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

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

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$3+2\times 6=15$$

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

$$4\times 3+6=18$$

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

$$15-18$$

Subtract 18 from 15.

$$-3$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$3-6+2\times 6-4\times 3$$

Simplify.

$$-3$$

Factor

$$-3$$