Solve {ccc}{10}&{13}&{-11}{0}&{-17}&{1}{-15}&{19}&{4}

$$\left| \begin{array} { c c c } { 10 } & { 13 } & { - 11 } \\ { 0 } & { - 17 } & { 1 } \\ { - 15 } & { 19 } & { 4 } \end{array} \right|$$

Evaluate

$$1740$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}10&13&-11\\0&-17&1\\-15&19&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}10&13&-11&10&13\\0&-17&1&0&-17\\-15&19&4&-15&19\end{matrix}\right)$$

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

$$10\left(-17\right)\times 4+13\left(-15\right)=-875$$

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

$$-15\left(-17\right)\left(-11\right)+19\times 10=-2615$$

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

$$-875-\left(-2615\right)$$

Subtract -2615 from -875.

$$1740$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}10&13&-11\\0&-17&1\\-15&19&4\end{matrix}\right))$$

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.

$$10det(\left(\begin{matrix}-17&1\\19&4\end{matrix}\right))-13det(\left(\begin{matrix}0&1\\-15&4\end{matrix}\right))-11det(\left(\begin{matrix}0&-17\\-15&19\end{matrix}\right))$$

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

$$10\left(-17\times 4-19\right)-13\left(-\left(-15\right)\right)-11\left(-\left(-15\left(-17\right)\right)\right)$$

Simplify.

$$10\left(-87\right)-13\times 15-11\left(-255\right)$$

Add the terms to obtain the final result.

$$1740$$

Factor

$$2^{2}\times 3\times 5\times 29$$