Solve {ccc}{1}&{2}&{3}{4}&{5}&{6}{7}&{-1}&{8}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { 3 } \\ { 4 } & { 5 } & { 6 } \\ { 7 } & { - 1 } & { 8 } \end{array} \right|$$

Evaluate

$$-51$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\4&5&6\\7&-1&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\4&5&6&4&5\\7&-1&8&7&-1\end{matrix}\right)$$

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

$$5\times 8+2\times 6\times 7+3\times 4\left(-1\right)=112$$

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

$$7\times 5\times 3-6+8\times 4\times 2=163$$

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

$$112-163$$

Subtract 163 from 112.

$$-51$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\4&5&6\\7&-1&8\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}5&6\\-1&8\end{matrix}\right))-2det(\left(\begin{matrix}4&6\\7&8\end{matrix}\right))+3det(\left(\begin{matrix}4&5\\7&-1\end{matrix}\right))$$

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

$$5\times 8-\left(-6\right)-2\left(4\times 8-7\times 6\right)+3\left(4\left(-1\right)-7\times 5\right)$$

Simplify.

$$46-2\left(-10\right)+3\left(-39\right)$$

Add the terms to obtain the final result.

$$-51$$

Factor

$$-51$$