Solve {lll}{29}&{2}&{14}{19}&{3}&{17}{39}&{8}&{38}

$$\left| \begin{array} { l l l } { 29 } & { 2 } & { 14 } \\ { 19 } & { 3 } & { 17 } \\ { 39 } & { 8 } & { 38 } \end{array} \right|$$

Evaluate

$$-266$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}29&2&14\\19&3&17\\39&8&38\end{matrix}\right))$$

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

$$\left(\begin{matrix}29&2&14&29&2\\19&3&17&19&3\\39&8&38&39&8\end{matrix}\right)$$

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

$$29\times 3\times 38+2\times 17\times 39+14\times 19\times 8=6760$$

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

$$39\times 3\times 14+8\times 17\times 29+38\times 19\times 2=7026$$

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

$$6760-7026$$

Subtract 7026 from 6760.

$$-266$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}29&2&14\\19&3&17\\39&8&38\end{matrix}\right))$$

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.

$$29det(\left(\begin{matrix}3&17\\8&38\end{matrix}\right))-2det(\left(\begin{matrix}19&17\\39&38\end{matrix}\right))+14det(\left(\begin{matrix}19&3\\39&8\end{matrix}\right))$$

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

$$29\left(3\times 38-8\times 17\right)-2\left(19\times 38-39\times 17\right)+14\left(19\times 8-39\times 3\right)$$

Simplify.

$$29\left(-22\right)-2\times 59+14\times 35$$

Add the terms to obtain the final result.

$$-266$$

Factor

$$-266$$