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.

  • 11
    $\begingroup$ Of course. A (square) matrix and its transpose have the same determinant. $\endgroup$
    – hardmath
    Apr 20, 2015 at 19:07
  • $\begingroup$ @hardmath elegant way to see that.Thanx $\endgroup$
    – manifold
    Mar 18, 2018 at 8:23

1 Answer 1


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.

  • $\begingroup$ There is a typo: I think that should be $a_{1,1}*a_{2,2}*\cdots*a_{n,n}$. $\endgroup$
    – Chrigi
    Feb 15, 2019 at 15:24
  • 1
    $\begingroup$ @chrigi Thanks. I have fixed that. $\endgroup$
    – Math Man
    Feb 16, 2019 at 22:14

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