Solve {ccc}{1}&{1}&{22}{5}&{-2}&{7}{2}&{-5}&{4}

$$\left| \begin{array} { c c c } { 1 } & { 1 } & { 22 } \\ { 5 } & { - 2 } & { 7 } \\ { 2 } & { - 5 } & { 4 } \end{array} \right|$$

Evaluate

$$-441$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&22\\5&-2&7\\2&-5&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&22&1&1\\5&-2&7&5&-2\\2&-5&4&2&-5\end{matrix}\right)$$

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

$$-2\times 4+7\times 2+22\times 5\left(-5\right)=-544$$

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

$$2\left(-2\right)\times 22-5\times 7+4\times 5=-103$$

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

$$-544-\left(-103\right)$$

Subtract -103 from -544.

$$-441$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&22\\5&-2&7\\2&-5&4\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-2&7\\-5&4\end{matrix}\right))-det(\left(\begin{matrix}5&7\\2&4\end{matrix}\right))+22det(\left(\begin{matrix}5&-2\\2&-5\end{matrix}\right))$$

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

$$-2\times 4-\left(-5\times 7\right)-\left(5\times 4-2\times 7\right)+22\left(5\left(-5\right)-2\left(-2\right)\right)$$

Simplify.

$$27-6+22\left(-21\right)$$

Add the terms to obtain the final result.

$$-441$$

Factor

$$-441$$