Solve {lll}{31}&{2}&{4}{29}&{1}&{2}{10}&{-1}&{1}

$$\left| \begin{array} { l l l } { 31 } & { 2 } & { 4 } \\ { 29 } & { 1 } & { 2 } \\ { 10 } & { - 1 } & { 1 } \end{array} \right|$$

Evaluate

$$-81$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}31&2&4\\29&1&2\\10&-1&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}31&2&4&31&2\\29&1&2&29&1\\10&-1&1&10&-1\end{matrix}\right)$$

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

$$31+2\times 2\times 10+4\times 29\left(-1\right)=-45$$

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

$$10\times 4-2\times 31+29\times 2=36$$

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

$$-45-36$$

Subtract 36 from -45.

$$-81$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}31&2&4\\29&1&2\\10&-1&1\end{matrix}\right))$$

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.

$$31det(\left(\begin{matrix}1&2\\-1&1\end{matrix}\right))-2det(\left(\begin{matrix}29&2\\10&1\end{matrix}\right))+4det(\left(\begin{matrix}29&1\\10&-1\end{matrix}\right))$$

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

$$31\left(1-\left(-2\right)\right)-2\left(29-10\times 2\right)+4\left(29\left(-1\right)-10\right)$$

Simplify.

$$31\times 3-2\times 9+4\left(-39\right)$$

Add the terms to obtain the final result.

$$-81$$

Factor

$$-81$$