Jacobian: Row Swapping & Factor Extracting

[terryfields]

how do we get from this line

||d(x,y,z)/d(p,q,r)||= ||sin(q)cos(r), pcos(q)cos(r), -psin(q)sin(r) end of line 1
line 2 sin(q)sin(r), pcos(q)sin(r), psin(q)cos(r) end of line 2
line 3 cos(q), -psin(q), 0||

to the next line where we take out cos(q) to get cos(q)||pcos(q)cos(r), -sin(q)sin(r) end of line 1
line 2 pcos(q)sin(r), psin(q)cos(r)||

i thought to take out a factor like taht you needed a line or a row of zeros where as here we don't have that?

Edit:read it wrong it has -(-psin(q)) of a different matrix after this so I'm guessing all they have done is swaped row 1 and 3 to make things easier, are there any restrictions to row swaping? and I'm guessing you can't column swap at all

 

[terryfields]

additional how to you get the determinant of pcos(a)cos(b) -sin(a)sin(b) line 1
line 2 pcos(a)sinb psin(a)cosb

to come out as p²sin(a)cos(a)??

 

[Dick]

Swapping two rows changes the sign of the determinant. Nothing else changes. The same with swapping columns.

 

[Dick]

additional how to you get the determinant of pcos(a)cos(b) -sin(a)sin(b) line 1
line 2 pcos(a)sinb psin(a)cosb

to come out as p²sin(a)cos(a)??

I think you missed a 'p'. Just use cos(b)^2+sin(b)^2=1.

 

[jambaugh]

how do we get from this line

||d(x,y,z)/d(p,q,r)||= ||sin(q)cos(r), pcos(q)cos(r), -psin(q)sin(r) end of line 1
line 2 sin(q)sin(r), pcos(q)sin(r), psin(q)cos(r) end of line 2
line 3 cos(q), -psin(q), 0||

Just so it's readable (and to practice my typesetting)...
$$\left|\frac{\partial(x,y,z)}{\partial(p,q,r)}\right|= <br /> \left|\begin{array}{ccc}{\sin(q)\cos(r) & p\cos(q)\cos(r) & -p\sin(q)\sin(r) &<br /> 2\sin(q)\sin(r) & p\cos(q)\sin(r) & p \sin(q)\cos(r) &<br /> 3\cos(q) & -p\sin(q) & 0&<br /> } \end{array}<br /> \right| =$$

$$=<br /> 3\cos(q)\left|\begin{array}{cc}{<br /> p\cos(q)\cos(r) & -p\sin(q)\sin(r) &<br /> p\cos(q)\sin(r) & p\cos(q)\sin(r) }<br /> \end{array}\right| <br /> -(-p\sin(q))\left| \begin{array}{cc}{<br /> \sin(q)\cos(r) & -p\sin(q)\sin(r) &<br /> 2\sin(q)\sin(r) & p\sin(q)\cos(r)<br /> }\end{array}<br /> \right| + 0$$

to the next line where we take out cos(q) to get cos(q)||pcos(q)cos(r), -sin(q)sin(r) end of line 1
line 2 pcos(q)sin(r), psin(q)cos(r)||

i thought to take out a factor like taht you needed a line or a row of zeros where as here we don't have that?

Edit:read it wrong it has -(-psin(q)) of a different matrix after this so I'm guessing all they have done is swaped row 1 and 3 to make things easier, are there any restrictions to row swaping? and I'm guessing you can't column swap at all

Swapping two rows or two columns will change the sign of the determinant. Multiplying a row or column by a value multiplies the determinant by that value.

What is happening here is cofactor expansion of a determinant. Pick any row or column (best the one with the most zeros) and for each non-zero term you either add or subtract that term times the submatrix you get when you eliminate both the row and column in which that term occurs. The add vs. subtract rule follows the pattern of checkerboard +/- starting with a + for the top left entry. The pattern is:

$$\left(\begin{array}{cccc}<br /> + & - & + \cdots \\<br /> - & + & - \cdots \\<br /> + & - & + \cdots\\<br /> \vdots & \vdots & \vdots \ddots<br /> \end{array}\right)$$

In this case we expand via the third row with its two non-zero entries. The first entry 3cos(q) ) is in a + position so we add it times the determinant of the matrix we get by eliminating column 1 and row 3.
To this we subtract (- position) the second term(-p sin(q) ) time the submatrix we get by eliminating column 2 and row 3. The zero term needn't be expanded but it would be zero times the submatrix we get by eliminating row 3 and column 3.

 

[terryfields]

thanks guys, off out now but will read through that later, from first glance i think all my problems are answered =)