Solve {ccc}{2}&{2}&{3}{4}&{6}&{-7}{7}&{2}&{3}

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

Evaluate

$$-160$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 6\times 3+2\left(-7\right)\times 7+3\times 4\times 2=-38$$

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

$$7\times 6\times 3+2\left(-7\right)\times 2+3\times 4\times 2=122$$

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

$$-38-122$$

Subtract 122 from -38.

$$-160$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\times 32-2\times 61+3\left(-34\right)$$

Add the terms to obtain the final result.

$$-160$$

Factor

$$-160$$