Solve {lll}{2}&{3}&{1}{4}&{7}&{5}{2}&{3}&{1}

$$\left| \begin{array} { l l l } { 2 } & { 3 } & { 1 } \\ { 4 } & { 7 } & { 5 } \\ { 2 } & { 3 } & { 1 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&1\\4&7&5\\2&3&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&1&2&3\\4&7&5&4&7\\2&3&1&2&3\end{matrix}\right)$$

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

$$2\times 7+3\times 5\times 2+4\times 3=56$$

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

$$56-56$$

Subtract 56 from 56.

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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).

$$det(\left(\begin{matrix}2&3&1\\4&7&5\\2&3&1\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}7&5\\3&1\end{matrix}\right))-3det(\left(\begin{matrix}4&5\\2&1\end{matrix}\right))+det(\left(\begin{matrix}4&7\\2&3\end{matrix}\right))$$

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

$$2\left(7-3\times 5\right)-3\left(4-2\times 5\right)+4\times 3-2\times 7$$

Simplify.

$$2\left(-8\right)-3\left(-6\right)-2$$

Add the terms to obtain the final result.

$$0$$

Factor

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