不太清楚这意味着什么。 “警告：Matrix的工作精度非常高。”
我有一个名为matrix bestM的3x4矩阵 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 我收到了警告 和C = [Inf Inf Inf]这是不对的，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);
答案 0 :(得分：5)
奇异矩阵可以被认为是等于零的矩阵，当你试图反转0时它会爆炸（变为无穷大），这就是你在这里得到的。用户1281385使用format命令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，格式的帮助命令的第一行是
格式不会影响MATLAB计算的完成方式。
答案 1 :(得分：1)
如找到here，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。
为什么我回答，是告诉你明确地使用inv几乎不是一个好主意。如果您需要相同的倒数几百次，那么它可能是值得的，但是，在大多数情况下，您对产品C感兴趣：
C = -inv(Q)*m
可以使用反斜杠运算符在Matlab中更准确，更快地计算：
C = -Q\m
输入help slash以获取更多相关信息。即使你碰巧发现自己确实需要明确的反转，我仍然建议你避免inv：
invQ = Q\eye(size(Q))
下面是一个小小的性能测试，用于演示极少数情况下显式反转可以派上用场的方法之一：
% 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
应该很清楚，明确的逆有其用途，但就像任何语言的goto构造一样 - 谨慎而明智地使用。