Solve {lll}{2}&{8}&{-3}{3}&{9}&{6}{6}&{1}&{0}

$$\left( \begin{array} { l l l } { 2 } & { 8 } & { - 3 } \\ { 3 } & { 9 } & { 6 } \\ { 6 } & { 1 } & { 0 } \end{array} \right)$$

Calculate Determinant

$$429$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&8&-3\\3&9&6\\6&1&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&8&-3&2&8\\3&9&6&3&9\\6&1&0&6&1\end{matrix}\right)$$

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

$$8\times 6\times 6-3\times 3=279$$

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

$$6\times 9\left(-3\right)+6\times 2=-150$$

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

$$279-\left(-150\right)$$

Subtract -150 from 279.

$$429$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&8&-3\\3&9&6\\6&1&0\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}9&6\\1&0\end{matrix}\right))-8det(\left(\begin{matrix}3&6\\6&0\end{matrix}\right))-3det(\left(\begin{matrix}3&9\\6&1\end{matrix}\right))$$

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

$$2\left(-6\right)-8\left(-6\times 6\right)-3\left(3-6\times 9\right)$$

Simplify.

$$2\left(-6\right)-8\left(-36\right)-3\left(-51\right)$$

Add the terms to obtain the final result.

$$429$$

Evaluate

$$\left(\begin{matrix}2&8&-3\\3&9&6\\6&1&0\end{matrix}\right)$$