Solve {rrr}{lambda}&{2}&{-1}{2}&{lambda}&{-1}{-1}&{-1}&{lambda}

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

Evaluate

$$\lambda ^{3}-6\lambda +4$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$\lambda \lambda \lambda +2\left(-1\right)\left(-1\right)-2\left(-1\right)=\lambda ^{3}+4$$

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

$$-\lambda \left(-1\right)-\left(-\lambda \right)+\lambda \times 2\times 2=6\lambda$$

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

$$\lambda ^{3}+4-6\lambda$$

Subtract 6\lambda from \lambda ^{3}+4.

$$\lambda ^{3}-6\lambda +4$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$\lambda \left(\lambda ^{2}-1\right)-2\left(2\lambda -1\right)-\left(\lambda -2\right)$$

Add the terms to obtain the final result.

$$\lambda ^{3}-6\lambda +4$$

Integrate w.r.t. λ

$$\frac{\lambda ^{4}}{4}-3\lambda ^{2}+4\lambda +С$$