Question

assignment 8: problem 3
(1 point)
find the determinant of the matrix
a=left\begin{array}{ccccc}8 & 9 & & cdots & 9 \\9 & 8 & 9 & cdots & 9 \\& & ddots & & \\9 & cdots & 9 & 8 & 9 \\9 & cdots & & 9 & 8end{array}
ight in mathbb{r}^{53 \times 53}
( operatorname{det}(a)= )

Explanation:

Step1: Use the formula for the determinant of a matrix of the form \(A=(a_{ij})\) where \(a_{ii}=8\) and \(a_{ij} = 9\) for \(i

eq j\)
Let \(A\) be an \(n\times n\) matrix (\(n = 53\) in this case). We can use the formula \(\det(A)=(8+(n - 1)\times9)(8 - 9)^{n - 1}\)

Step2: Substitute \(n = 53\) into the formula

First, calculate \(8+(n - 1)\times9\):
\(8+(53 - 1)\times9=8+52\times9=8 + 468=476\)
Second, calculate \((8 - 9)^{n - 1}\):
\((8 - 9)^{53-1}=(- 1)^{52}=1\)

Answer:

\(476\)