Solve {ccc}{2}&{1}&{3}{-2}&{0}&{5}{2}&{3}&{19}

$$\left( \begin{array} { c c c } { 2 } & { 1 } & { 3 } \\ { - 2 } & { 0 } & { 5 } \\ { 2 } & { 3 } & { 19 } \end{array} \right)$$

Calculate Determinant

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&1&3\\-2&0&5\\2&3&19\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&1&3&2&1\\-2&0&5&-2&0\\2&3&19&2&3\end{matrix}\right)$$

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

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

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

$$3\times 5\times 2+19\left(-2\right)=-8$$

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

$$-8-\left(-8\right)$$

Subtract -8 from -8.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&1&3\\-2&0&5\\2&3&19\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}0&5\\3&19\end{matrix}\right))-det(\left(\begin{matrix}-2&5\\2&19\end{matrix}\right))+3det(\left(\begin{matrix}-2&0\\2&3\end{matrix}\right))$$

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

$$2\left(-3\times 5\right)-\left(-2\times 19-2\times 5\right)+3\left(-2\right)\times 3$$

Simplify.

$$2\left(-15\right)-\left(-48\right)+3\left(-6\right)$$

Add the terms to obtain the final result.

$$0$$

Evaluate

$$\left(\begin{matrix}2&1&3\\-2&0&5\\2&3&19\end{matrix}\right)$$