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

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

Evaluate

$$3\left(k-\gamma \right)$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-k+i+2\gamma =i+2\gamma -k$$

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

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

Subtract -k+i+2\gamma from i-\gamma +2k.

$$3k-3\gamma$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-\gamma \times 3+k\times 3$$

Add the terms to obtain the final result.

$$3k-3\gamma$$