91Ó°ÊÓ

Let's assume two generic invertible 2x2 matrices A and B. It means their determinant should not be equal to zero: $$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad B = \begin{pmatrix} p & q \\ r & s \end{pmatrix}, \quad \det(A) \neq 0, \quad \det(B) \neq 0 $$

We need to find the sum of matrices A and B, which we denote as C: $$ C = A + B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} p & q \\ r & s \end{pmatrix} = \begin{pmatrix} a + p & b + q \\ c + r & d + s \end{pmatrix} $$

For the matrix C to be non-invertible, its determinant must be equal to zero. Hence, we can form the determinant equation and try to simplify it, while ensuring to meet the initial conditions: $$ \det(C) = (a + p)(d + s) - (b + q)(c + r) = 0 $$

Now we need to choose the matrices A and B such that their determinants are non-zero, but the determinant of their sum (A+B) is equal to zero. One such example can be: $$ A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \\ B = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} $$ Following the initial conditions: $$ \det(A) = (1)(2) - (0)(0) = 2 \neq 0 \\ \det(B) = (-1)(-1) - (1)(1) = 1 \neq 0 \\ C = A + B = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} + \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \\ \det(C) = (0)(1) - (1)(1) = -1 $$ With these example matrices A and B, we have satisfied all the conditions given in the exercise: A and B are invertible, their sum (A+B) is not equal to the zero matrix, and (A+B) is not invertible.