HP 50g Graphing Calculator User Manual
Page 365
Page 11-38
2 Y \#3#1@RCIJ
Finally, we eliminate the –1/16 from position (1,2) by using:
16 Y # 2#1@RCIJ
We now have an identity matrix in the portion of the augmented matrix
corresponding to the original coefficient matrix A, thus we can proceed to
obtain the solution while accounting for the row and column exchanges coded
in the permutation matrix P. We identify the unknown vector x, the modified
independent vector b’ and the permutation matrix P as:
The solution is given by P
⋅x=b’, or
Which results in
Step-by-step calculator procedure for solving linear systems
The example we just worked is, of course, the step-by-step, user-driven
procedure to use full pivoting for Gauss-Jordan elimination solution of linear
equation systems. You can see the step-by-step procedure used by the
calculator to solve a system of equations, without user intervention, by setting
the step-by-step option in the calculator’s CAS, as follows:
1
-1/16
0
33/16
0
1
0
0
1
0
-1
0
0
1
0
0
1
1
1
0
0
1
0
0
2
0 1 0
0
1
0
-1
0 0 1
0
0
1
1
1 0 0
.
0
0
1
1
0
0
0
1
0
,
1
1
2
'
,
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
P
b
x
Z
Y
X
.
1
1
3
0
0
1
1
0
0
0
1
0
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⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
=
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⎥
⎥
⎦
⎤
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⎢
⎢
⎣
⎡
⋅
⎥
⎥
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⎢
⎢
⎣
⎡
Z
Y
X
.
1
1
3
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
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−
=
⎥
⎥
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⎦
⎤
⎢
⎢
⎢
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⎡
X
Z
Y