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.
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$\begingroup$ Are you familiar with row reduction? $\endgroup$Michelle– Michelle2020-10-05 10:42:28 +00:00Commented Oct 5, 2020 at 10:42
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2 Answers
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Hint : $$det(A)=det(A_1)\cdot det(A_2) \cdot \cdot \cdot \cdot det(A_k)$$
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1$\begingroup$ Proving the hint is harder than using it here. $\endgroup$nicomezi– nicomezi2020-10-05 08:33:04 +00:00Commented Oct 5, 2020 at 8:33
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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.
answered Oct 5, 2020 at 11:42