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

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

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$4\times 5+2\times 2+2\times 3=30$$

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

$$4+3\times 2+5\times 2\times 2=30$$

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

$$30-30$$

Subtract 30 from 30.

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Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$14-2\times 8+2$$

Add the terms to obtain the final result.

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Factor

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