Solve {ccc}{1}&{1}&{k}{1}&{1}&{3}{2}&{k}&{-1}

$$\left( \begin{array} { c c c } { 1 } & { 1 } & { k } \\ { 1 } & { 1 } & { 3 } \\ { 2 } & { k } & { - 1 } \end{array} \right)$$

Calculate Determinant

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

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&k\\1&1&3\\2&k&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&k&1&1\\1&1&3&1&1\\2&k&-1&2&k\end{matrix}\right)$$

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

$$-1+3\times 2+kk=k^{2}+5$$

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

$$2k+k\times 3-1=5k-1$$

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

$$k^{2}+5-\left(5k-1\right)$$

Subtract 5k-1 from 5+k^{2}.

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

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&k\\1&1&3\\2&k&-1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}1&3\\k&-1\end{matrix}\right))-det(\left(\begin{matrix}1&3\\2&-1\end{matrix}\right))+kdet(\left(\begin{matrix}1&1\\2&k\end{matrix}\right))$$

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

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

Simplify.

$$-3k-1-\left(-7\right)+k\left(k-2\right)$$

Add the terms to obtain the final result.

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

Evaluate

$$\left(\begin{matrix}1&1&k\\1&1&3\\2&k&-1\end{matrix}\right)$$