Solve {lll}{4}&{5}&{6}{1}&{9}&{7}{4}&{9}&{5}

$$\left( \begin{array} { l l l } { 4 } & { 5 } & { 6 } \\ { 1 } & { 9 } & { 7 } \\ { 4 } & { 9 } & { 5 } \end{array} \right)$$

Calculate Determinant

$$-119$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&5&6\\1&9&7\\4&9&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&5&6&4&5\\1&9&7&1&9\\4&9&5&4&9\end{matrix}\right)$$

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

$$4\times 9\times 5+5\times 7\times 4+6\times 9=374$$

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

$$4\times 9\times 6+9\times 7\times 4+5\times 5=493$$

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

$$374-493$$

Subtract 493 from 374.

$$-119$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&5&6\\1&9&7\\4&9&5\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}9&7\\9&5\end{matrix}\right))-5det(\left(\begin{matrix}1&7\\4&5\end{matrix}\right))+6det(\left(\begin{matrix}1&9\\4&9\end{matrix}\right))$$

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

$$4\left(9\times 5-9\times 7\right)-5\left(5-4\times 7\right)+6\left(9-4\times 9\right)$$

Simplify.

$$4\left(-18\right)-5\left(-23\right)+6\left(-27\right)$$

Add the terms to obtain the final result.

$$-119$$

Evaluate

$$\left(\begin{matrix}4&5&6\\1&9&7\\4&9&5\end{matrix}\right)$$