Solve {lll}{2}&{1}&{3}{5}&{2}&{1}{3}&{8}&{6}

$$\left( \begin{array} { l l l } { 2 } & { 1 } & { 3 } \\ { 5 } & { 2 } & { 1 } \\ { 3 } & { 8 } & { 6 } \end{array} \right)$$

Calculate Determinant

$$83$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&1&3\\5&2&1\\3&8&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&1&3&2&1\\5&2&1&5&2\\3&8&6&3&8\end{matrix}\right)$$

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

$$2\times 2\times 6+3+3\times 5\times 8=147$$

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

$$3\times 2\times 3+8\times 2+6\times 5=64$$

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

$$147-64$$

Subtract 64 from 147.

$$83$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&1&3\\5&2&1\\3&8&6\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}2&1\\8&6\end{matrix}\right))-det(\left(\begin{matrix}5&1\\3&6\end{matrix}\right))+3det(\left(\begin{matrix}5&2\\3&8\end{matrix}\right))$$

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

$$2\left(2\times 6-8\right)-\left(5\times 6-3\right)+3\left(5\times 8-3\times 2\right)$$

Simplify.

$$2\times 4-27+3\times 34$$

Add the terms to obtain the final result.

$$83$$

Evaluate

$$\left(\begin{matrix}2&1&3\\5&2&1\\3&8&6\end{matrix}\right)$$