $$ \begin{aligned} \mathrm{PV} &= \frac{a}{1+r} + \frac{a}{(1+r)^{2}} + \cdots\\ &\qquad + \frac{a}{(1+r)^{n}}\\ &= a\,\frac{1-(1+r)^{-n}}{r}\,. \end{aligned} $$ A loan of $L$ repaid by $n$ such payments is the statement that these are the same sum of money seen from the two ends: $$ \begin{aligned} &a\,\frac{1-(1+r)^{-n}}{r} = L\\ &\qquad\Longleftrightarrow\qquad a = L\,\frac{r}{1-(1+r)^{-n}}\,. \end{aligned} $$

If the second-order partial derivatives of $f$ are continuous, then $$ \begin{aligned} &f''_{12} = f''_{21}\,, \qquad\text{that is}\\ &\frac{\partial^{2}f}{\partial y\,\partial x} = \frac{\partial^{2}f}{\partial x\,\partial y}\,. \end{aligned} $$ This holds for essentially every function met in economics.

1. Crowded contours mean steep ground. Neighbouring contours differ by a fixed rise, so the less horizontal distance between them, the faster the climb.
2. Which way is uphill is settled by the labels, not by the shape of the curves.
3. Two level curves at different heights never cross. A point cannot be at two heights at once.

If $f$ is differentiable and $x=g(t)$, $y=h(t)$ are differentiable, then $$ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}\,. \tag{9.1} $$ One contribution per variable, and they add.

Dividing (9.1) through by $z$, $$ \frac{\dot z}{z} = \mathrm{El}_{x}z\cdot\frac{\dot x}{x} + \mathrm{El}_{y}z\cdot\frac{\dot y}{y}\,, $$ where a dot denotes $d/dt$. The growth rate of $z$ is the elasticity-weighted sum of the growth rates of $x$ and $y$.

Along a level curve of $f$, wherever $f'_{2}\ne0$, $$ \frac{dy}{dx} = -\,\frac{f'_{1}}{f'_{2}}\,. \tag{9.2} $$

Let $f$ have a maximum or a minimum at an interior point $(x_{0},y_{0})$ of its domain, and let both partial derivatives exist there. Then $$ \begin{aligned} &f'_{1}(x_{0},y_{0}) = 0 \\ &\qquad\text{and}\qquad f'_{2}(x_{0},y_{0}) = 0\,. \end{aligned} $$

Let $f$ have continuous second-order partial derivatives on a convex set $S$. Then $f$ is concave on $S$ if and only if, at every point of $S$, $$ \begin{aligned} &f''_{11}\le0\,, \\ &f''_{22}\le0\,, \\ &f''_{11}f''_{22} - \bigl(f''_{12}\bigr)^{2} \ge 0\,. \end{aligned} \tag{10.2} $$ Reversing the first two inequalities, and keeping the third, characterises convexity.

If $f$ is concave on a convex set $S$ and $(x_{0},y_{0})$ is an interior critical point, then $(x_{0},y_{0})$ is a maximum point for $f$ over the whole of $S$. If $f$ is convex, such a point is a minimum over the whole of $S$.

An interior local extreme point of a differentiable $f$ is a critical point: $f'_{1}=f'_{2}=0$ there.

With $A,B,C$ as above:

1. if $A<0$ and $AC-B^{2}>0$, then $(x_{0},y_{0})$ is a strict local maximum point;
2. if $A>0$ and $AC-B^{2}>0$, then $(x_{0},y_{0})$ is a strict local minimum point;
3. if $AC-B^{2}<0$, then $(x_{0},y_{0})$ is a saddle point;
4. if $AC-B^{2}=0$, the test gives no information.

If $f$ is continuous on a nonempty compact set $S\subseteq{\mathbb R}^{2}$, then there exist points $(a,b)$ and $(c,d)$ in $S$ with $$ \begin{aligned} &f(a,b)\le f(x,y)\le f(c,d)\\ &\qquad\text{for all }(x,y)\in S\,. \end{aligned} $$

At a solution of $\max u(x,y)$ subject to $px+qy=m$ with $u'_{y}\ne0$, $$ \frac{u'_{x}(x,y)}{u'_{y}(x,y)} = \frac{p}{q}\,. $$ Equivalently, the slope $-u'_{x}/u'_{y}$ of the level curve of $u$ equals the slope $-p/q$ of the budget line: at the optimum the budget line is tangent to an indifference curve.

Under the assumptions of Definition 11.20, $$ \frac{df^{*}(c)}{dc} = \lambda(c)\,. $$ In particular, for a small change $dc$ in the constraint level, $$ f^{*}(c+dc) - f^{*}(c) \;\approx\; \lambda(c)\,dc\,. $$

Let $\mathbf{A},\mathbf{B},\mathbf{C}$ be $m\times n$ matrices, let $\alpha,\beta$ be scalars, and let $\mathbf{0}$ be the $m\times n$ zero matrix, all of whose entries are $0$. Then $$ \begin{array}{l} (\mathbf{A}+\mathbf{B})+\mathbf{C}=\mathbf{A}+(\mathbf{B}+\mathbf{C})\,, \\ \mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}\,,\\[4pt] \mathbf{A}+\mathbf{0}=\mathbf{A}\,, \\ \mathbf{A}+(-\mathbf{A})=\mathbf{0}\,,\\[4pt] (\alpha+\beta)\mathbf{A}=\alpha\mathbf{A}+\beta\mathbf{A}\,, \\ \alpha(\mathbf{A}+\mathbf{B})=\alpha\mathbf{A}+\alpha\mathbf{B}\,. \end{array} $$

For $n$-vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ and a scalar $\alpha$:

1. $\mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a}$;
2. $(\alpha\mathbf{a})\cdot\mathbf{b} =\mathbf{a}\cdot(\alpha\mathbf{b}) =\alpha(\mathbf{a}\cdot\mathbf{b})$;
3. $\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) =\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\cdot\mathbf{c}$;
4. $\mathbf{a}\cdot\mathbf{a}\ge 0$, with equality if and only if $\mathbf{a}=\mathbf{0}$.

Let $\mathbf{A}$ be the coefficient matrix of the system (12.1), let $\mathbf{x}$ be the column vector of unknowns and $\mathbf{b}$ the column vector of constant terms. Then (12.1) says exactly $$ \mathbf{A}\mathbf{x}=\mathbf{b}\,. $$

Let $\mathbf{A},\mathbf{B},\mathbf{C}$ be matrices and $\alpha$ a scalar. Whenever the orders are such that the products written down exist, $$ \begin{array}{l} (\mathbf{A}\mathbf{B})\mathbf{C} = \mathbf{A}(\mathbf{B}\mathbf{C})\,,\\ \quad\text{(associative law)}\\[4pt] \mathbf{A}(\mathbf{B}+\mathbf{C}) = \mathbf{A}\mathbf{B}+\mathbf{A}\mathbf{C}\,,\\ \quad\text{(left distributive law)}\\[4pt] (\mathbf{A}+\mathbf{B})\mathbf{C} = \mathbf{A}\mathbf{C}+\mathbf{B}\mathbf{C}\,,\\ \quad\text{(right distributive law)}\\[4pt] (\alpha\mathbf{A})\mathbf{B} = \mathbf{A}(\alpha\mathbf{B}) = \alpha(\mathbf{A}\mathbf{B})\,. \end{array} $$

If $\mathbf{A}$ is $m\times n$ then $\mathbf{A}\mathbf{I}_{n}=\mathbf{A}$ and $\mathbf{I}_{m}\mathbf{A}=\mathbf{A}$. In particular $\mathbf{A}\mathbf{I}=\mathbf{I}\mathbf{A}=\mathbf{A}$ for every square $\mathbf{A}$, and $\mathbf{I}$ is the only matrix with this property.

For matrices $\mathbf{A},\mathbf{B}$ and a scalar $\alpha$, whenever the operations written down are defined, $$ \begin{aligned} (\mathbf{A}^{\top})^{\top}&=\mathbf{A}\,,\\ (\mathbf{A}+\mathbf{B})^{\top}&=\mathbf{A}^{\top}+\mathbf{B}^{\top}\,,\\ (\alpha\mathbf{A})^{\top}&=\alpha\mathbf{A}^{\top}\,,\\ (\mathbf{A}\mathbf{B})^{\top}&=\mathbf{B}^{\top}\mathbf{A}^{\top}\,. \end{aligned} $$

An elementary row operation produces an equivalent system: the set of solutions is unchanged.

A system of linear equations has either no solution, exactly one solution, or infinitely many. There is no fourth possibility, and Gaussian elimination decides which occurs:

• a row $(0\;\cdots\;0\mid c)$ with $c\ne0$ — no solution;
• otherwise, a leading entry in every one of the $n$ unknown columns — exactly one solution;
• otherwise — infinitely many, with one degree of freedom for each column without a leading entry.

Suppose $\mathbf{s}$ is one solution of $\mathbf{A}\mathbf{x}=\mathbf{b}$. Then the solutions of $\mathbf{A}\mathbf{x}=\mathbf{b}$ are exactly the vectors $\mathbf{s}+\mathbf{h}$, where $\mathbf{h}$ is a solution of the corresponding homogeneous system $\mathbf{A}\mathbf{x}=\mathbf{0}$. In words: $$ \begin{aligned} &\text{all solutions}\\ &=\text{one particular solution}\\ &\quad+\text{all homogeneous solutions}\,. \end{aligned} $$

Write $\mathbf{A}_{j}$ for the matrix $\mathbf{A}$ with its $j$th column replaced by $\mathbf{b}$. If $|\mathbf{A}|\ne0$, the system $\mathbf{A}\mathbf{x}=\mathbf{b}$ has exactly one solution, namely $$ \begin{aligned} x_{1}&=\frac{|\mathbf{A}_{1}|}{|\mathbf{A}|} =\frac{\begin{vmatrix} b_{1} & a_{12}\\ b_{2} & a_{22}\end{vmatrix}} {|\mathbf{A}|}\,, \\ x_{2}&=\frac{|\mathbf{A}_{2}|}{|\mathbf{A}|} =\frac{\begin{vmatrix} a_{11} & b_{1}\\ a_{21} & b_{2}\end{vmatrix}} {|\mathbf{A}|}\,. \end{aligned} $$

With $\mathbf{A}_{j}$ again denoting $\mathbf{A}$ with column $j$ replaced by $\mathbf{b}$: if $|\mathbf{A}|\ne0$ then $\mathbf{A}\mathbf{x}=\mathbf{b}$ has exactly one solution, and $$ x_{j}=\frac{|\mathbf{A}_{j}|}{|\mathbf{A}|}\,,\qquad j=1,2,3\,. $$

For square matrices $\mathbf{A},\mathbf{B}$ of the same order, $$ \begin{aligned} |\mathbf{A}\mathbf{B}| &= |\mathbf{A}|\cdot|\mathbf{B}|\,, \\ |\mathbf{A}^{\top}| &= |\mathbf{A}|\,, \\ |\mathbf{I}_{n}| &= 1\,. \end{aligned} $$

A matrix has at most one inverse.

A square matrix $\mathbf{A}$ has an inverse if and only if $|\mathbf{A}|\ne0$. A square matrix is called singular when $|\mathbf{A}|=0$ and non-singular when $|\mathbf{A}|\ne0$; so invertible and non-singular mean the same thing.

If $|\mathbf{A}|=ad-bc\ne0$, then $$ \begin{aligned} &\mathbf{A}=\begin{pmatrix} a & b\\ c & d\end{pmatrix} \quad\Longrightarrow\quad \\ &\mathbf{A}^{-1}=\frac{1}{ad-bc} \begin{pmatrix} d & -b\\ -c & a\end{pmatrix}\,. \end{aligned} $$ Swap the diagonal entries, change the sign of the other two, divide by the determinant.

For square $\mathbf{A}$ and $\mathbf{X}$ of the same order, $$ \begin{aligned} &\mathbf{A}\mathbf{X}=\mathbf{I} \;\Longleftrightarrow\; \mathbf{X}=\mathbf{A}^{-1}\\ &\;\Longleftrightarrow\; \mathbf{X}\mathbf{A}=\mathbf{I}\,. \end{aligned} $$

Let $\mathbf{A},\mathbf{B}$ be invertible of order $n$ and $c\ne0$ a scalar. Then $\mathbf{A}^{-1}$, $\mathbf{A}\mathbf{B}$, $\mathbf{A}^{\top}$ and $c\mathbf{A}$ are all invertible, and $$ \begin{aligned} (\mathbf{A}^{-1})^{-1}&=\mathbf{A}\,,\\ (\mathbf{A}\mathbf{B})^{-1}&=\mathbf{B}^{-1}\mathbf{A}^{-1}\,,\\ (\mathbf{A}^{\top})^{-1}&=(\mathbf{A}^{-1})^{\top}\,,\\ (c\mathbf{A})^{-1}&=c^{-1}\mathbf{A}^{-1}\,. \end{aligned} $$

Let $\mathbf{A}$ be of order $n$ with $|\mathbf{A}|\ne0$. Then $$ \begin{aligned} &\mathbf{A}\mathbf{X}=\mathbf{B} \iff \mathbf{X}=\mathbf{A}^{-1}\mathbf{B} \\ &\quad(\mathbf{B}\text{ with }n\text{ rows})\,, \\ &\mathbf{Y}\mathbf{A}=\mathbf{B} \iff \mathbf{Y}=\mathbf{B}\mathbf{A}^{-1} \\ &\quad(\mathbf{B}\text{ with }n\text{ columns})\,. \end{aligned} $$ In particular $\mathbf{A}\mathbf{x}=\mathbf{b}$ has the unique solution $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}$, for every right-hand side $\mathbf{b}$.

For a square matrix $\mathbf{A}$ of order $n$, the following say the same thing:

1. $|\mathbf{A}|\ne0$;
2. $\mathbf{A}$ has an inverse;
3. $\mathbf{A}\mathbf{x}=\mathbf{b}$ has exactly one solution for every $\mathbf{b}\in{\mathbb R}^{n}$;
4. Gaussian elimination on $\mathbf{A}$ produces a leading entry in every column.

Let $\mathbf{A}$ be $n\times n$.

• Has $\mathbf{A}\mathbf{x}=\mathbf{b}$ exactly one solution? Compute $|\mathbf{A}|$. Non-zero: yes, for every $\mathbf{b}$. Zero: not for any $\mathbf{b}$ — and the determinant says nothing further.
• What are the solutions? Eliminate. This works whatever $|\mathbf{A}|$ is, and whether or not the system is square.
• How many are there, when $|\mathbf{A}|=0$? Eliminate and read the last row: $(0\;\cdots\;0\mid 0)$ gives infinitely many, $(0\;\cdots\;0\mid c)$ with $c\ne0$ gives none.
• What is $\mathbf{A}^{-1}$? Order $2$: the formula of Fact 14.20. Any order: Method 15.3.