Numerics and Linear Algebra

Prerequisites: Calculus

Machine and arbitrary precision

0.1 + 0.2
InputForm[0.1 + 0.2]
0.1 + 0.2 == 0.3
$MachinePrecision
Precision[N[Pi, 30]]

0.3
0.30000000000000004
True
15.9546
30.

Machine numbers are ordinary 64-bit floats (about 16 digits), so 0.1 + 0.2 isn't exactly 0.3 — InputForm shows every digit.
== on approximate numbers tolerates differences in the last few binary digits, so the comparison is True. Other languages would say False.
Numbers created with N[expr, n] are arbitrary-precision and track their precision through calculations.

N[Exp[Pi Sqrt[163]], 35]

262537412640768743.99999999999925007

With machine precision this famous number looks like an integer; 35 digits reveal that it isn't.

Vectors are lists; matrices are lists of rows.

m = {{1, 2}, {3, 4}};
Det[m]
Inverse[m]
m . {1, 1}
m . m
m * m
Transpose[m]
LinearSolve[m, {5, 11}]

-2
{{-2, 1}, {3/2, -1/2}}
{3, 7}
{{7, 10}, {15, 22}}
{{1, 4}, {9, 16}}
{{1, 3}, {2, 4}}
{1, 2}

. (Dot) is matrix multiplication; * multiplies element by element.
LinearSolve[m, b] solves m . x == b — better than computing the inverse.
MatrixForm[m] displays a matrix as a grid (for display only — don't store it in a variable you compute with).

Eigenvalues[{{2, 1}, {1, 2}}]
Eigenvectors[{{2, 1}, {1, 2}}]
IdentityMatrix[2]

{3, 1}
{{1, 1}, {-1, 1}}
{{1, 0}, {0, 1}}

Variance[{1, 2, 3, 4}]
data = {{1, 2.1}, {2, 3.9}, {3, 6.2}, {4, 7.8}};
lm = LinearModelFit[data, x, x];
Normal[lm]
lm["RSquared"]
lm[5]

5/3
0.15 + 1.94 x
0.995661
9.85

Variance and StandardDeviation use the sample (n − 1) formula.
LinearModelFit returns a fitted model object: lm["RSquared"], lm["ParameterTable"], and lm[x0] for predictions. NonlinearModelFit and FindFit fit any formula.

Mixing exact and approximate numbers in a huge symbolic computation — exact arithmetic on big expressions can be slow. Use N early when you only need numbers.
Using * for matrix multiplication.