Solve {ccc}{i}&{j}&{k}{1}&{0}&{-2}{0}&{1}&{-1}

$$\left| \begin{array} { c c c } { i } & { j } & { k } \\ { 1 } & { 0 } & { - 2 } \\ { 0 } & { 1 } & { - 1 } \end{array} \right|$$

Evaluate

$$j+k+2i$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&j&k\\1&0&-2\\0&1&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&j&k&i&j\\1&0&-2&1&0\\0&1&-1&0&1\end{matrix}\right)$$

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

$$k=k$$

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

$$-2i-j=-2i-j$$

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

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

Subtract -2i-j from k.

$$j+k+2i$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&j&k\\1&0&-2\\0&1&-1\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}0&-2\\1&-1\end{matrix}\right))-jdet(\left(\begin{matrix}1&-2\\0&-1\end{matrix}\right))+kdet(\left(\begin{matrix}1&0\\0&1\end{matrix}\right))$$

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

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

Simplify.

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

Add the terms to obtain the final result.

$$j+k+2i$$