Solve {lll}{5}&{4}&{3}{2}&{5}&{1}{3}&{9}&{2}

$$\left( \begin{array} { l l l } { 5 } & { 4 } & { 3 } \\ { 2 } & { 5 } & { 1 } \\ { 3 } & { 9 } & { 2 } \end{array} \right)$$

Calculate Determinant

$$10$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&4&3\\2&5&1\\3&9&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&4&3&5&4\\2&5&1&2&5\\3&9&2&3&9\end{matrix}\right)$$

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

$$5\times 5\times 2+4\times 3+3\times 2\times 9=116$$

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

$$3\times 5\times 3+9\times 5+2\times 2\times 4=106$$

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

$$116-106$$

Subtract 106 from 116.

$$10$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&4&3\\2&5&1\\3&9&2\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}5&1\\9&2\end{matrix}\right))-4det(\left(\begin{matrix}2&1\\3&2\end{matrix}\right))+3det(\left(\begin{matrix}2&5\\3&9\end{matrix}\right))$$

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

$$5\left(5\times 2-9\right)-4\left(2\times 2-3\right)+3\left(2\times 9-3\times 5\right)$$

Simplify.

$$5-4+3\times 3$$

Add the terms to obtain the final result.

$$10$$

Evaluate

$$\left(\begin{matrix}5&4&3\\2&5&1\\3&9&2\end{matrix}\right)$$