Hello, DivideBy0!
Here's a primitive way to solve it . . .
Let X .= .| a b |A = [4 3]
. . . [2 1]
If AX = [3 4]
. . . . . .[1 6]
Find X.
. . . - . . .| c d |
Then we have: . | 4 3 | . | a b | -= -| 3 4 |
. . . . . . . . . . . .| 2 1 | . | c d | . . . | 1 6 |
Hence: . | 4a + 3c . 4b + 3d | -= -| 3 4 |
. . - . - . | 2a + .c . .2b + -d | . . . | 1 6 |
And we have two system of equations:
. . 4a + 3c .= .3 . . . . 4b + 3d .= .4
. . 2a + .c -= -1 . . . . 2b + . d .= .6
Solve by your favorite method and get:
. . a = 0, .b = 7, .c = 1, .d = -8
Hello, singular!
You left out the fraction . . .
. . . . . . | a b |
Given: . | . - .|
. . . . . . | c d |
. . . . - . . . . . . .| . . d . . . . .-b . - |
. . . . - . . . . . . .| -------- . --------- |
. . . . - . . . . . . .| ad - bc . ad - bc .|
The inverse is: . | . . . . . . . . . . . . |
. . . . - . . . . . . .|. . -c . . . . . a . . .|
. . . . - . . . . . . .| -------- . --------- |
. . . . - . . . . . . .| ad - bc . ad - bc .|
In baby-talk:
. . Switch the elements on the main (southeast) diagonal.
. . Change the signs of the elements on the other diagonal.
. . Divide all elements by the determinant of the matrix.