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Expert-verified Found in: Page 290 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Use Exercise 31 to find${\mathbit{d}}{\mathbit{e}}{\mathbit{t}}\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 2& 3& 4& 5\\ 1& 4& 9& 16& 25\\ 1& 8& 27& 64& 125\\ 1& 16& 81& 256& 625\end{array}\right]$Do not use technology.

Therefore, the determinant of the given matrix is given by,

$detA=288$.

See the step by step solution

## Step 1: By using Vandermonde’s Formula.

For $n=1$, we have a $2×2$ matrix

$A=\left[\begin{array}{cc}1& 1\\ {a}_{0}& {a}_{1}\end{array}\right]$

Using Vandermonde's formula, we have

$\underset{}{\prod _{i>j}\left({a}_{i}-{a}_{j}\right)={a}_{1}-{a}_{0}=detA}$.

## Step 2: To find the determinant.

Let ${a}_{0}=1,{a}_{1}=2,{a}_{2}=3,{a}_{3}=4,{a}_{4}=5$.

Now we can use Vandermonde's formula to find this determinant.

We compute,

$\begin{array}{l}detA=\prod _{\frac{i,j-1}{i>j}}\left({a}_{i}-{a}_{j}\right)\\ detA=\left(2-1\right)\left(3-1\right)\left(4-1\right)\left(5-1\right)\left(3-2\right)\left(4-2\right)\left(5-2\right)\left(4-3\right)\left(5-3\right)\left(5-4\right)\\ detA=288\end{array}$ ### Want to see more solutions like these? 