Solve {ccc}{14}&{14}&{2}{22}&{12}&{4}{15}&{5}&{5}

$$\left| \begin{array} { c c c } { 14 } & { 14 } & { 2 } \\ { 22 } & { 12 } & { 4 } \\ { 15 } & { 5 } & { 5 } \end{array} \right|$$

Evaluate

$$-280$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}14&14&2\\22&12&4\\15&5&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}14&14&2&14&14\\22&12&4&22&12\\15&5&5&15&5\end{matrix}\right)$$

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

$$14\times 12\times 5+14\times 4\times 15+2\times 22\times 5=1900$$

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

$$15\times 12\times 2+5\times 4\times 14+5\times 22\times 14=2180$$

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

$$1900-2180$$

Subtract 2180 from 1900.

$$-280$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}14&14&2\\22&12&4\\15&5&5\end{matrix}\right))$$

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.

$$14det(\left(\begin{matrix}12&4\\5&5\end{matrix}\right))-14det(\left(\begin{matrix}22&4\\15&5\end{matrix}\right))+2det(\left(\begin{matrix}22&12\\15&5\end{matrix}\right))$$

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

$$14\left(12\times 5-5\times 4\right)-14\left(22\times 5-15\times 4\right)+2\left(22\times 5-15\times 12\right)$$

Simplify.

$$14\times 40-14\times 50+2\left(-70\right)$$

Add the terms to obtain the final result.

$$-280$$

Factor

$$-280$$