y
P
A
R
ex
R
R
/2
ey
x
A
BC
BC
R
/2
3
2
R
0
0
z
F
ABCP
=
++−=
∑
∑
00
/2 /2 0
0
x
y
M
AR Pe BR CR
M
=
−− − =
=
∑
∑
0
0
3/2 3/2 0
z
x
M
Pe CR BR
=
+−=
∑
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Solving Systems of Linear Equations: Homogeneous and Non-Homogeneous Cases, Study notes of Engineering
A step-by-step solution for solving systems of linear equations using matrix algebra. Both homogeneous and non-homogeneous systems, as well as the concept of the weighted system. It also discusses the regions of tipping between variables when the weight is zero.
Typology: Study notes
Pre 2010
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y
P
A ex R (^) R^ R/
ey
x
A B^
C
B^
C R/2 3 R^2 0
0
F^ z A^ B
C^
P = + +^
−^ =
∑ ∑^
(^00)
/ 2^
/ 2^
0
0 M^ x y AR^
Pe^
BR^
CR
M
= −^
−^
−^
=
=
∑ ∑^
(^00) 3 / 2
3 / 2
0
M^ z x Pe^
CR^
BR = +
−^
=
solving for A,B,C
P^ ⎡^
y y x P A^
R^
e R P B^
R^
e^
e ⎡^ R
=^
⎡^
=^
−^
⎣^
y^
x y^
x
R P C^
R^
e^
e ⎣^ R
⎡^
=^
−^
⎣^
Note: if e
= ex^
=0 theny^
P^3
A^
B^
C
=^
=^
If the table is homogeneous and has a weight W then
y
W^
P
A^
R^
e ⎡^ R
=^
+^
If^ the table is homogeneous and has a weight W then
y^
x
R W^
P
B^
R^
e^
e
R W^
⎣^ P
⎡^
=^
+^
−^
⎣^
y^
x
W^
P
C^
R^
e^
e ⎡^ R
=^
+^
−^
⎣^
If the weight is nonzero then we need a larger distance to make A = 0
y