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

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

Calculate Determinant

$$-59$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 8+3\times 5+4\times 6\times 4=127$$

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

$$5\times 8\times 4+4\times 2+6\times 3=186$$

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

$$127-186$$

Subtract 186 from 127.

$$-59$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\times 4-3+4\left(-16\right)$$

Add the terms to obtain the final result.

$$-59$$

Evaluate

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