HP 50G, Руководство пользователя

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Page 11-27
Compare these three solutions with the ones calculated with the numerical
solver.
Solution with the inverse matrix
The solution to the system A
⋅
x = b , where A is a square matrix is x = A
-1
⋅
b .
This results from multiplying the first equation by A
-1
, i.e., A
-1
⋅
A
⋅
x = A
-1
⋅
b . By
definition, A
-1
⋅
A = I , thus we write I
⋅
x = A
-1
⋅
b . Also, I
⋅
x = x , thus, we have, x
= A
-1
⋅
b .
For the example used earlier, namely,
2x
1
+ 3x
2
–5x
3
= 13,
x
1
– 3x
2
+ 8x
3
= -13,
2x
1
– 2x
2
+ 4x
3
= -6,
we can find the solution in the calculator as follows:
which is the same result found earlier.
Solution by “division” of matrices
While the operation of division is not defined for matrices, we can use the
calculator’s
/
key to “divide” vector b by matrix A to solve for x in the matrix
equation A
⋅
x = b . This is an arbitrary extension of the algebraic division
operation to matrices, i.e., from A
⋅
x = b , we dare to write x = b / A
(Mathematicians would cringe if they see this!) This, of course is interpreted as
(1/
A )
⋅
b = A
-1
⋅
b , which is the same as using the inverse of A as in the previous
section.