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

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

Calculate Determinant

$$5$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 2+5\times 4+2\times 3\times 3=42$$

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

$$4\times 2\times 2+3\times 2+3\times 5=37$$

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

$$42-37$$

Subtract 37 from 42.

$$5$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-1\right)-5\left(-1\right)+2$$

Add the terms to obtain the final result.

$$5$$

Evaluate

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