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

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

Evaluate

$$28$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$3\left(-4\right)\times 3+5-3\times 2\left(-1\right)=-25$$

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

$$3-\left(-25\right)$$

Subtract -25 from 3.

$$28$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$7-\left(-\left(-21\right)\right)+3\times 14$$

Add the terms to obtain the final result.

$$28$$

Factor

$$2^{2}\times 7$$