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Question 1: Why is the matrix with a zero on the right diagonal non-invertible? See pic below. Strang 4th edition enter image description here

A second question is: why are the non-diagonal cells NOT required to be non-zeros also?

I'm looking for an answer without determinants. That is $A^{-1}$ undoes what A does.

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    A zero on the diagonal says nothing about invertibility unless the matrix is triangular. – 2017-01-08
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    Doesn't make much sense to start with "A second question is..." when you don't even link to the first question. – 2017-01-08
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    The title is the first question – 2017-01-08
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    How does this have two upvotes? OP: Please edit your question so it is comprehensible. – 2017-01-08
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    @WizardProgramming You forgot to mention it's a *diagonal* matrix. `why are the non-diagonal cells NOT required to be non-zeros also` Because they *are* all $0$ by the definition of a diagonal matrix. – 2017-01-08
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    @dxiv, what is the answer to question 1 then ? – 2017-01-08
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    @WizardProgramming Suppose WLOG that $d_1=0$ then the top row of $A$ is all $0$s. Now multiply $A$ by any matrix $B$ and show that the top-left element in the product will be $0$. Compare that to the identity matrix, and conclude that $A \cdot B \ne I$ for $\forall B\,$. Done. – 2017-01-08

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