# Revision history [back]

If you ask for the type of $f$, you get <ππ’ππ'ππππ.ππ’ππππππ.ππ‘ππππππππ.π΄π‘ππππππππ'>: apparently the substitution into symbolic expressions is not overloaded to matrix-valued functions. However, with

var('x y')
f = matrix(SR, [[sin(x)],[cos(x)]])
type(f)


you get <ππ’ππ'ππππ.ππππππ‘.ππππππ‘β―ππ’ππππππβ―πππππ.πΌπππππ‘β―ππ’ππππππβ―πππππ'>, and

f(x=x+y)


gives the expected result.

If you ask for the type of $f$, you get <ππ’ππ'ππππ.ππ’ππππππ.ππ‘ππππππππ.π΄π‘ππππππππ'>: apparently the substitution into symbolic expressions is not overloaded to matrix-valued functions. However, with

var('x y')
f = matrix(SR, [[sin(x)],[cos(x)]])
type(f)


you get <ππ’ππ'ππππ.ππππππ‘.ππππππ‘β―ππ’ππππππβ―πππππ.πΌπππππ‘β―ππ’ππππππβ―πππππ'>, and

f(x=x+y)


gives the expected result.

Alternatively, try with:

def f(x): return matrix(SR, [[sin(x)],[cos(x)]])


to obtain an object $f$ of type 'function'.