Solve {ccc}{-1}&{3}&{1}{0}&{5}&{4}{1}&{-2}&{6}

$$\left( \begin{array} { c c c } { - 1 } & { 3 } & { 1 } \\ { 0 } & { 5 } & { 4 } \\ { 1 } & { - 2 } & { 6 } \end{array} \right)$$

Calculate Determinant

$$-31$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-1&3&1\\0&5&4\\1&-2&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}-1&3&1&-1&3\\0&5&4&0&5\\1&-2&6&1&-2\end{matrix}\right)$$

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

$$-5\times 6+3\times 4=-18$$

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

$$5-2\times 4\left(-1\right)=13$$

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

$$-18-13$$

Subtract 13 from -18.

$$-31$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-1&3&1\\0&5&4\\1&-2&6\end{matrix}\right))$$

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.

$$-det(\left(\begin{matrix}5&4\\-2&6\end{matrix}\right))-3det(\left(\begin{matrix}0&4\\1&6\end{matrix}\right))+det(\left(\begin{matrix}0&5\\1&-2\end{matrix}\right))$$

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

$$-\left(5\times 6-\left(-2\times 4\right)\right)-3\left(-4\right)-5$$

Simplify.

$$-38-3\left(-4\right)-5$$

Add the terms to obtain the final result.

$$-31$$

Evaluate

$$\left(\begin{matrix}-1&3&1\\0&5&4\\1&-2&6\end{matrix}\right)$$