Solve {lll}{1}&{2}&{3}{2}&{5}&{6}{7}&{3}&{1}

$$\left( \begin{array} { l l l } { 1 } & { 2 } & { 3 } \\ { 2 } & { 5 } & { 6 } \\ { 7 } & { 3 } & { 1 } \end{array} \right)$$

Calculate Determinant

$$-20$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\2&5&6\\7&3&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\2&5&6&2&5\\7&3&1&7&3\end{matrix}\right)$$

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

$$5+2\times 6\times 7+3\times 2\times 3=107$$

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

$$7\times 5\times 3+3\times 6+2\times 2=127$$

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

$$107-127$$

Subtract 127 from 107.

$$-20$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\2&5&6\\7&3&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}5&6\\3&1\end{matrix}\right))-2det(\left(\begin{matrix}2&6\\7&1\end{matrix}\right))+3det(\left(\begin{matrix}2&5\\7&3\end{matrix}\right))$$

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

$$5-3\times 6-2\left(2-7\times 6\right)+3\left(2\times 3-7\times 5\right)$$

Simplify.

$$-13-2\left(-40\right)+3\left(-29\right)$$

Add the terms to obtain the final result.

$$-20$$

Evaluate

$$\left(\begin{matrix}1&2&3\\2&5&6\\7&3&1\end{matrix}\right)$$