Solve {ccc}{2}&{0}&{1}{1}&{-4}&{-1}{-1}&{8}&{3}

$$\left| \begin{array} { c c c } { 2 } & { 0 } & { 1 } \\ { 1 } & { - 4 } & { - 1 } \\ { - 1 } & { 8 } & { 3 } \end{array} \right|$$

Evaluate

$$-4$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&0&1\\1&-4&-1\\-1&8&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&0&1&2&0\\1&-4&-1&1&-4\\-1&8&3&-1&8\end{matrix}\right)$$

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

$$2\left(-4\right)\times 3+8=-16$$

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

$$-\left(-4\right)+8\left(-1\right)\times 2=-12$$

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

$$-16-\left(-12\right)$$

Subtract -12 from -16.

$$-4$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&0&1\\1&-4&-1\\-1&8&3\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}-4&-1\\8&3\end{matrix}\right))+det(\left(\begin{matrix}1&-4\\-1&8\end{matrix}\right))$$

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

$$2\left(-4\times 3-8\left(-1\right)\right)+8-\left(-\left(-4\right)\right)$$

Simplify.

$$2\left(-4\right)+4$$

Add the terms to obtain the final result.

$$-4$$

Factor

$$-4$$