Solve {lll}{1}&{1}&{1}{2}&{5}&{6}{4}&{25}&{36}

$$\left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 2 } & { 5 } & { 6 } \\ { 4 } & { 25 } & { 36 } \end{array} \right|$$

Evaluate

$$12$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}1&1&1\\2&5&6\\4&25&36\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\2&5&6&2&5\\4&25&36&4&25\end{matrix}\right)$$

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

$$5\times 36+6\times 4+2\times 25=254$$

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

$$4\times 5+25\times 6+36\times 2=242$$

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

$$254-242$$

Subtract 242 from 254.

$$12$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}1&1&1\\2&5&6\\4&25&36\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}5&6\\25&36\end{matrix}\right))-det(\left(\begin{matrix}2&6\\4&36\end{matrix}\right))+det(\left(\begin{matrix}2&5\\4&25\end{matrix}\right))$$

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

$$5\times 36-25\times 6-\left(2\times 36-4\times 6\right)+2\times 25-4\times 5$$

Simplify.

$$30-48+30$$

Add the terms to obtain the final result.

$$12$$

Factor

$$2^{2}\times 3$$