Solve {ccc}{2}&{-1}&{5}{3}&{1}&{5}{4}&{-1}&{9}=

$$\left| \begin{array} { c c c } { 2 } & { - 1 } & { 5 } \\ { 3 } & { 1 } & { 5 } \\ { 4 } & { - 1 } & { 9 } \end{array} \right| =$$

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-1&5\\3&1&5\\4&-1&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-1&5&2&-1\\3&1&5&3&1\\4&-1&9&4&-1\end{matrix}\right)$$

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

$$2\times 9-5\times 4+5\times 3\left(-1\right)=-17$$

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

$$4\times 5-5\times 2+9\times 3\left(-1\right)=-17$$

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

$$-17-\left(-17\right)$$

Subtract -17 from -17.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-1&5\\3&1&5\\4&-1&9\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}1&5\\-1&9\end{matrix}\right))-\left(-det(\left(\begin{matrix}3&5\\4&9\end{matrix}\right))\right)+5det(\left(\begin{matrix}3&1\\4&-1\end{matrix}\right))$$

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

$$2\left(9-\left(-5\right)\right)-\left(-\left(3\times 9-4\times 5\right)\right)+5\left(3\left(-1\right)-4\right)$$

Simplify.

$$2\times 14-\left(-7\right)+5\left(-7\right)$$

Add the terms to obtain the final result.

$$0$$

Factor

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