Solve {ccc}{5}&{-1}&{9}{3}&{-3}&{20}{1}&{1}&{2}

$$\left| \begin{array} { c c c } { 5 } & { - 1 } & { 9 } \\ { 3 } & { - 3 } & { 20 } \\ { 1 } & { 1 } & { 2 } \end{array} \right|$$

Evaluate

$$-90$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&-1&9\\3&-3&20\\1&1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&-1&9&5&-1\\3&-3&20&3&-3\\1&1&2&1&1\end{matrix}\right)$$

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

$$5\left(-3\right)\times 2-20+9\times 3=-23$$

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

$$-3\times 9+20\times 5+2\times 3\left(-1\right)=67$$

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

$$-23-67$$

Subtract 67 from -23.

$$-90$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&-1&9\\3&-3&20\\1&1&2\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}-3&20\\1&2\end{matrix}\right))-\left(-det(\left(\begin{matrix}3&20\\1&2\end{matrix}\right))\right)+9det(\left(\begin{matrix}3&-3\\1&1\end{matrix}\right))$$

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

$$5\left(-3\times 2-20\right)-\left(-\left(3\times 2-20\right)\right)+9\left(3-\left(-3\right)\right)$$

Simplify.

$$5\left(-26\right)-\left(-\left(-14\right)\right)+9\times 6$$

Add the terms to obtain the final result.

$$-90$$

Factor

$$-90$$