Solve {ccc}{2}&{-1}&{-3}{-2}&{1}&{4}{1}&{3}&{0}

$$\left( \begin{array} { c c c } { 2 } & { - 1 } & { - 3 } \\ { - 2 } & { 1 } & { 4 } \\ { 1 } & { 3 } & { 0 } \end{array} \right)$$

Calculate Determinant

$$-7$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-1&-3\\-2&1&4\\1&3&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-1&-3&2&-1\\-2&1&4&-2&1\\1&3&0&1&3\end{matrix}\right)$$

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

$$-4-3\left(-2\right)\times 3=14$$

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

$$-3+3\times 4\times 2=21$$

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

$$14-21$$

Subtract 21 from 14.

$$-7$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-1&-3\\-2&1&4\\1&3&0\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}1&4\\3&0\end{matrix}\right))-\left(-det(\left(\begin{matrix}-2&4\\1&0\end{matrix}\right))\right)-3det(\left(\begin{matrix}-2&1\\1&3\end{matrix}\right))$$

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

$$2\left(-3\times 4\right)-\left(-\left(-4\right)\right)-3\left(-2\times 3-1\right)$$

Simplify.

$$2\left(-12\right)-\left(-\left(-4\right)\right)-3\left(-7\right)$$

Add the terms to obtain the final result.

$$-7$$

Evaluate

$$\left(\begin{matrix}2&-1&-3\\-2&1&4\\1&3&0\end{matrix}\right)$$