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

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

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 4+4\times 3\times 3+5\times 2=54$$

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

$$3\times 2\times 5+3+4\times 2\times 4=65$$

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

$$54-65$$

Subtract 65 from 54.

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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).

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

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.

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

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

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

Simplify.

$$5-4\left(-1\right)+5\left(-4\right)$$

Add the terms to obtain the final result.

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Factor

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