Solve {ccc}{1^{2}}&{3^{2}}&{5^{2}}{2^{2}}&{5^2}&{15^2}{3^2}&{8}&{17^2}

$$\left| \begin{array} { c c c } { 1 ^ { 2 } } & { 3 ^ { 2 } } & { 5 ^ { 2 } } \\ { 2 ^ { 2 } } & { 5 ^ { 2 } } & { 15 ^ { 2 } } \\ { 3 ^ { 2 } } & { 8 } & { 17 ^ { 2 } } \end{array} \right|$$

Evaluate

$$8421$$

Steps Using Sarrus' Rule

Calculate 1 to the power of 2 and get 1.

$$det(\left(\begin{matrix}1&3^{2}&5^{2}\\2^{2}&5^{2}&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 3 to the power of 2 and get 9.

$$det(\left(\begin{matrix}1&9&5^{2}\\2^{2}&5^{2}&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 5 to the power of 2 and get 25.

$$det(\left(\begin{matrix}1&9&25\\2^{2}&5^{2}&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 2 to the power of 2 and get 4.

$$det(\left(\begin{matrix}1&9&25\\4&5^{2}&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 5 to the power of 2 and get 25.

$$det(\left(\begin{matrix}1&9&25\\4&25&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 15 to the power of 2 and get 225.

$$det(\left(\begin{matrix}1&9&25\\4&25&225\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 3 to the power of 2 and get 9.

$$det(\left(\begin{matrix}1&9&25\\4&25&225\\9&8&17^{2}\end{matrix}\right))$$

Calculate 17 to the power of 2 and get 289.

$$det(\left(\begin{matrix}1&9&25\\4&25&225\\9&8&289\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}1&9&25&1&9\\4&25&225&4&25\\9&8&289&9&8\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$25\times 289+9\times 225\times 9+25\times 4\times 8=26250$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$9\times 25\times 25+8\times 225+289\times 4\times 9=17829$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$26250-17829$$

Subtract 17829 from 26250.

$$8421$$

Steps Using Expansion by Minors

Calculate 1 to the power of 2 and get 1.

$$det(\left(\begin{matrix}1&3^{2}&5^{2}\\2^{2}&5^{2}&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 3 to the power of 2 and get 9.

$$det(\left(\begin{matrix}1&9&5^{2}\\2^{2}&5^{2}&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 5 to the power of 2 and get 25.

$$det(\left(\begin{matrix}1&9&25\\2^{2}&5^{2}&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 2 to the power of 2 and get 4.

$$det(\left(\begin{matrix}1&9&25\\4&5^{2}&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 5 to the power of 2 and get 25.

$$det(\left(\begin{matrix}1&9&25\\4&25&15^{2}\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 15 to the power of 2 and get 225.

$$det(\left(\begin{matrix}1&9&25\\4&25&225\\3^{2}&8&17^{2}\end{matrix}\right))$$

Calculate 3 to the power of 2 and get 9.

$$det(\left(\begin{matrix}1&9&25\\4&25&225\\9&8&17^{2}\end{matrix}\right))$$

Calculate 17 to the power of 2 and get 289.

$$det(\left(\begin{matrix}1&9&25\\4&25&225\\9&8&289\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$det(\left(\begin{matrix}25&225\\8&289\end{matrix}\right))-9det(\left(\begin{matrix}4&225\\9&289\end{matrix}\right))+25det(\left(\begin{matrix}4&25\\9&8\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$25\times 289-8\times 225-9\left(4\times 289-9\times 225\right)+25\left(4\times 8-9\times 25\right)$$

Simplify.

$$5425-9\left(-869\right)+25\left(-193\right)$$

Add the terms to obtain the final result.

$$8421$$

Factor

$$3\times 7\times 401$$