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

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

Calculate Determinant

$$19$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$1+3\times 2\times 4=25$$

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

$$2\times 3=6$$

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

$$25-6$$

Subtract 6 from 25.

$$19$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$1-3\left(-6\right)$$

Add the terms to obtain the final result.

$$19$$

Evaluate

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