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

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

Evaluate

$$7$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\left(-1\right)\times 3+2+4\times 3\times 2=20$$

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

$$-4+2\times 2\times 2+3\times 3=13$$

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

$$20-13$$

Subtract 13 from 20.

$$7$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-7\right)-7+4\times 7$$

Add the terms to obtain the final result.

$$7$$

Factor

$$7$$