Solve {rrr}{1}&{-1}&{2}{3}&{lambda}&{2}{5}&{3}&{6}

$$\left| \begin{array} { r r r } { 1 } & { - 1 } & { 2 } \\ { 3 } & { \lambda } & { 2 } \\ { 5 } & { 3 } & { 6 } \end{array} \right|$$

Evaluate

$$20-4\lambda$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&-1&2\\3&\lambda &2\\5&3&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&-1&2&1&-1\\3&\lambda &2&3&\lambda \\5&3&6&5&3\end{matrix}\right)$$

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

$$\lambda \times 6-2\times 5+2\times 3\times 3=6\lambda +8$$

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

$$5\lambda \times 2+3\times 2+6\times 3\left(-1\right)=10\lambda -12$$

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

$$6\lambda +8-\left(10\lambda -12\right)$$

Subtract 10\lambda -12 from 6\lambda +8.

$$20-4\lambda$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&-1&2\\3&\lambda &2\\5&3&6\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}\lambda &2\\3&6\end{matrix}\right))-\left(-det(\left(\begin{matrix}3&2\\5&6\end{matrix}\right))\right)+2det(\left(\begin{matrix}3&\lambda \\5&3\end{matrix}\right))$$

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

$$\lambda \times 6-3\times 2-\left(-\left(3\times 6-5\times 2\right)\right)+2\left(3\times 3-5\lambda \right)$$

Simplify.

$$6\lambda -6-\left(-8\right)+2\left(9-5\lambda \right)$$

Add the terms to obtain the final result.

$$20-4\lambda$$

Integrate w.r.t. λ

$$С-2\lambda ^{2}+20\lambda$$