Solve {ccc}{2}&{4}&{-1}{3}&{11}&{-2}{3}&{11}&{4}

$$\left| \begin{array} { c c c } { 2 } & { 4 } & { - 1 } \\ { 3 } & { 11 } & { - 2 } \\ { 3 } & { 11 } & { 4 } \end{array} \right|$$

Evaluate

$$60$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&4&-1\\3&11&-2\\3&11&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&4&-1&2&4\\3&11&-2&3&11\\3&11&4&3&11\end{matrix}\right)$$

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

$$2\times 11\times 4+4\left(-2\right)\times 3-3\times 11=31$$

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

$$3\times 11\left(-1\right)+11\left(-2\right)\times 2+4\times 3\times 4=-29$$

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

$$31-\left(-29\right)$$

Subtract -29 from 31.

$$60$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&4&-1\\3&11&-2\\3&11&4\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}11&-2\\11&4\end{matrix}\right))-4det(\left(\begin{matrix}3&-2\\3&4\end{matrix}\right))-det(\left(\begin{matrix}3&11\\3&11\end{matrix}\right))$$

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

$$2\left(11\times 4-11\left(-2\right)\right)-4\left(3\times 4-3\left(-2\right)\right)-\left(3\times 11-3\times 11\right)$$

Simplify.

$$2\times 66-4\times 18$$

Add the terms to obtain the final result.

$$60$$

Factor

$$2^{2}\times 3\times 5$$