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

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

Evaluate

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

Steps Using Sarrus' Rule

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

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

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

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

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

$$-i+j+k=j+k-i$$

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

$$k+i-j=i+k-j$$

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

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

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

$$2j-2i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$2j-2i$$