The formula this page uses
det [[a, b], [c, d]] = ad − bc A⁻¹ = (1 / det A) · [[d, −b], [−c, a]]
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det [[a, b], [c, d]] = ad − bc A⁻¹ = (1 / det A) · [[d, −b], [−c, a]]
The system
It means the matrix squashes space flat: a square of area 1 gets mapped onto a line of area 0. Because you cannot un-flatten it, there is no inverse, and the system either has no solutions or infinitely many depending on b.
Each entry of the product is a row of the first matrix paired term by term with a column of the second. That pairing only works if the row and the column have the same length, which is why an m×n can only multiply an n×p.
It carries the same information in a shorter form. A holds only the coefficients, so you must keep b beside it. The two zero-determinant rows in the table above share the identical matrix and differ only in b, yet one has no solutions and the other has infinitely many.
No, and for anything bigger than 2×2 you usually should not. Row-reducing the augmented matrix takes fewer operations and keeps working when the determinant turns out to be zero, whereas the inverse method simply breaks.
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