不太清楚这意味着什么。“警告：矩阵对于工作精度来说是独一无二的。”
我有一个 3x4 矩阵，称为矩阵 bestM 矩阵 Q 是 bestM 的 3x3，矩阵 m 是 bestM 的最后一列
我想做 C = -Q * 矩阵 m 的逆矩阵，我得到那个警告和 C =[Inf Inf Inf] 这是不正确的，因为我正在计算世界上的相机中心
bestM = [-0.0031 -0.0002 0.0005 0.9788;
-0.0003 -0.0006 0.0028 0.2047;
-0.0000 -0.0000 0.0000 0.0013];
Q = bestM(1:3,1:3);
m = bestM(:,4);
X = inv(Q);
C = -X*m;
disp(C);
可以将奇异矩阵视为等效于零的矩阵，当您尝试反转 0 时，它会爆炸（变为无穷大），这就是您在这里得到的。用户 1281385 使用 format 命令来提高精度是绝对错误的；format 命令用于更改显示给您的格式。事实上，格式帮助命令的第一行说
格式不影响 MATLAB 计算的完成方式。
如此处发现，奇异矩阵是没有逆矩阵的矩阵。正如 dvreed77 已经指出的那样，您可以将其视为矩阵的 1/0。
为什么我要回答，是要告诉你inv明确使用几乎从来都不是一个好主意。如果您需要几百次相同的逆，这可能是值得的，但是，在大多数情况下，您对产品感兴趣C：
C = -inv(Q)*m
在 Matlab 中使用反斜杠运算符可以更准确、更快地计算：
C = -Q\m
键入help slash以获取更多信息。即使您碰巧发现自己确实需要明确地反转，我仍然建议您避免inv：
invQ = Q\eye(size(Q))
下面是一个小性能测试，以演示显式逆可以派上用场的极少数情况之一：
% This test will demonstrate the one case I ever encountered where
% an explicit inverse proved useful. Unfortunately, I cannot disclose
% the full details without breaking the law, but roughly, it came down
% to this: The (large) design matrix A, a result of a few hundred
% co-registrated images, needed to be used to solve several thousands
% of systems, where the result matrices b came from processing the
% images one-by-one.
%
% That means the same design matrix was re-used thousands of times, to
% solve thousands of systems at a time. To add to the fun, the images
% were also complex-valued, but I'll leave that one out of consideration
% for now :)
clear; clc
% parameters for this demo
its = 1e2;
sz = 2e3;
Bsz = 2e2;
% initialize design matrix
A = rand(sz);
% initialize cell-array to prevent allocating memory from consuming
% unfair amounts of time in the first loop.
% Also, initialize them, NOT copy them (as in D=C,E=D), because Matlab
% follows a lazy copy-on-write scheme, which would influence the results
C = {cellfun(@(~) zeros(sz,Bsz), cell(its,1), 'uni', false) zeros(its,1)};
D = {cellfun(@(~) zeros(sz,Bsz), cell(its,1), 'uni', false) zeros(its,1)};
E = {cellfun(@(~) zeros(sz,Bsz), cell(its,1), 'uni', false) zeros(its,1)};
% The impact of rand() is the same in both loops, so it has no
% effect, it just gives a longer total run time. Still, we do the
% rand explicitly to *include* the indexing operation in the test.
% Also, caching will most definitely influence the results, because
% any compiler (JIT), even without optimizations, might recognize the
% easy performance gain when the code computes the same array over and
% over again. It probably will, but we have no control over when and
% wherethat happens. So, we prevent that from happening at all, by
% re-initializing b at every iteration.
% The assignment to cell is a necessary part of the demonstration;
% it is the desired output of the whole calculation. Assigning to cell
% instead of overwriting 'ans' takes some time, which is to be included
% in the demonstration, again for cache reasons: the extra time is now
% guaranteed to be equal in both loops, so it really does not matter --
% only the total run time will be affected.
% Direct computation
start = tic;
for ii = 1:its
b = rand(sz,Bsz);
C{ii,1} = A\b;
C{ii,2} = max(max(abs( A*C{ii,1}-b )));
end
time0 = toc(start);
[max([C{:,2}]) mean([C{:,2}]) std([C{:,2}])]
% LU factorization (everyone's
start = tic;
[L,U,P] = lu(A, 'vector');
for ii = 1:its
b = rand(sz,Bsz);
D{ii,1} = U\(L\b(P,:));
D{ii,2} = max(max(abs( A*D{ii,1}-b )));
end
time1 = toc(start);
[max([D{:,2}]) mean([D{:,2}]) std([D{:,2}])]
% explicit inv
start = tic;
invA = A\eye(size(A)); % NOTE: DON'T EVER USE INV()!
for ii = 1:its
b = rand(sz,Bsz);
E{ii,1} = invA*b;
E{ii,2} = max(max(abs( A*E{ii,1}-b )));
end
time2 = toc(start);
[max([E{:,2}]) mean([E{:,2}]) std([E{:,2}])]
speedup0_1 = (time0/time1-1)*100
speedup1_2 = (time1/time2-1)*100
speedup0_2 = (time0/time2-1)*100
结果：
% |Ax-b|
1.0e-12 * % max. mean st.dev.
0.1121 0.0764 0.0159 % A\b
0.1167 0.0784 0.0183 % U\(L\b(P,;))
0.0968 0.0845 0.0078 % invA*b
speedup0_1 = 352.57 % percent
speedup1_2 = 12.86 % percent
speedup0_2 = 410.80 % percent
应该清楚的是，显式逆有它的用途，但就像goto任何语言中的构造一样——谨慎而明智地使用它。