Solve {lll}{1}&{2}&{1}{1}&{k}&{1}{2}&{-1}&{-1}

$$\left( \begin{array} { l l l } { 1 } & { 2 } & { 1 } \\ { 1 } & { k } & { 1 } \\ { 2 } & { - 1 } & { - 1 } \end{array} \right)$$

Calculate Determinant

$$6-3k$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$2k-1-2=2k-3$$

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

$$3-k-\left(2k-3\right)$$

Subtract 2k-3 from -k+3.

$$6-3k$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$k\left(-1\right)-\left(-1\right)-2\left(-1-2\right)-1-2k$$

Simplify.

$$1-k-2\left(-3\right)-2k-1$$

Add the terms to obtain the final result.

$$6-3k$$

Evaluate

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