Question

我想在Python的Numpy中获得矩阵X和Adjugate（X）的点积，我尝试了从here找到的两种方法。两种方法的佐剂给出相同的答案，但是当我执行点积运算时，给出了不同的答案。这是代码：

stylingdivs:{'childDiv1','childDiv4'}

以下是输出：

<div class="parentDiv">
  <div class="childDiv1342">
    First Div Child
  </div>
  <div class="childDiv2244">
    Second Div Child
  </div>
  <div class="childDiv3342">
    Third Div Child
  </div>
  <div class="childDiv4324">
    Fourth Div Child
  </div>
  `<div class="childDiv5324">
    Fifth Div Child
  </div>`
  `<div class="childDiv6324">
    Sixth Div Child
  </div>`
  `<div class="childDiv7323">
    Seventh Div Child
  </div>`
</div>

然后我尝试通过将它们与X的逆矩阵进行比较来检查它，这次X的所有逆都给出相同的值
代码：

#first method
def CM(A):
    row, col = A.shape
    minor = np.zeros((row-1,col-1))
    cofactor = list()

    for r in range(row):
        for c in range(col):
            minor[:r,:c] = A[:r,:c]
            minor[r:,:c] = A[r+1:,:c]
            minor[:r,c:] = A[:r,c+1:]
            minor[r:,c:] = A[r+1:,c+1:]
            cofactor.append(np.linalg.det(minor)*(-1)**(r+c))

    return np.array(cofactor).reshape(3,3)

#second method
def CM1(A):
    return np.linalg.inv(A).T*np.linalg.det(A)

#define x
X = np.array([[-3,2,-5],[-1,0,-2],[3,-4,1]])
print("Output:\n")
print("X =\n", X)

#the adjugate from both method give same answer
print("\nAdjugate(X) =\n",CM(X).T)
print("\nAdj1(X) =\n",CM1(X).T)

#but when I perform dot product, different answers were given
print("\nX dot Adjugate(X) =\n",X.dot(CM(X).T))
print("\nX dot Adj1(X) =\n",X.dot(CM1(X).T))

输出：

Output:

X =
 [[-3  2 -5]
 [-1  0 -2]
 [ 3 -4  1]]

Adjugate(X) =
 [[-8. 18. -4.]
 [-5. 12. -1.]
 [ 4. -6.  2.]]

Adj1(X) =
 [[-8. 18. -4.]
 [-5. 12. -1.]
 [ 4. -6.  2.]]

X dot Adjugate(X) =
 [[-6.00000000e+00  1.42108547e-14  0.00000000e+00]
 [-1.77635684e-15 -6.00000000e+00  0.00000000e+00]
 [ 1.06581410e-14 -1.42108547e-14 -6.00000000e+00]]

X dot Adj1(X) =
 [[-6.00000000e+00  6.21724894e-15  2.22044605e-15]
 [ 1.77635684e-15 -6.00000000e+00  8.88178420e-16]
 [-4.44089210e-15 -6.21724894e-15 -6.00000000e+00]]

有人可以向我解释为什么吗？

块状，相同值的输入得到不同的点积

0 个答案: