Question: One about flipping coins. Coin A has 0.6 probability of being heads, Coin B has a 0.3 probability of being heads and Coin C has a 0.1 probability of heads. Given that a coin is flipped and lands on heads, what is the probability that when it is flipped again it will also be heads.
I'm not sure how I can apply Bayes' rule here.

I did a Monte Carole simulation the answer come out to be around 0.46, I am not sure how to give a analytical answer.

Answer: 
(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.
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
$$
\det (A^2)=\left(\det(A)\right)^2
$$