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

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

Evaluate

$$47$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3+4\times 4\times 4=67$$

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

$$4\times 2+4\times 3=20$$

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

$$67-20$$

Subtract 20 from 67.

$$47$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-5-3\times 4+4\times 16$$

Add the terms to obtain the final result.

$$47$$

Factor

$$47$$