Solve {ccc}{2}&{-5}&{7}{5}&{2}&{1}{9}&{0}&{2}

$$\left| \begin{array} { c c c } { 2 } & { - 5 } & { 7 } \\ { 5 } & { 2 } & { 1 } \\ { 9 } & { 0 } & { 2 } \end{array} \right|$$

Evaluate

$$-113$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-5&7\\5&2&1\\9&0&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-5&7&2&-5\\5&2&1&5&2\\9&0&2&9&0\end{matrix}\right)$$

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

$$2\times 2\times 2-5\times 9=-37$$

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

$$9\times 2\times 7+2\times 5\left(-5\right)=76$$

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

$$-37-76$$

Subtract 76 from -37.

$$-113$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-5&7\\5&2&1\\9&0&2\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}2&1\\0&2\end{matrix}\right))-\left(-5det(\left(\begin{matrix}5&1\\9&2\end{matrix}\right))\right)+7det(\left(\begin{matrix}5&2\\9&0\end{matrix}\right))$$

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

$$2\times 2\times 2-\left(-5\left(5\times 2-9\right)\right)+7\left(-9\times 2\right)$$

Simplify.

$$2\times 4-\left(-5\right)+7\left(-18\right)$$

Add the terms to obtain the final result.

$$-113$$

Factor

$$-113$$