Solve {ccc}{4}&{-3}&{6}{8}&{6}&{5}{0}&{3}&{-1}

$$\left( \begin{array} { c c c } { 4 } & { - 3 } & { 6 } \\ { 8 } & { 6 } & { 5 } \\ { 0 } & { 3 } & { - 1 } \end{array} \right)$$

Calculate Determinant

$$36$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&-3&6\\8&6&5\\0&3&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&-3&6&4&-3\\8&6&5&8&6\\0&3&-1&0&3\end{matrix}\right)$$

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

$$4\times 6\left(-1\right)+6\times 8\times 3=120$$

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

$$3\times 5\times 4-8\left(-3\right)=84$$

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

$$120-84$$

Subtract 84 from 120.

$$36$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&-3&6\\8&6&5\\0&3&-1\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}6&5\\3&-1\end{matrix}\right))-\left(-3det(\left(\begin{matrix}8&5\\0&-1\end{matrix}\right))\right)+6det(\left(\begin{matrix}8&6\\0&3\end{matrix}\right))$$

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

$$4\left(6\left(-1\right)-3\times 5\right)-\left(-3\times 8\left(-1\right)\right)+6\times 8\times 3$$

Simplify.

$$4\left(-21\right)-\left(-3\left(-8\right)\right)+6\times 24$$

Add the terms to obtain the final result.

$$36$$

Evaluate

$$\left(\begin{matrix}4&-3&6\\8&6&5\\0&3&-1\end{matrix}\right)$$