Solve {rrr}{1}&{1}&{1}{1}&{4}&{-1}{-1}&{8}&{3}

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

Evaluate

$$30$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\times 3-\left(-1\right)+8=21$$

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

$$-4+8\left(-1\right)+3=-9$$

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

$$21-\left(-9\right)$$

Subtract -9 from 21.

$$30$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$20-2+12$$

Add the terms to obtain the final result.

$$30$$

Factor

$$2\times 3\times 5$$