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

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

Evaluate

$$49$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 6\times 3-3\times 2-4\times 5=10$$

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

$$2\times 6\left(-4\right)-3\times 2+3\times 5=-39$$

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

$$10-\left(-39\right)$$

Subtract -39 from 10.

$$49$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\times 21-21-4\left(-7\right)$$

Add the terms to obtain the final result.

$$49$$

Factor

$$7^{2}$$