Solve {rrr}{1}&{-4}&{-3}{1}&{-5}&{-3}{-1}&{6}&{4}

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

Calculate Determinant

$$-1$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-5\times 4-4\left(-3\right)\left(-1\right)-3\times 6=-50$$

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

$$-\left(-5\right)\left(-3\right)+6\left(-3\right)+4\left(-4\right)=-49$$

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

$$-50-\left(-49\right)$$

Subtract -49 from -50.

$$-1$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-1$$

Evaluate

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