Solve {ccc}{1}&{2}&{3}{2}&{5}&{7}{-2}&{-4}&{-5}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { 3 } \\ { 2 } & { 5 } & { 7 } \\ { - 2 } & { - 4 } & { - 5 } \end{array} \right|$$

Evaluate

$$1$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\2&5&7\\-2&-4&-5\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\2&5&7&2&5\\-2&-4&-5&-2&-4\end{matrix}\right)$$

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

$$5\left(-5\right)+2\times 7\left(-2\right)+3\times 2\left(-4\right)=-77$$

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

$$-2\times 5\times 3-4\times 7-5\times 2\times 2=-78$$

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

$$-77-\left(-78\right)$$

Subtract -78 from -77.

$$1$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\2&5&7\\-2&-4&-5\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}5&7\\-4&-5\end{matrix}\right))-2det(\left(\begin{matrix}2&7\\-2&-5\end{matrix}\right))+3det(\left(\begin{matrix}2&5\\-2&-4\end{matrix}\right))$$

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

$$5\left(-5\right)-\left(-4\times 7\right)-2\left(2\left(-5\right)-\left(-2\times 7\right)\right)+3\left(2\left(-4\right)-\left(-2\times 5\right)\right)$$

Simplify.

$$3-2\times 4+3\times 2$$

Add the terms to obtain the final result.

$$1$$

Factor

$$1$$