Solve {ccc}{1}&{3}&{1}{2}&{0}&{4}{-1}&{-3}&{-3}

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

Calculate Determinant

$$12$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 4\left(-1\right)+2\left(-3\right)=-18$$

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

$$-3\times 4-3\times 2\times 3=-30$$

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

$$-18-\left(-30\right)$$

Subtract -30 from -18.

$$12$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$12-3\left(-2\right)-6$$

Add the terms to obtain the final result.

$$12$$

Evaluate

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