1/X & X^-1 are they the same?

i'm about to write something that involves alot of reciprocals.

i want to know is 1/X the same as X^-1?

mathematically its the same, but which does matlab prefers? (that can make it simulate faster)

or are they really the same?

Is X scalar or matrix?

If X is scalar, you should use the dot operators, ./ and .^

I do not know at the moment which would be faster or more accurate (if either.)

Answer 1

If X is a square matrix, 1/X shall be inv(X) since Matlab can't recognize 1/X for a matrix. Moreover, inv(X) is just the same for a square matrix as X^(-1).

For large matrix, the function inv() is well optimized by matlab and it costs less time than X^(-1). While for small matrix, their computational costs are comparable.

Answer 2

Hi Raymond,

if it's scalars take Walter's advice on ./ and .^ (should be no measurable difference between those). If X is a matrix, it depends, what you need the reziprocals for. For solving linear equations? In this case / (and \) are much preferable to ^(-1): / and \ solve linear systems, ^(-1) computes the inverse (which is a way to solve linear equations but a bad (unstable) one).

Titus

Answer 3

 clear all, clc

 ver
% ---------------------------------------------------------------------------
% MATLAB Version 7.13.0.564 (R2011b)
% OperatingSystem: MicrosoftWindows7 Version6.1 (Build7601: ServicePack1)
% Java VM Version: Java 1.6.0_17-b04 with Sun Microsystems Inc. 
% Java HotSpot(TM) Client VM mixed mode
% ---------------------------------------------------------------------------

 X = randn(100);

 for j = 1:5

    tic
    for i = 1:1e5
        invX1 = eye(100)/X;
    end
    time1(j) = toc              % 47.9   46.8   47.3   46.5   46.9

    tic
    for i = 1:1e5
        invX2 = X^(-1);
    end
    time2(j) = toc            %  62.6   62.8   62.6   61.9   62.3

    for i = 1:1e5
        invX3 = inv(X);
    end
    time3(j) = toc             %  98.7   96.3   96.5   97.8   97.2

 end

Hope this helps.

Greg

maxabs(invX2-invX1), maxabs(invX3-invX1), maxabs(invX2-invX3)

ans = 2.6645e-014

ans = 2.6645e-014

ans = 0

Greg

Answer 4

The division is faster than the power operator. If you are in doubt, test it:

x = rand(1, 1e6);
tic;
for i = 1:length(x)
  x(i) = 1 / x(i);
end
toc;

tic;
for i = 1:length(x)
  x(i) = x(i) ^ -1;
end
toc;

Of course it would be faster to process the complete array at once in this example:

tic;
x = 1 ./ x;
toc;

tic;
x = x .^ -1;
toc;