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

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

Evaluate

$$-3i+2k-5j$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

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

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

$$-3i+2k-5j$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-3i-j\times 5+k\times 2$$

Add the terms to obtain the final result.

$$-3i+2k-5j$$