Solve {ccc}{2}&{-1}&{1}{1}&{2}&{-3}{3}&{lambda}&{5}

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

Evaluate

$$7\left(\lambda +4\right)$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 2\times 5-\left(-3\times 3\right)+\lambda =\lambda +29$$

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

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

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

$$\lambda +29-\left(1-6\lambda \right)$$

Subtract 1-6\lambda from 29+\lambda .

$$7\lambda +28$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(3\lambda +10\right)-\left(-14\right)+\lambda -6$$

Add the terms to obtain the final result.

$$7\lambda +28$$

Integrate w.r.t. λ

$$\frac{7\lambda ^{2}}{2}+28\lambda +С$$