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

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

Evaluate

$$5j+k+2i$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-i+j\times 3\times 2+k\left(-1\right)=6j-k-i$$

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

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

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

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

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

$$5j+k+2i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2i-j\left(-5\right)+k$$

Add the terms to obtain the final result.

$$5j+k+2i$$