Solve {lll}{2}&{3}&{4}{1}&{7}&{8}{9}&{1}&{2}=

$$\left( \begin{array} { l l l } { 2 } & { 3 } & { 4 } \\ { 1 } & { 7 } & { 8 } \\ { 9 } & { 1 } & { 2 } \end{array} \right) =$$

Calculate Determinant

$$-26$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&4\\1&7&8\\9&1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&4&2&3\\1&7&8&1&7\\9&1&2&9&1\end{matrix}\right)$$

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

$$2\times 7\times 2+3\times 8\times 9+4=248$$

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

$$9\times 7\times 4+8\times 2+2\times 3=274$$

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

$$248-274$$

Subtract 274 from 248.

$$-26$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&4\\1&7&8\\9&1&2\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}7&8\\1&2\end{matrix}\right))-3det(\left(\begin{matrix}1&8\\9&2\end{matrix}\right))+4det(\left(\begin{matrix}1&7\\9&1\end{matrix}\right))$$

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

$$2\left(7\times 2-8\right)-3\left(2-9\times 8\right)+4\left(1-9\times 7\right)$$

Simplify.

$$2\times 6-3\left(-70\right)+4\left(-62\right)$$

Add the terms to obtain the final result.

$$-26$$

Evaluate

$$\left(\begin{matrix}2&3&4\\1&7&8\\9&1&2\end{matrix}\right)$$