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Solve symbolic matrix cell

asked 2019-03-19 14:44:52 -0600

I'm trying to take a cell from a matrix and solve for a variable symbolically. When I run this, I get the error:

TypeError: The first argument must be a symbolic expression or a list of symbolic expressions.

This is the code I am trying to execute:

var('a','b','c','x','y','z')
Rz = matrix([[cos(a),-sin(a),0,0],[sin(a),cos(a),0,0],[0,0,1,0],[0,0,0,1]])
Ry = matrix([[cos(b),0,sin(b),0],[0,1,0,0],[-sin(b),0,cos(b),0],[0,0,0,1]])
Rx = matrix([[1,0,0,0],[0,cos(c),-sin(c),0],[0,sin(c),cos(c),0],[0,0,0,1]])
P = matrix([[x],[y],[z],[1]])

Rzyxp = Rz * Ry * Rx * P
show(Rzyxp)

eqn2 = Rzyxp[2]
show(eqn2)

result1 = solve(eqn2==0, b)
show(result1)

Any help would be appreciated

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answered 2019-03-20 14:31:51 -0600

Emmanuel Charpentier gravatar image

nbrion's answer's usefulness can be slightly enhanced with a few assumptions:

 sage: with assuming(b>-pi/2,b<pi/2): ((eqn2[0]).solve(b)[0]/cos(b)).trig_reduce()
....: .arctan().trig_expand()
b == arctan(z*cos(c)/x + y*sin(c)/x)

HTH,

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answered 2019-03-20 11:33:44 -0600

nbruin gravatar image

Rzyxp is a matrix, so Rzyxp[2] is its third row: a vector; albeit a one-dimensional one. In order to get the entry itself, you need to index with two indices:

solve(Rzyxp[2,0],b)

does not give an error (but possibly not a very useful answer either).

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Asked: 2019-03-19 14:44:52 -0600

Seen: 62 times

Last updated: Mar 20