Does a lower triangular matrix have a determinant that is equal to the product of the elements in the diagonal similar to an upper triangular matrix.
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13Of course. A (square) matrix and its transpose have the same determinant. – hardmath Apr 20 '15 at 19:07
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@hardmath elegant way to see that.Thanx – manifold Mar 18 '18 at 08:23
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Related – Rodrigo de Azevedo May 24 '23 at 06:04
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The matrix looks like this: $$\begin{bmatrix} a_{1,1}&0&\cdots&\cdots&0 \\b_{1,2}&a_{2,2}&\ddots&&\vdots \\ \vdots&\ddots&\ddots&\ddots&\vdots \\ \vdots&&\ddots&\ddots&0 \\ b_{1,n}&\cdots&\cdots&b_{n-1,n}&a_{n,n} \end{bmatrix}$$ The determinant can be written as the sum of the product of the elements in the top row with their associated minors: so this determinant would be: $$a_{1,1}*\begin{bmatrix} a_{2,2}&0&\cdots&0 \\b_{2,3}&a_{3,3}&\ddots&\vdots \\ \vdots&\ddots&\ddots&0 \\ b_{2,n}&\cdots&b_{n-1,n}&a_{n,n} \end{bmatrix}$$ This action eliminates the first column, $b_{1,x}$ and we are left with a similar matrix to find the determinant of. Reducing this one likewise we are left with $a_{1,1}*a_{2,2}*$(a matrix of remaining rows and columns) and so on until we have the product $a_{1,1}*a_{2,2}*\cdots*a_{n,n}$. The product of the diagonal.
Math Man
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There is a typo: I think that should be $a_{1,1}a_{2,2}\cdots*a_{n,n}$. – Chrigi Feb 15 '19 at 15:24
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