X2 matrix A with real entries. We want to prove that A^-1 is also a 2x2 matrix with real entries.

To prove this, let's start by assuming that A is a non-singular 2x2 matrix with real entries. This means that A has an inverse, denoted as A^-1, which satisfies the equation AA^-1 = A^-1A = I, where I is the identity matrix.

We can write A as:

A = [a b]
[c d]

where a, b, c, and d are real numbers.

Since A is non-singular, its determinant is non-zero. The determinant of A is given by:

det(A) = ad - bc

Since A is non-singular, we have det(A) ≠ 0. This implies that ad - bc ≠ 0, which further implies that either ad ≠ bc or ad = bc ≠ 0.

Now, let's find the inverse of A. The inverse of A can be found using the formula:

A^-1 = (1/det(A)) * adj(A)

where adj(A) is the adjugate of A.

For a 2x2 matrix, the adjugate is given by:

adj(A) = [d -b]
[-c a]

Using this formula, we can calculate the adjugate:

adj(A) = [d -b]
[-c a]

Now, let's calculate the inverse of A:

A^-1 = (1/det(A)) * adj(A)
= (1/(ad - bc)) * [d -b]
[-c a]

To prove that A^-1 is also a 2x2 matrix with real entries, we need to show that all the entries of A^-1 are real numbers.

Since a, b, c, and d are real numbers, and ad - bc ≠ 0, we have:

A^-1 = (1/(ad - bc)) * [d -b]
[-c a]

The entries of A^-1 are given by:

A^-1 = [(1/(ad - bc)) * d, (1/(ad - bc)) * -b]
[(1/(ad - bc)) * -c, (1/(ad - bc)) * a]

Since ad - bc ≠ 0, the denominator (ad - bc) is a non-zero real number. Therefore, the entries of A^-1 are all real numbers.

Hence, we have proved that if A is a non-singular 2x2 matrix with real entries, then its inverse A^-1 is also a 2x2 matrix with real entries.