Solve {lll}{2}&{-1}&{5}{3}&{1}&{-2}{1}&{4}&{6}

$$\left| \begin{array} { l l l } { 2 } & { - 1 } & { 5 } \\ { 3 } & { 1 } & { - 2 } \\ { 1 } & { 4 } & { 6 } \end{array} \right|$$

Evaluate

$$103$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-1&5\\3&1&-2\\1&4&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-1&5&2&-1\\3&1&-2&3&1\\1&4&6&1&4\end{matrix}\right)$$

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

$$2\times 6-\left(-2\right)+5\times 3\times 4=74$$

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

$$5+4\left(-2\right)\times 2+6\times 3\left(-1\right)=-29$$

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

$$74-\left(-29\right)$$

Subtract -29 from 74.

$$103$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-1&5\\3&1&-2\\1&4&6\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}1&-2\\4&6\end{matrix}\right))-\left(-det(\left(\begin{matrix}3&-2\\1&6\end{matrix}\right))\right)+5det(\left(\begin{matrix}3&1\\1&4\end{matrix}\right))$$

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

$$2\left(6-4\left(-2\right)\right)-\left(-\left(3\times 6-\left(-2\right)\right)\right)+5\left(3\times 4-1\right)$$

Simplify.

$$2\times 14-\left(-20\right)+5\times 11$$

Add the terms to obtain the final result.

$$103$$

Factor

$$103$$