Matlab中subs函数

Question

syms x y;
subs(x*y, {x, y}, {[0 1; -1 0], [1 -1; -2 1]})
x与y是'*'，不是'.*'啊,为什么会得到这个结果
ans = 0 -1
2 0

Answer 1

subs是matlab里用于替换的函数，syms定义参数，
subs(x*y, {x, y}, {[0 1; -1 0], [1 -1; -2 1]})，意思是用{[0 1; -1 0]里的各个元素分别替换x，用 [1 -1; -2 1]里的各个元素替换y，分别做积，“；”起到换行作用。比如用0替换x，1替换y，做积得0；由1*（-1）=-1，（-1）*（-2）=2，0*1=0

Answer 2

这个不是矩阵乘法，就是将X,Y 对应代入相乘得到的表达式，0*1，1*-1；-1*-2,0*1 得到结果

Answer 3

呃，是么

噢。这是符号计算啊，跟数值计算不是同一个玩意儿

matlab里符号计算是调用符号计算引擎，一般是maple或者matlab_symbolic（就是mupad），而.*是matlab数值计算里才有的符号

Answer 4

这里subs的替换换就是元素替换，没有矩阵替换后执行矩阵运算

你要进行矩阵运算可以这样写

>> x=sym([0 1;-1 0]);
>> y=sym([1 -1;-2 1]);
>> x*y

ans =

[ -2, 1]
[ -1, 1]

追问

我知道这些，我的问题是为什么会出现这个结果？

追答

subs的替换就是元素替换，就算是带入矩阵也是执行元素运算，subs程序就是这样写的啊
SUBS Symbolic substitution.
SUBS(S) replaces all the variables in the symbolic expression S with
values obtained from the calling function, or the MATLAB workspace.

SUBS(S,NEW) replaces the free symbolic variable in S with NEW.
SUBS(S,OLD,NEW) replaces OLD with NEW in the symbolic expression S.
OLD is a symbolic variable, a string representing a variable name, or
a string (quoted) expression. NEW is a symbolic or numeric variable
or expression.

If OLD and NEW are vectors or arrays of the same size, each element
of OLD is replaced by the corresponding element of NEW. If S and OLD
are scalars and NEW is an array or cell array, the scalars are expanded
to produce an array result. If NEW is a cell array of numeric matrices,
the substitutions are performed elementwise (i.e., subs(x*y,{x,y},{A,B})
returns A.*B when A and B are numeric).

If SUBS(S,OLD,NEW) does not change S, then SUBS(S,NEW,OLD) is tried.
This provides backwards compatibility with previous versions and
eliminates the need to remember the order of the arguments.
SUBS(S,OLD,NEW,0) does not switch the arguments if S does not change.

Multiple Substitutions:
subs(cos(a)+sin(b),{a,b},[sym('alpha'),2]) or
subs(cos(a)+sin(b),{a,b},{sym('alpha'),2}) returns
cos(alpha)+sin(2)

Scalar Expansion Case:
subs(exp(a*t),'a',-magic(2)) returns

[ exp(-t), exp(-3*t)]
[ exp(-4*t), exp(-2*t)]

Multiple Scalar Expansion:
subs(x*y,{x,y},{[0 1;-1 0],[1 -1;-2 1]}) returns
[ 0, -1]
[ 2, 0]

Answer 5

这个结果对的啊