我在Julia的矩阵乘法中犯了什么错误? [英] What mistake did I make in this matrix multiplication in Julia?

问题描述

在朱莉娅中:

In  [1]: M1 = [1 3 4;
              45 64 33;
              456 3 454;]

Out [1]: 3x3 Array{Int64,2}:
   1   3    4
  45  64   33
 456   3  454

In  [2]: M1 * inv(M1)

Out [2]: 3x3 Array{Float64,2}:
  1.0           6.93889e-18  -8.67362e-19
  0.0           1.0          -2.08167e-17
 -1.42109e-14  -8.88178e-16   1.0        

M1 * inv(M1)应该通过定义来获得恒等矩阵.怎么了?

M1 * inv(M1) is supposed to get the Identity matrix by definition. What's wrong?

我在Matlab中尝试过同样的事情:

I tried the same thing in Matlab:

>> M1 = [1 3 4;
         45 64 33;
        456 3 454;]
M1 =
     1     3     4
    45    64    33
   456     3   454
>> inv(M1)
ans =
  -0.280088987764182   0.013057987135465   0.001518595540939
   0.052057842046719   0.013251438796731  -0.001421869710306
   0.280978865406007  -0.013203075881414   0.000686753397495
>> M1 * inv(M1)
ans =
   1.000000000000000   0.000000000000000  -0.000000000000000
                   0   1.000000000000000  -0.000000000000000
  -0.000000000000014  -0.000000000000001   1.000000000000000
>> 

Matlab在这里返回正确的结果.我想朱莉娅在这里不会犯错.那么我的计算/符号出了什么问题?

Matlab returns the right result here. I guess Julia will not make a mistake here. So what's wrong with my calculation / notation?

修改

该问题是由浮点结果中的位数引起的.我应该问过，如何在Julia中设置结果数字的精度?

The problem is caused by number of digits in floating point result. I should have asked, how to set result digits precision in Julia?

推荐答案

Julia和Matlab实际上给出了相同的结果 (例如，在两种情况下，左下元素均为-1.4e-14): 它不是完全相同的单位矩阵，因为浮点算法不精确.

Julia and Matlab actually give the same result (for instance, the bottom-left element is -1.4e-14 in both cases): it is not exactly the identity matrix because floating point arithmetic is not exact.

您可以在显示结果之前显式舍入结果.

You can explicitly round the result before displaying it.

M1 = [
  1    3   4;
  45  64  33;
  456  3 454
]
round( M1 * inv(M1), 6 )
# 3x3 Array{Float64,2}:
#  1.0   0.0  -0.0
#  0.0   1.0  -0.0
#  0.0  -0.0   1.0

如果您想要确切的结果，也可以使用有理.

If you want an exact result, you can also use rationals.

M1 = [
  1//1  3   4;
  45   64  33;
  456   3 454
]
M1 * inv(M1)
# 3x3 Array{Rational{Int64},2}:
#  1//1  0//1  0//1
#  0//1  1//1  0//1
#  0//1  0//1  1//1

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