Solve {lll}{3}&{6}&{7}{4}&{3}&{5}{2}&{7}&{4}

$$\left( \begin{array} { l l l } { 3 } & { 6 } & { 7 } \\ { 4 } & { 3 } & { 5 } \\ { 2 } & { 7 } & { 4 } \end{array} \right)$$

Calculate Determinant

$$49$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3\times 4+6\times 5\times 2+7\times 4\times 7=292$$

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

$$2\times 3\times 7+7\times 5\times 3+4\times 4\times 6=243$$

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

$$292-243$$

Subtract 243 from 292.

$$49$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\left(-23\right)-6\times 6+7\times 22$$

Add the terms to obtain the final result.

$$49$$

Evaluate

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