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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
This results from multiplying the first equation by A
definition, A
-1
⋅A = I, thus we write I⋅x = A
-1
x = A
⋅ b.
For the example used earlier, namely,
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
-1
, i.e., A
-1
⋅b. Also, I⋅x = x, thus, we have,
2x
+ 3x
–5x
= 13,
1
2
3
x
– 3x
+ 8x
= -13,
1
2
3
2x
– 2x
+ 4x
= -6,
1
2
3
-1
⋅ b.
-1
⋅A⋅x = A
-1
⋅b. By
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