Solve {ccc}{43}&{1}&{6}{35}&{7}&{4}{17}&{3}&{2}

$$\left| \begin{array} { c c c } { 43 } & { 1 } & { 6 } \\ { 35 } & { 7 } & { 4 } \\ { 17 } & { 3 } & { 2 } \end{array} \right|$$

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}43&1&6\\35&7&4\\17&3&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}43&1&6&43&1\\35&7&4&35&7\\17&3&2&17&3\end{matrix}\right)$$

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

$$43\times 7\times 2+4\times 17+6\times 35\times 3=1300$$

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

$$17\times 7\times 6+3\times 4\times 43+2\times 35=1300$$

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

$$1300-1300$$

Subtract 1300 from 1300.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}43&1&6\\35&7&4\\17&3&2\end{matrix}\right))$$

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.

$$43det(\left(\begin{matrix}7&4\\3&2\end{matrix}\right))-det(\left(\begin{matrix}35&4\\17&2\end{matrix}\right))+6det(\left(\begin{matrix}35&7\\17&3\end{matrix}\right))$$

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

$$43\left(7\times 2-3\times 4\right)-\left(35\times 2-17\times 4\right)+6\left(35\times 3-17\times 7\right)$$

Simplify.

$$43\times 2-2+6\left(-14\right)$$

Add the terms to obtain the final result.

$$0$$

Factor

$$0$$