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

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

Evaluate

$$51$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\left(-4\right)\left(-1\right)+5\times 5\times 3+3=94$$

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

$$3\left(-4\right)+3\times 5\times 4-5=43$$

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

$$94-43$$

Subtract 43 from 94.

$$51$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$4\left(-11\right)-5\left(-16\right)+15$$

Add the terms to obtain the final result.

$$51$$

Factor

$$3\times 17$$