Consider the following 2x2 matrix, using the Cayley -Hamilton theorem show that if det(A) does not equal 0...

Consider the following 2x2 matrix, using the Cayley -Hamilton theorem show that if det(A) does not equal 0, then A^-1 = (tr(A) I - A) / (det(A))

The matrix A =
[a b]
[c d]

thanks !

The characteristic equation for A is given by |A - λI| = 0.

For our 2 x 2 matrix, this yields
|a-λ b|
|c d-λ| = 0

Expanding,
(a - λ)(d - λ) - bc = 0
==> λ^2 - (a + d)λ + (ad - bc) = 0.

However, tr(A) = a + d, and det(A) = ad - bc.

Hence, the characteristic equation for A is
λ^2 - tr(A) λ + det(A) = 0.
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Now, we invoke the Cayley-Hamilton Theorem, which states that A satisfies its own characteristic equation. So, we obtain A^2 - tr(A) A + det(A) I = 0.

Finally, we solve for A^(-1), which we can do, because det(A) is nonzero.
A^2 - tr(A) A + det(A) I = 0.
==> tr(A) A - A^2 = det(A) I
==> A (tr(A) I - A) = det(A) I
==> A (tr(A) I - A)/det(A) = I
==> A^(-1) = (tr(A) I - A)/det(A).

I hope this helps!