Solve {ccc}{3}&{4}&{6}{1}&{-1}&{-3}{1}&{2}&{3}

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

Evaluate

$$3$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-6+2\left(-3\right)\times 3+3\times 4=-12$$

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

$$-9-\left(-12\right)$$

Subtract -12 from -9.

$$3$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$3$$

Factor

$$3$$