Area. The matrix [[a, b], [c, d]] sends the unit square to a parallelogram with sides u = (a, c) (the first column, where [1, 0] lands) and v = (b, d) (the second column, where [0, 1] lands). Draw it inside the rectangle of width a + b and height c + d. Cut away everything that is not the parallelogram: two triangles of area ½ac, two of area ½bd, and two small rectangles of area bc. What remains:
(a+b)(c+d) − ac − bd − 2bc = ad − bc
So the determinant is the area of the image of the unit square,
with a minus sign when orientation flips. Zero area = a crushed
dimension = no inverse.
Inverse. We want X with A @ X = I. Try X = (1/det) · [[d, −b], [−c, a]] and multiply it out:
A @ [[d, −b], [−c, a]] = [[ad − bc, 0], [0, ad − bc]] = (ad − bc) · I
divide by det: A @ ( [[d, −b], [−c, a]] / det ) = I ✓
Recipe: swap the diagonal, negate the off-diagonal, divide by det.
A = [[1, 2], [3, 4]], det = −2:
A⁻¹ = [[4, −2], [−3, 1]] / (−2) = [[−2, 1], [1.5, −0.5]]
Dividing by zero is impossible — the algebraic face of “singular matrices have no inverse”.