Solve {lll}{2}&{3}&{1}{1}&{5}&{0}{7}&{1}&{2}

$$\left( \begin{array} { l l l } { 2 } & { 3 } & { 1 } \\ { 1 } & { 5 } & { 0 } \\ { 7 } & { 1 } & { 2 } \end{array} \right)$$

Calculate Determinant

$$-20$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&1\\1&5&0\\7&1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&1&2&3\\1&5&0&1&5\\7&1&2&7&1\end{matrix}\right)$$

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

$$2\times 5\times 2+1=21$$

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

$$7\times 5+2\times 3=41$$

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

$$21-41$$

Subtract 41 from 21.

$$-20$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&1\\1&5&0\\7&1&2\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}5&0\\1&2\end{matrix}\right))-3det(\left(\begin{matrix}1&0\\7&2\end{matrix}\right))+det(\left(\begin{matrix}1&5\\7&1\end{matrix}\right))$$

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

$$2\times 5\times 2-3\times 2+1-7\times 5$$

Simplify.

$$2\times 10-3\times 2-34$$

Add the terms to obtain the final result.

$$-20$$

Evaluate

$$\left(\begin{matrix}2&3&1\\1&5&0\\7&1&2\end{matrix}\right)$$