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

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

Calculate Determinant

$$-12$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 6=12$$

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

$$6\times 4=24$$

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

$$12-24$$

Subtract 24 from 12.

$$-12$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-6\times 4+2\times 6$$

Simplify.

$$-24+2\times 6$$

Add the terms to obtain the final result.

$$-12$$

Evaluate

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