Solve {lll}{2}&{3}&{5}{0}&{2}&{4}{7}&{1}&{3}

$$\left| \begin{array} { l l l } { 2 } & { 3 } & { 5 } \\ { 0 } & { 2 } & { 4 } \\ { 7 } & { 1 } & { 3 } \end{array} \right|$$

Evaluate

$$18$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&5\\0&2&4\\7&1&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&5&2&3\\0&2&4&0&2\\7&1&3&7&1\end{matrix}\right)$$

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

$$2\times 2\times 3+3\times 4\times 7=96$$

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

$$7\times 2\times 5+4\times 2=78$$

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

$$96-78$$

Subtract 78 from 96.

$$18$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&5\\0&2&4\\7&1&3\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}2&4\\1&3\end{matrix}\right))-3det(\left(\begin{matrix}0&4\\7&3\end{matrix}\right))+5det(\left(\begin{matrix}0&2\\7&1\end{matrix}\right))$$

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

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

Simplify.

$$2\times 2-3\left(-28\right)+5\left(-14\right)$$

Add the terms to obtain the final result.

$$18$$

Factor

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