Prove that block diagonal matrix is invertible

Question

Suppose that $A_{1}, A_{2}, A_{3}, \dotsc, A_{k} $ are square matrixes. Define: $$ A = \begin{bmatrix} A_1 & 0 & \cdots & 0 \\ 0 & A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & A_k \end{bmatrix} $$ prove that $ A $ is invertible if and only if $A_{1}, A_{2}, A_{3}, \dotsc, A_{k} $ are invertible.

Answer 1

Hint : $$det(A)=det(A_1)\cdot det(A_2) \cdot \cdot \cdot \cdot det(A_k)$$

Answer 2

Hint: a really easy way to show that a matrix is invertible is to write down the inverse explicitly. For instance, the matrix $$ \pmatrix{20 & 0 \\ 0 & \pi} $$ is invertible, and the proof is that $$ \pmatrix{\frac{1}{20} & 0 \\ 0 & \frac{1}{\pi}} $$ is its inverse, which I can show by multiplying them. Can you think of something you'd multiply the block matrix $$ \pmatrix{ A_1 & 0 \\ 0 & A_2 } $$ by to get the identity? Perhaps you might want to work this out by inversion-by-hand in the case where $$ A_1 = \pmatrix{6} $$ and $$ A_2 = \pmatrix{1 & 3 \\ 1 & 4} $$ for instance, and see whether you notice anything.