Solving Linear Systems and Matrix Algebra: Overview

Topic 1 established that

A\mathbf{x}

is a linear combination of

A

's columns. This topic answers the natural next question: given

A

and a target

\mathbf{b}

, how do you actually find the

\mathbf{x}

with

A\mathbf{x}=\mathbf{b}

? The answer — Gaussian elimination — is the single most-used algorithm in applied linear algebra, run trillions of times a day inside every risk system, pricing engine, and portfolio optimiser that solves a linear system. This topic builds it from scratch, packages the bookkeeping into matrix algebra, and shows that elimination is itself a factorisation,

A=LU

, that is faster to reuse than an inverse matrix.

By the end of this topic you should be able to:

solve small systems by hand using forward elimination and back substitution, tracking pivots;
perform matrix addition, multiplication, and transposition, and know which properties (associativity, distributivity) hold and which (commutativity) generally fail;
compute a matrix inverse via Gauss–Jordan elimination and state the conditions for invertibility;
factor $A=LU$ (with row exchanges when needed, $PA=LU$ ) and explain why this factorisation is the practical way to solve $A\mathbf{x}=\mathbf{b}$ .

In this lesson

Contents
The big picture

The full lesson is for members.

Read the full lesson

New to SelfMadeQuant.com? See membership options or browse the Linear Algebra, Optimisation and Numerical Methods syllabus.