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The problem is to find the determinant of a given matrix: $$ A = \begin{pmatrix} 7 & 4 & 0 & ... & 0 \\ 3 & 7 & 4 & ... & 0 \\ 0 & 3 & 7 & ... & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & ... & 7 \end{pmatrix}. $$ I am able to express the determinant in recurrence relation $ a_{n}=7a_{n-1} - 12a_{n-2} $. Solving this, I get the formula: $ a_{n} = 4\cdot 4^{n} - 3\cdot3^{n} ,\, n\geq 1$. It's good, but I'm more intrested in row/column play. I tried to add everything to the first column, so I get a bunch of 14's, I can make them turn to 1, but it does nothing. Maybe I'm missing something.

Ben Grossmann
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saddysaw
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    @DietrichBurde Note that the Wikipedia solution (as well as the accepted answer on the post you found) coincide with the solution that OP already obtained, for which they want an alternative approach. – Ben Grossmann Dec 15 '21 at 15:03
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    @BenGrossmann Yes, you are right. But I think that also alternative approaches have been posted here as answers already. It is a very popular question in this area. This post gives the steps behind the recursion, which is also helpful and is in fact using "row/column play", as requested. – Dietrich Burde Dec 15 '21 at 15:04

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