Solve {lll}{1}&{3}&{5}{4}&{2}&{0}{1}&{10}&{12}

$$\left| \begin{array} { l l l } { 1 } & { 3 } & { 5 } \\ { 4 } & { 2 } & { 0 } \\ { 1 } & { 10 } & { 12 } \end{array} \right|$$

Evaluate

$$70$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&5\\4&2&0\\1&10&12\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&5&1&3\\4&2&0&4&2\\1&10&12&1&10\end{matrix}\right)$$

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

$$2\times 12+5\times 4\times 10=224$$

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

$$2\times 5+12\times 4\times 3=154$$

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

$$224-154$$

Subtract 154 from 224.

$$70$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&5\\4&2&0\\1&10&12\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}2&0\\10&12\end{matrix}\right))-3det(\left(\begin{matrix}4&0\\1&12\end{matrix}\right))+5det(\left(\begin{matrix}4&2\\1&10\end{matrix}\right))$$

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

$$2\times 12-3\times 4\times 12+5\left(4\times 10-2\right)$$

Simplify.

$$24-3\times 48+5\times 38$$

Add the terms to obtain the final result.

$$70$$

Factor

$$2\times 5\times 7$$