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Exercise Prove that

[ a b ]
c d

is invertible ⇔ ad - bc ≠ 0.

Answer Denote the matrix by A.

In the lecture, we have already shown that ad - bc ≠ 0 is necessary for A to be invertible.

Conversely, assume ad - bc ≠ 0. Then either a ≠ 0 or c ≠ 0. If a ≠ 0, then the operation -a-1c[row 1] + [row 2] on A gives us

R = [ a b ]
0 d - a-1cb

Since a ≠ 0 and d - a-1cb = a-1(ad - bc) ≠ 0, all rows and columns of R are pivot. By this criterion, A is invertible. If c ≠ 0, then the operations [row 1] ↔ [row 2], -c-1a[row 1] + [row 2] on A give us

R = [ c d ]
0 b - c-1ad

Since c ≠ 0 and b - c-1ad = -c-1(ad - BC) ≠ 0, all rows and columns of R are pivot. By this criterion, A is invertible.