Solve {ccc}{1}&{3}&{5}{0}&{1}&{2}{0}&{-3}&{6}=

$$\left( \begin{array} { c c c } { 1 } & { 3 } & { 5 } \\ { 0 } & { 1 } & { 2 } \\ { 0 } & { - 3 } & { 6 } \end{array} \right) =$$

Calculate Determinant

$$12$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&5\\0&1&2\\0&-3&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&5&1&3\\0&1&2&0&1\\0&-3&6&0&-3\end{matrix}\right)$$

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

$$6=6$$

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

$$-3\times 2=-6$$

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

$$6-\left(-6\right)$$

Subtract -6 from 6.

$$12$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&5\\0&1&2\\0&-3&6\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}1&2\\-3&6\end{matrix}\right))-3det(\left(\begin{matrix}0&2\\0&6\end{matrix}\right))+5det(\left(\begin{matrix}0&1\\0&-3\end{matrix}\right))$$

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

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

Simplify.

$$12$$

Evaluate

$$\left(\begin{matrix}1&3&5\\0&1&2\\0&-3&6\end{matrix}\right)$$