Solve {ccc}{1}&{4}&{1}{(-1)}&{2}&{1}{2}&{3}&{2}

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

Calculate Determinant

$$10$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 2+4\times 2-3=9$$

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

$$2\times 2+3+2\left(-1\right)\times 4=-1$$

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

$$9-\left(-1\right)$$

Subtract -1 from 9.

$$10$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$1-4\left(-4\right)-7$$

Add the terms to obtain the final result.

$$10$$

Evaluate

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