HP 48gII - Manual (Page 352)

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