Solve {lll}{i}&{J}&{k}{3}&{1}&{1}{1}&{-1}&{3}

$$\left| \begin{array} { l l l } { i } & { J } & { k } \\ { 3 } & { 1 } & { 1 } \\ { 1 } & { - 1 } & { 3 } \end{array} \right|$$

Evaluate

$$4i-4k-8J$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&J&k\\3&1&1\\1&-1&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&J&k&i&J\\3&1&1&3&1\\1&-1&3&1&-1\end{matrix}\right)$$

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

$$3i+J+k\times 3\left(-1\right)=J-3k+3i$$

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

$$k-i+3\times 3J=9J+k-i$$

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

$$J-3k+3i-\left(9J+k-i\right)$$

Subtract k-i+9J from 3i+J-3k.

$$4i-4k-8J$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&J&k\\3&1&1\\1&-1&3\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}1&1\\-1&3\end{matrix}\right))-Jdet(\left(\begin{matrix}3&1\\1&3\end{matrix}\right))+kdet(\left(\begin{matrix}3&1\\1&-1\end{matrix}\right))$$

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

$$i\left(3-\left(-1\right)\right)-J\left(3\times 3-1\right)+k\left(3\left(-1\right)-1\right)$$

Simplify.

$$4i-J\times 8+k\left(-4\right)$$

Add the terms to obtain the final result.

$$4i-4k-8J$$