Solve {rrr}{4}&{-1}&{0}{5}&{7}&{6}{2}&{1}&{3}

$$\left| \begin{array} { r r r } { 4 } & { - 1 } & { 0 } \\ { 5 } & { 7 } & { 6 } \\ { 2 } & { 1 } & { 3 } \end{array} \right|$$

Evaluate

$$63$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&-1&0\\5&7&6\\2&1&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&-1&0&4&-1\\5&7&6&5&7\\2&1&3&2&1\end{matrix}\right)$$

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

$$4\times 7\times 3-6\times 2=72$$

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

$$6\times 4+3\times 5\left(-1\right)=9$$

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

$$72-9$$

Subtract 9 from 72.

$$63$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&-1&0\\5&7&6\\2&1&3\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}7&6\\1&3\end{matrix}\right))-\left(-det(\left(\begin{matrix}5&6\\2&3\end{matrix}\right))\right)$$

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

$$4\left(7\times 3-6\right)-\left(-\left(5\times 3-2\times 6\right)\right)$$

Simplify.

$$4\times 15-\left(-3\right)$$

Add the terms to obtain the final result.

$$63$$

Factor

$$3^{2}\times 7$$