HP 49g+, Руководство пользователя, Страница: 348 / 862

Page 11-26
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