Solve {ccc}{1}&{-1}&{4}{2}&{-3}&{1}{3}&{-5}&{-1}

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

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$-3\left(-1\right)-3+4\times 2\left(-5\right)=-40$$

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

$$3\left(-3\right)\times 4-5-2\left(-1\right)=-39$$

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

$$-40-\left(-39\right)$$

Subtract -39 from -40.

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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).

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

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.

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

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

$$-3\left(-1\right)-\left(-5\right)-\left(-\left(2\left(-1\right)-3\right)\right)+4\left(2\left(-5\right)-3\left(-3\right)\right)$$

Simplify.

$$8-\left(-\left(-5\right)\right)+4\left(-1\right)$$

Add the terms to obtain the final result.

$$-1$$

Factor

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