Solve {ccc}{i}&{j}&{k}{2}&{2}&{-1}{3}&{8}&{1}

$$\left| \begin{array} { c c c } { i } & { j } & { k } \\ { 2 } & { 2 } & { - 1 } \\ { 3 } & { 8 } & { 1 } \end{array} \right|$$

Evaluate

$$10i+10k-5j$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&j&k\\2&2&-1\\3&8&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&j&k&i&j\\2&2&-1&2&2\\3&8&1&3&8\end{matrix}\right)$$

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

$$2i+j\left(-1\right)\times 3+k\times 2\times 8=2i+16k-3j$$

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

$$3\times 2k+8\left(-i\right)+2j=2j+6k-8i$$

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

$$2i+16k-3j-\left(2j+6k-8i\right)$$

Subtract 6k-8i+2j from 2i-3j+16k.

$$10i+10k-5j$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&j&k\\2&2&-1\\3&8&1\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}2&-1\\8&1\end{matrix}\right))-jdet(\left(\begin{matrix}2&-1\\3&1\end{matrix}\right))+kdet(\left(\begin{matrix}2&2\\3&8\end{matrix}\right))$$

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

$$i\left(2-8\left(-1\right)\right)-j\left(2-3\left(-1\right)\right)+k\left(2\times 8-3\times 2\right)$$

Simplify.

$$10i-j\times 5+k\times 10$$

Add the terms to obtain the final result.

$$10i+10k-5j$$