Solve {ccc}{2}&{5}&{1}{3}&{1}&{2}{-2}&{1}&{0}

$$\left( \begin{array} { c c c } { 2 } & { 5 } & { 1 } \\ { 3 } & { 1 } & { 2 } \\ { - 2 } & { 1 } & { 0 } \end{array} \right)$$

Calculate Determinant

$$-19$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&5&1\\3&1&2\\-2&1&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&5&1&2&5\\3&1&2&3&1\\-2&1&0&-2&1\end{matrix}\right)$$

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

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

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

$$-2+2\times 2=2$$

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

$$-17-2$$

Subtract 2 from -17.

$$-19$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&5&1\\3&1&2\\-2&1&0\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}1&2\\1&0\end{matrix}\right))-5det(\left(\begin{matrix}3&2\\-2&0\end{matrix}\right))+det(\left(\begin{matrix}3&1\\-2&1\end{matrix}\right))$$

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-19$$

Evaluate

$$\left(\begin{matrix}2&5&1\\3&1&2\\-2&1&0\end{matrix}\right)$$