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

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

Calculate Determinant

$$-6$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3+2\times 5\times 3+3\times 2=45$$

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

$$3+2\times 5\times 3+3\times 3\times 2=51$$

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

$$45-51$$

Subtract 51 from 45.

$$-6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-6$$

Evaluate

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