Question

Find the determinant of the nxn matrix A with 5's on the diagonal, l's above the diagonal, and O's below the diagonal. det (A) =

Answer 1

Answer:

n times 5

Step-by-step explanation:

A matrix Anxn of this way is called an upper triangular matrix. It can be proved that the determinant of this kind of matrix is  

$a_{11}+a_{22}+...+a_{nn}$

In this case, it would be 5+5+...+5 (n times) = n times 5

We are going to develop each determinant by the first column taking as pivot points the elements of the diagonal

$det\left[\begin{array}{cccc}5&a_{12}&a_{13}...&a_{1n}\\0&5&a_{23}...&a_{2n}\\...&...&...&...\\0&0&0&5\end{array}\right] =5+det\left[\begin{array}{ccc}5&a_{23}...&a_{2n}\\0&5&a_{3n}\\...&...&...\\0&0&5\end{array}\right]=5+5+...+det\left[\begin{array}{cc}5&a_{n-1,n}\\0&5\end{array}\right]=5+5+...+5+5\;(n\;times)$