Solve {ccc}{5}&{1}&{14}{0}&{10}&{2}{0}&{0}&{-1}

$$\left| \begin{array} { c c c } { 5 } & { 1 } & { 14 } \\ { 0 } & { 10 } & { 2 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right|$$

Evaluate

$$-50$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&1&14\\0&10&2\\0&0&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&1&14&5&1\\0&10&2&0&10\\0&0&-1&0&0\end{matrix}\right)$$

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

$$5\times 10\left(-1\right)=-50$$

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

$$\text{true}$$

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

$$-50$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&1&14\\0&10&2\\0&0&-1\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}10&2\\0&-1\end{matrix}\right))-det(\left(\begin{matrix}0&2\\0&-1\end{matrix}\right))+14det(\left(\begin{matrix}0&10\\0&0\end{matrix}\right))$$

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

$$5\times 10\left(-1\right)$$

Simplify.

$$5\left(-10\right)$$

Add the terms to obtain the final result.

$$-50$$

Factor

$$-50$$