91Ó°ÊÓ

First, given that AB is invertible, there exists a matrix C such that: \( (AB)C = I_n \) and \( C(AB) = I_n \) where \(I_n\) is the identity matrix. Now, let's focus on the equation \( (AB)C = I_n \). We have the equation: \( A(BC) = I_n \) Since the right-hand side is the identity matrix, we can say that BC is the right inverse of A. If we now prove that BC is also the left inverse of A, we can conclude that A is invertible. To do that, we will use the hint. Consider an arbitrary square matrix M such that MB = I. Then, we have: \( M(AB) = (MB)A = (I_n)A = A \) So, now, we have MA = A and our goal is to show that M = BC. Now focus on the equation \( C(AB) = I_n \): \( (CB)A = I_n \) Since the right-hand side is the identity matrix, we can say that CB is the right inverse of A. We just need to show that it is also the left inverse of A. Let's multiply both sides of the equation \( (CB)A = I_n \) by M: \( M(CB)A = MI_n \) Then we have: \( (MC)(BA) = A \) Now, compare this to the equation \( MA = A \). Since their right-hand sides are equal, we can say their left-hand sides are equal as well: \( (MC)(BA) = MI_n \) Hence, we have: \( M = MC \) From this, we can conclude that BC = M, and thus, A is invertible. Following a similar line of reasoning, we can also prove that B is invertible.

Consider a 2×1 matrix A and a 1×2 matrix B given by: \( A = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, B = \begin{bmatrix} 1 & 0 \end{bmatrix} \) Their product is given by: \( AB = \begin{bmatrix} 1*1 & 1*0 \\ 0*1 & 0*0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \) Now, let's find the inverse of AB: \( (AB)^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}^{-1} \) Since AB is not a full rank matrix, its inverse does not exist. Thus, AB is not invertible. Now, let's look at an example where the product of nonsquare matrices is invertible: Consider a 2×1 matrix A and a 1×2 matrix B given by: \( A = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, B = \begin{bmatrix} 1 & 1 \end{bmatrix} \) Their product is given by: \( AB = \begin{bmatrix} 1*1 & 1*1 \\ 1*1 & 1*1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \) Now, let's find the inverse of AB: \( (AB)^{-1} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) Since the inverse of AB exists, AB is invertible. However, by definition, the square matrices A and B are not invertible, as they are not square matrices. In conclusion, we have shown that a product of nonsquare matrices can be invertible, even though the factors, by definition, are not.