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

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

Evaluate

$$3\left(j+i\right)$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

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

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

$$3j+3i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3i-j\left(-3\right)$$

Add the terms to obtain the final result.

$$3j+3i$$