Solve {ccc}{1}&{3}&{-5}{2}&{-4}&{7}{6}&{8}&{3}

$$\left( \begin{array} { c c c } { 1 } & { 3 } & { - 5 } \\ { 2 } & { - 4 } & { 7 } \\ { 6 } & { 8 } & { 3 } \end{array} \right)$$

Calculate Determinant

$$-160$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&-5\\2&-4&7\\6&8&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&-5&1&3\\2&-4&7&2&-4\\6&8&3&6&8\end{matrix}\right)$$

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

$$-4\times 3+3\times 7\times 6-5\times 2\times 8=34$$

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

$$6\left(-4\right)\left(-5\right)+8\times 7+3\times 2\times 3=194$$

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

$$34-194$$

Subtract 194 from 34.

$$-160$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&-5\\2&-4&7\\6&8&3\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-4&7\\8&3\end{matrix}\right))-3det(\left(\begin{matrix}2&7\\6&3\end{matrix}\right))-5det(\left(\begin{matrix}2&-4\\6&8\end{matrix}\right))$$

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

$$-4\times 3-8\times 7-3\left(2\times 3-6\times 7\right)-5\left(2\times 8-6\left(-4\right)\right)$$

Simplify.

$$-68-3\left(-36\right)-5\times 40$$

Add the terms to obtain the final result.

$$-160$$

Evaluate

$$\left(\begin{matrix}1&3&-5\\2&-4&7\\6&8&3\end{matrix}\right)$$