Solve {lll}{1}&{1}&{1}{2}&{3}&{-1}{4}&{9}&{1}

$$\left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 2 } & { 3 } & { - 1 } \\ { 4 } & { 9 } & { 1 } \end{array} \right|$$

Evaluate

$$12$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&1\\2&3&-1\\4&9&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\2&3&-1&2&3\\4&9&1&4&9\end{matrix}\right)$$

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

$$3-4+2\times 9=17$$

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

$$4\times 3+9\left(-1\right)+2=5$$

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

$$17-5$$

Subtract 5 from 17.

$$12$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&1\\2&3&-1\\4&9&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}3&-1\\9&1\end{matrix}\right))-det(\left(\begin{matrix}2&-1\\4&1\end{matrix}\right))+det(\left(\begin{matrix}2&3\\4&9\end{matrix}\right))$$

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

$$3-9\left(-1\right)-\left(2-4\left(-1\right)\right)+2\times 9-4\times 3$$

Simplify.

$$12-6+6$$

Add the terms to obtain the final result.

$$12$$

Factor

$$2^{2}\times 3$$