Solve {lll}{1}&{2}&{3}{3}&{6}&{7}{9}&{12}&{17}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 3 } \\ { 3 } & { 6 } & { 7 } \\ { 9 } & { 12 } & { 17 } \end{array} \right|$$

Evaluate

$$-12$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&3\\3&6&7\\9&12&17\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&3&1&2\\3&6&7&3&6\\9&12&17&9&12\end{matrix}\right)$$

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

$$6\times 17+2\times 7\times 9+3\times 3\times 12=336$$

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

$$9\times 6\times 3+12\times 7+17\times 3\times 2=348$$

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

$$336-348$$

Subtract 348 from 336.

$$-12$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&3\\3&6&7\\9&12&17\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}6&7\\12&17\end{matrix}\right))-2det(\left(\begin{matrix}3&7\\9&17\end{matrix}\right))+3det(\left(\begin{matrix}3&6\\9&12\end{matrix}\right))$$

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

$$6\times 17-12\times 7-2\left(3\times 17-9\times 7\right)+3\left(3\times 12-9\times 6\right)$$

Simplify.

$$18-2\left(-12\right)+3\left(-18\right)$$

Add the terms to obtain the final result.

$$-12$$

Factor

$$-12$$