Solve {lll}{1}&{2}&{3}{2}&{3}&{4}{4}&{3}&{5}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 3 } \\ { 2 } & { 3 } & { 4 } \\ { 4 } & { 3 } & { 5 } \end{array} \right|$$

Evaluate

$$-3$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\2&3&4\\4&3&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\2&3&4&2&3\\4&3&5&4&3\end{matrix}\right)$$

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

$$3\times 5+2\times 4\times 4+3\times 2\times 3=65$$

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

$$4\times 3\times 3+3\times 4+5\times 2\times 2=68$$

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

$$65-68$$

Subtract 68 from 65.

$$-3$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\2&3&4\\4&3&5\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}3&4\\3&5\end{matrix}\right))-2det(\left(\begin{matrix}2&4\\4&5\end{matrix}\right))+3det(\left(\begin{matrix}2&3\\4&3\end{matrix}\right))$$

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

$$3\times 5-3\times 4-2\left(2\times 5-4\times 4\right)+3\left(2\times 3-4\times 3\right)$$

Simplify.

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

Add the terms to obtain the final result.

$$-3$$

Factor

$$-3$$