Gaussian elimination

This is the second post in a series of posts about matrix methods.
We want to solve \(\mathbf{A}\mathbf{x}=\mathbf{b}\), where \(\mathbf{A}\) is a (known) matrix, \(\mathbf{b}\) is a (known) vector, and \(\mathbf{x}\) is an unknown vector.
This matrix system can be thought of as a way of representing simultaneous equations. For example, the following matrix problem and system of simultaneous equations are equivalent.
\begin{align*} \begin{pmatrix}2&1\\3&1\end{pmatrix}\mathbf{x}&=\begin{pmatrix}3\\4\end{pmatrix} &&\quad&& \begin{array}{r} 2x+2y=3&\\ 3x+2y=4& \end{array} \end{align*}
The simultaneous equations here would usually be solved by adding or subtracting the equations together. In this example, subtracting the first equation from the second gives \(x=1\). From there, it is not hard to find that \(y=\frac12\).
One approach to solving \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is to find the inverse matrix \(\mathbf{A}^{-1}\), and use \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}\). In this post, we use Gaussian elimination—a method that closely resembles the simultaneous equation method—to find \(\mathbf{A}^{-1}\).

Gaussian elimination

As an example, we will use Gaussian elimination to find the inverse of the matrix
$$\begin{pmatrix} 1&-2&4\\ -2&3&-2\\ -2&2&2 \end{pmatrix}.$$
First, write the matrix with an identity matrix next to it.
$$\left(\begin{array}{ccc|ccc} 1&-2&4&1&0&0\\ -2&3&-2&0&1&0\\ -2&2&2&0&0&1 \end{array}\right)$$
Our aim is then to use row operations to change the matrix on the left of the vertical line into the identity matrix, as the matrix on the right will then be the inverse. We are allowed to use two row operations: we can multiply a row by a scalar; or we can add a multiple of a row to another row. These operations closely resemble the steps used to solve simultaneous equations.
We will get the matrix to the left of the vertical line to be the identity in a systematic manner: our first aim is to get the first column to read 1, 0, 0. We already have the 1; to get the 0s, add 2 times the first row to both the second and third rows.
$$\left(\begin{array}{ccc|ccc} 1&-2&4&1&0&0\\ 0&-1&6&2&1&0\\ 0&-2&10&2&0&1 \end{array}\right)$$
Our next aim is to get the second column to read 0, 1, 0. To get the 1, we multiply the second row by -1.
$$\left(\begin{array}{ccc|ccc} 1&-2&4&1&0&0\\ 0&1&-6&-2&-1&0\\ 0&-2&10&2&0&1 \end{array}\right)$$
To get the 0s, we add 2 times the second row to both the first and third rows.
$$\left(\begin{array}{ccc|ccc} 1&0&-8&-3&-2&0\\ 0&1&-6&-2&-1&0\\ 0&0&-2&-2&-2&1 \end{array}\right)$$
Our final aim is to get the third column to read 0, 0, 1. To get the 1, we multiply the third row by -½.
$$\left(\begin{array}{ccc|ccc} 1&0&-8&-3&-2&0\\ 0&1&-6&-2&-1&0\\ 0&0&1&1&1&-\tfrac{1}{2} \end{array}\right)$$
To get the 0s, we add 8 and 6 times the third row to the first and second rows (respectively).
$$\left(\begin{array}{ccc|ccc} 1&0&0&5&6&-4\\ 0&1&0&4&5&-3\\ 0&0&1&1&1&-\tfrac{1}{2} \end{array}\right)$$
We have the identity on the left of the vertical bar, so we can conclude that
$$\begin{pmatrix} 1&-2&4\\ -2&3&-2\\ -2&2&2 \end{pmatrix}^{-1} = \begin{pmatrix} 5&6&-4\\ 4&5&-3\\ 1&1&-\tfrac{1}{2} \end{pmatrix}.$$

How many operations

This method can be used on matrices of any size. We can imagine doing this with an \(n\times n\) matrix and look at how many operations the method will require, as this will give us an idea of how long this method would take for very large matrices. Here, we count each use of \(+\), \(-\), \(\times\) and \(\div\) as a (floating point) operation (often called a flop).
Let's think about what needs to be done to get the \(i\)th column of the matrix equal to 0, ..., 0, 1, 0, ..., 0.
First, we need to divide everything in the \(i\)th row by the value in the \(i\)th row and \(i\)th column. The first \(i-1\) entries in the column will already be 0s though, so there is no need to divide these. This leaves \(n-(i-1)\) entries that need to be divided, so this step takes \(n-(i-1)\), or \(n+1-i\) operations.
Next, for each other row (let's call this the \(j\)th row), we add or subtract a multiple of the \(i\)th row from the \(j\)th row. (Again the first \(i-1\) entries can be ignored as they are 0.) Multiplying the \(i\)th row takes \(n+1-i\) operations, then adding/subtracting takes another \(n+1-i\) operations. This needs to be done for \(n-1\) rows, so takes a total of \(2(n-1)(n+1-i)\) operations.
After these two steps, we have finished with the \(i\)th column, in a total of \((2n-1)(n+1-i)\) operations.
We have to do this for each \(i\) from 1 to \(n\), so the total number of operations to complete Gaussian elimination is
$$ (2n-1)(n+1-1) + (2n-1)(n+1-2) +...+ (2n-1)(n+1-n) $$
This simplifies to $$\tfrac12n(2n-1)(n+1)$$ or $$n^3+\tfrac12n^2-\tfrac12n.$$
The highest power of \(n\) is \(n^3\), so we say that this algorithm is an order \(n^3\) algorithm, often written \(\mathcal{O}(n^3)\). We focus on the highest power of \(n\) as if \(n\) is very large, \(n^3\) will be by far the largest number in the expression, so gives us an idea of how fast/slow this algorithm will be for large matrices.
\(n^3\) is not a bad start—it's far better than \(n^4\), \(n^5\), or \(2^n\)—but there are methods out there that will take less than \(n^3\) operations. We'll see some of these later in this series.
Previous post in series
Matrix multiplication
This is the second post in a series of posts about matrix methods.
Next post in series
Inverting a matrix

Similar posts

Inverting a matrix
Matrix multiplication
Happy τ+e-6 Approximation Day!
A surprising fact about quadrilaterals


Comments in green were written by me. Comments in blue were not written by me.
 Add a Comment 

I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li>
To prove you are not a spam bot, please type "l" then "i" then "n" then "e" then "a" then "r" in the box below (case sensitive):


Show me a random blog post

Jul 2020

Happy τ+e-6 Approximation Day!

May 2020

A surprising fact about quadrilaterals
Interesting tautologies

Mar 2020

Log-scaled axes

Feb 2020

PhD thesis, chapter ∞
PhD thesis, chapter 5
PhD thesis, chapter 4
PhD thesis, chapter 3
Inverting a matrix
PhD thesis, chapter 2

Jan 2020

PhD thesis, chapter 1
Gaussian elimination
Matrix multiplication
Christmas (2019) is over
▼ show ▼
▼ show ▼
▼ show ▼
▼ show ▼
▼ show ▼
▼ show ▼
▼ show ▼
▼ show ▼


game show probability ucl game of life fractals sorting games bodmas sound matrix of minors rugby numerical analysis oeis video games flexagons data visualisation realhats bempp cambridge computational complexity go mathsjam braiding talking maths in public weak imposition bubble bobble sobolev spaces geometry palindromes squares mathslogicbot graphs christmas matrix multiplication binary javascript graph theory approximation wave scattering sport programming golden spiral propositional calculus chebyshev polynomials harriss spiral pythagoras inverse matrices draughts tmip london dates the aperiodical manchester exponential growth probability signorini conditions matt parker interpolation finite element method frobel electromagnetic field arithmetic football hannah fry curvature light big internet math-off news stickers dragon curves boundary element methods royal institution london underground chalkdust magazine pac-man chess coins data craft accuracy pi approximation day a gamut of games folding paper menace wool hats christmas card logic triangles simultaneous equations dataset estimation phd gerry anderson convergence ternary quadrilaterals speed matrix of cofactors error bars determinants golden ratio php platonic solids twitter advent calendar final fantasy people maths raspberry pi matrices cross stitch noughts and crosses weather station latex logs plastic ratio machine learning statistics national lottery books trigonometry inline code preconditioning countdown nine men's morris reuleaux polygons geogebra reddit python asteroids manchester science festival mathsteroids tennis pizza cutting puzzles rhombicuboctahedron european cup map projections martin gardner hexapawn gaussian elimination folding tube maps radio 4 royal baby pi world cup captain scarlet misleading statistics


Show me a random blog post
▼ show ▼
© Matthew Scroggs 2012–2020