How is |A| != 0 shown when A^-1 exists?

|A||A^-1| = |I| = 1, so neither determinant can be zero

Because A^-1 = adj A, so |A| = 0 is impossible

Which identity gives the inverse when |A| != 0?

A(adj A) = (adj A)A = |A| I, so A^-1 = adj A / |A|

A + adj A = |A| I, so A^-1 = |A| adj A

If B and C are inverses, B = B(AC) = (BA)C = C

Because the determinant of the inverse is 1

Because adj A is unique, B must equal C only if A is symmetric

Short answer

A square matrix A has an inverse iff |A| ≠ 0. Necessity: |A||A⁻¹| = 1 ⇒ |A| ≠ 0. Sufficiency: if |A| ≠ 0, A⁻¹ = adj(A)/|A| exists.

Prove that the necessary and sufficient condition for a square matrix A to possess an inverse is that A is non-singular.

A square matrix A of order n has an inverse if and only if |A| ≠ 0 (A is non-singular). We prove both directions.

Tool used: for any square matrix, A(adj A) = (adj A)A = |A| Iₙ, where adj A is the transpose of the matrix of cofactors.

Let A⁻¹ exist, so AA⁻¹ = I. Taking determinants and using |AB| = |A||B|:

|A| · |A⁻¹| = |I| = 1

The product of the two numbers is 1, so neither can be zero. Hence |A| ≠ 0.

Let |A| ≠ 0. From A(adj A) = (adj A)A = |A| I, divide by the non-zero scalar |A|:

A · (adj A / |A|) = (adj A / |A|) · A = I

So the matrix B = adj A / |A| satisfies AB = BA = I. By definition B is the inverse of A, so A⁻¹ = adj A / |A| exists.

If B and C are both inverses, B = BI = B(AC) = (BA)C = IC = C.

Thus A has an inverse if and only if A is non-singular. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)