Solve {ccc}{2}&{4}&{5}{7}&{3}&{6}{1}&{8}&{9}

$$\left| \begin{array} { c c c } { 2 } & { 4 } & { 5 } \\ { 7 } & { 3 } & { 6 } \\ { 1 } & { 8 } & { 9 } \end{array} \right|$$

Evaluate

$$-5$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&4&5\\7&3&6\\1&8&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&4&5&2&4\\7&3&6&7&3\\1&8&9&1&8\end{matrix}\right)$$

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

$$2\times 3\times 9+4\times 6+5\times 7\times 8=358$$

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

$$3\times 5+8\times 6\times 2+9\times 7\times 4=363$$

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

$$358-363$$

Subtract 363 from 358.

$$-5$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&4&5\\7&3&6\\1&8&9\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}3&6\\8&9\end{matrix}\right))-4det(\left(\begin{matrix}7&6\\1&9\end{matrix}\right))+5det(\left(\begin{matrix}7&3\\1&8\end{matrix}\right))$$

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

$$2\left(3\times 9-8\times 6\right)-4\left(7\times 9-6\right)+5\left(7\times 8-3\right)$$

Simplify.

$$2\left(-21\right)-4\times 57+5\times 53$$

Add the terms to obtain the final result.

$$-5$$

Factor

$$-5$$