Solve {ccc}{1}&{-1}&{3}{2}&{2}&{4}{3}&{1}&{7}

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

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 7-4\times 3+3\times 2=8$$

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

$$3\times 2\times 3+4+7\times 2\left(-1\right)=8$$

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

$$8-8$$

Subtract 8 from 8.

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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).

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

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.

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

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

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

Simplify.

$$10-\left(-2\right)+3\left(-4\right)$$

Add the terms to obtain the final result.

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Factor

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