Solve {lll}{1}&{5}&{4}{2}&{7}&{2}{3}&{6}&{3}

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

Evaluate

$$-27$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$7\times 3+5\times 2\times 3+4\times 2\times 6=99$$

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

$$3\times 7\times 4+6\times 2+3\times 2\times 5=126$$

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

$$99-126$$

Subtract 126 from 99.

$$-27$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$9+4\left(-9\right)$$

Add the terms to obtain the final result.

$$-27$$

Factor

$$-27$$