Solve {lll}{7}&{4}&{2}{8}&{11}&{13}{4}&{7}&{9}

$$\left| \begin{array} { l l l } { 7 } & { 4 } & { 2 } \\ { 8 } & { 11 } & { 13 } \\ { 4 } & { 7 } & { 9 } \end{array} \right|$$

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}7&4&2\\8&11&13\\4&7&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}7&4&2&7&4\\8&11&13&8&11\\4&7&9&4&7\end{matrix}\right)$$

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

$$7\times 11\times 9+4\times 13\times 4+2\times 8\times 7=1013$$

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

$$4\times 11\times 2+7\times 13\times 7+9\times 8\times 4=1013$$

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

$$1013-1013$$

Subtract 1013 from 1013.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}7&4&2\\8&11&13\\4&7&9\end{matrix}\right))$$

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.

$$7det(\left(\begin{matrix}11&13\\7&9\end{matrix}\right))-4det(\left(\begin{matrix}8&13\\4&9\end{matrix}\right))+2det(\left(\begin{matrix}8&11\\4&7\end{matrix}\right))$$

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

$$7\left(11\times 9-7\times 13\right)-4\left(8\times 9-4\times 13\right)+2\left(8\times 7-4\times 11\right)$$

Simplify.

$$7\times 8-4\times 20+2\times 12$$

Add the terms to obtain the final result.

$$0$$

Factor

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