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

Asked by raymond on 18 Jun 2012
Latest activity Commented on by Greg Heath on 18 Jun 2012

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?

## 1 Comment

Walter Roberson on 18 Jun 2012

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.)

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Answer by Rui Zhao on 18 Jun 2012

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 by Titus Edelhofer on 18 Jun 2012

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 by Greg Heath on 18 Jun 2012
` 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

## 1 Comment

Greg Heath on 18 Jun 2012

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

ans = 2.6645e-014

ans = 2.6645e-014

ans = 0

Greg

Answer by Jan Simon on 18 Jun 2012

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;
```