How to prove that If A is invertible then $A^{-1}$ and $A^2$ are invertible. True or false (with a counterexample if false and a reason if true):
(a) A $4 \times 4$ matrix with a row of zeros is not invertible.
(b) Every matrix with $1's$ down the main diagonal is invertible.
(c) If A is invertible then $A^{-1}$ and $A^2$ are invertible.  
Are my answers correct, and how do I prove my answer for (c)?
For (a) I think it's true because if you have a row of zeroes equal to a number not zero, then that does not make sense, making the matrix false, if the row of zeroes equals zero then it would simply become a $3\times4$ matrix.  
For (b) I think it's false because you can have a square matrix full of $1's$ and we know that it isn't invertible because if a matrix has the same column or row twice then it is not an inverse.  
For (c) I think the answer is true because if $A$ is invertible then isn't there a rule that states that $A^{-1}$ and $A^2$ are also invertile? How would I prive that?
 A: (a) Yes this is true, but you cannot think of a $4 \times 4$ matrix with a row of all zeroes as a $3 \times 4$ matrix. But the determinant of any matrix with an all zero row is zero, hence the matrix is not invertible.
(b) Correct
(c) follows straight from the definition. A matrix $A$ is said to be invertible if there exists a matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. Note by this definition, $A^{-1}$ is also invertible with inverse $A$. To show that $A^2$ is invertible, just note that
$$A^2(A^{-1})^2 = AAA^{-1}A^{-1} = AIA^{-1} = AA^{-1} = I.$$
You can check that $(A^{-1})^2A^2 = I$ as well.
A: The answer for (b) is good. (If you want to be very precise you'll have to state that you exclude the case of $1 \times 1$ matrices, but often this goes without saying.)
The answer for (a) might be good idea, but not expressed too well. 
A way to argue this is that $A$ invertible, means that $Ax = b$ has  a solution whatever the $b$. 
But this is false in this case, as the result of $Ax_0$ will always yield a vector with $0$ in the row that is the $0$-row in the matrix. 
For (c). Just show that there is a matrix $C$ such that $A^{-1}C=CA^{-1}=I$ and likewise there is a matrix $D$ such that $A^2D = D A^2 = I$. (In the first case take $C = A$ and in the second $D = (A^{-1})^2$.)
A: For (a) : your answer is correct but the motivation is wrong. A $4 \times 4$matrix with a row of zeros is a truly $4 \times 4$ matrix but its determinat is null, and we know that 

a matrix is invertible if and only if its determinat is not null. 

You can use this fact also:
For (b) : Your answer is correct because the determinant is an alternating multilinear function of the rows ( or columns) of the matrix, so it is null if two rows (or columns) are equal.
for (c) : Use the fact that, for two matrices $A,B$ we have: $\det(AB)=\det(A) \det(B)$. This means that we have:
$$
\det(A^{-1})\det (A)=\det(A^{-1}A)=\det (I)=1 \quad \Rightarrow \quad \det(A^{-1})=\det(A)^{-1}
$$ 
and
$$
\det (A^2)=\left(\det(A)\right)^2
$$
