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

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

Answer 2

The determinant of the matrix $A$ is $5^{n}$, where $n$ is the size of the square matrix.

According to the statement, we have a matrix $A$ with the following characteristics:

$A = \left[\begin{array}{ccccc}5&1&\dots&1&1\\0&5&\dots&1&1\\ \vdots&\vdots &\ddots &\vdots &\vdots\\0&0&\dots&5&1\\0&0&\dots &0&5\end{array}\right]$, $A \in \mathbb{R}^{n\times n}$

We notice that is an upper diagonal matrix and according to linear algebra, its determinant is the product of all elements from main diagonal. Hence, the determinant of the matrix is:

$\det (A) = 5^{n}$

The determinant of the matrix $A$ is $5^{n}$, where $n$ is the size of the square matrix.

We kindly invite to check this question on determinants: https://brainly.com/question/4470545