Solve {ccc}{2}&{-3}&{1}{-1}&{5}&{7}{2}&{2}&{2}

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

Evaluate

$$-68$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 5\times 2-3\times 7\times 2-2=-24$$

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

$$2\times 5+2\times 7\times 2+2\left(-1\right)\left(-3\right)=44$$

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

$$-24-44$$

Subtract 44 from -24.

$$-68$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$2\left(5\times 2-2\times 7\right)-\left(-3\left(-2-2\times 7\right)\right)+-2-2\times 5$$

Simplify.

$$2\left(-4\right)-\left(-3\left(-16\right)\right)-12$$

Add the terms to obtain the final result.

$$-68$$

Factor

$$-68$$