本文整理匯總了Python中sympy.matrices.MatrixSymbol.as_explicit方法的典型用法代碼示例。如果您正苦於以下問題：Python MatrixSymbol.as_explicit方法的具體用法？Python MatrixSymbol.as_explicit怎麽用？Python MatrixSymbol.as_explicit使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類sympy.matrices.MatrixSymbol的用法示例。
在下文中一共展示了MatrixSymbol.as_explicit方法的3個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Python代碼示例。
示例1: test_MatrixSymbol_determinant
# 需要導入模塊: from sympy.matrices import MatrixSymbol [as 別名]
# 或者: from sympy.matrices.MatrixSymbol import as_explicit [as 別名]
def test_MatrixSymbol_determinant():
A = MatrixSymbol('A', 4, 4)
assert A.as_explicit().det() == A[0, 0]*A[1, 1]*A[2, 2]*A[3, 3] - \
A[0, 0]*A[1, 1]*A[2, 3]*A[3, 2] - A[0, 0]*A[1, 2]*A[2, 1]*A[3, 3] + \
A[0, 0]*A[1, 2]*A[2, 3]*A[3, 1] + A[0, 0]*A[1, 3]*A[2, 1]*A[3, 2] - \
A[0, 0]*A[1, 3]*A[2, 2]*A[3, 1] - A[0, 1]*A[1, 0]*A[2, 2]*A[3, 3] + \
A[0, 1]*A[1, 0]*A[2, 3]*A[3, 2] + A[0, 1]*A[1, 2]*A[2, 0]*A[3, 3] - \
A[0, 1]*A[1, 2]*A[2, 3]*A[3, 0] - A[0, 1]*A[1, 3]*A[2, 0]*A[3, 2] + \
A[0, 1]*A[1, 3]*A[2, 2]*A[3, 0] + A[0, 2]*A[1, 0]*A[2, 1]*A[3, 3] - \
A[0, 2]*A[1, 0]*A[2, 3]*A[3, 1] - A[0, 2]*A[1, 1]*A[2, 0]*A[3, 3] + \
A[0, 2]*A[1, 1]*A[2, 3]*A[3, 0] + A[0, 2]*A[1, 3]*A[2, 0]*A[3, 1] - \
A[0, 2]*A[1, 3]*A[2, 1]*A[3, 0] - A[0, 3]*A[1, 0]*A[2, 1]*A[3, 2] + \
A[0, 3]*A[1, 0]*A[2, 2]*A[3, 1] + A[0, 3]*A[1, 1]*A[2, 0]*A[3, 2] - \
A[0, 3]*A[1, 1]*A[2, 2]*A[3, 0] - A[0, 3]*A[1, 2]*A[2, 0]*A[3, 1] + \
A[0, 3]*A[1, 2]*A[2, 1]*A[3, 0]示例2: test_as_explicit_nonsquare_symbol
# 需要導入模塊: from sympy.matrices import MatrixSymbol [as 別名]
# 或者: from sympy.matrices.MatrixSymbol import as_explicit [as 別名]
def test_as_explicit_nonsquare_symbol():
X = MatrixSymbol('X', 2, 3)
assert MatPow(X, 1).as_explicit() == X.as_explicit()
for r in [0, 2, S.Half, S.Pi]:
raises(ShapeError, lambda: MatPow(X, r).as_explicit())示例3: test_as_explicit_symbol
# 需要導入模塊: from sympy.matrices import MatrixSymbol [as 別名]
# 或者: from sympy.matrices.MatrixSymbol import as_explicit [as 別名]
def test_as_explicit_symbol():
X = MatrixSymbol('X', 2, 2)
assert MatPow(X, 0).as_explicit() == ImmutableMatrix(Identity(2))
assert MatPow(X, 1).as_explicit() == X.as_explicit()
assert MatPow(X, 2).as_explicit() == (X.as_explicit())**2