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

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

Calculate Determinant

$$21$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\times 7+4\times 8\times 3+2\times 2\times 6=155$$

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

$$3\times 5\times 2+6\times 8+7\times 2\times 4=134$$

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

$$155-134$$

Subtract 134 from 155.

$$21$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-13-4\left(-10\right)+2\left(-3\right)$$

Add the terms to obtain the final result.

$$21$$

Evaluate

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