Solve {ccc}{6}&{-2}&{3}{1}&{1}&{1}{4}&{3}&{-7}

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

Evaluate

$$-85$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$6\left(-7\right)-2\times 4+3\times 3=-41$$

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

$$4\times 3+3\times 6-7\left(-2\right)=44$$

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

$$-41-44$$

Subtract 44 from -41.

$$-85$$

Steps Using Expansion by Minors

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

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

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.

$$6det(\left(\begin{matrix}1&1\\3&-7\end{matrix}\right))-\left(-2det(\left(\begin{matrix}1&1\\4&-7\end{matrix}\right))\right)+3det(\left(\begin{matrix}1&1\\4&3\end{matrix}\right))$$

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

$$6\left(-7-3\right)-\left(-2\left(-7-4\right)\right)+3\left(3-4\right)$$

Simplify.

$$6\left(-10\right)-\left(-2\left(-11\right)\right)+3\left(-1\right)$$

Add the terms to obtain the final result.

$$-85$$

Factor

$$-85$$