Solve operatorname{det}{lll}{2}&{3}&{4}{6}&{0}&{1}{3}&{1}&{2}

$$\operatorname { det } \left( \begin{array} { l l l } { 2 } & { 3 } & { 4 } \\ { 6 } & { 0 } & { 1 } \\ { 3 } & { 1 } & { 2 } \end{array} \right)$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&4\\6&0&1\\3&1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&4&2&3\\6&0&1&6&0\\3&1&2&3&1\end{matrix}\right)$$

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

$$3\times 3+4\times 6=33$$

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

$$2+2\times 6\times 3=38$$

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

$$33-38$$

Subtract 38 from 33.

$$-5$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&4\\6&0&1\\3&1&2\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}0&1\\1&2\end{matrix}\right))-3det(\left(\begin{matrix}6&1\\3&2\end{matrix}\right))+4det(\left(\begin{matrix}6&0\\3&1\end{matrix}\right))$$

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

$$2\left(-1\right)-3\left(6\times 2-3\right)+4\times 6$$

Simplify.

$$2\left(-1\right)-3\times 9+4\times 6$$

Add the terms to obtain the final result.

$$-5$$

Factor

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