To prove that $A^{-1}$ (the inverse of matrix $A$) exists if and only if $A^{-T}$ (the inverse of the transpose of $A$, also known as the transpose of $A^{-1}$) exists, we can break the proof into two parts: the "if" part and the "only if" part.

Part 1: If $A^{-1}$ exists, then $A^{-T}$ exists.

1. Given: $A^{-1}$ exists.

   • This means there is a matrix $A^{-1}$ such that: $A A^{-1} = A^{-1} A = I$ where $I$ is the identity matrix.

2. Transpose both sides:
   Since transposing a matrix product reverses the order of multiplication and transposes each matrix: $(A A^{-1})^T = I^T$ $(A^{-1})^T A^T = I$ But $I^T = I$, since the identity matrix is symmetric: $(A^{-1})^T A^T = I$ This shows that $(A^{-1})^T$ is a left-inverse of $A^T$.

3. Similarly, transpose the other multiplication:
   $(A^{-1} A)^T = I^T$ $A^T (A^{-1})^T = I$ This shows that $(A^{-1})^T$ is also a right-inverse of $A^T$.

4. Conclusion:
   Since $(A^{-1})^T$ is both a left-inverse and a right-inverse of $A^T$, it must be the inverse of $A^T$. Thus, the inverse of $A^T$, which is denoted by $A^{-T}$, exists and is equal to $(A^{-1})^T$.

Therefore, if $A^{-1}$ exists, then $A^{-T}$ exists.

Part 2: If $A^{-T}$ exists, then $A^{-1}$ exists.

1. Given: $A^{-T}$ exists.
   • This means there is a matrix