Solve {ccc}{1}&{-1}&{3}{-1}&{2}&{lambda}{0}&{5}&{0}

$$\left( \begin{array} { c c c } { 1 } & { - 1 } & { 3 } \\ { - 1 } & { 2 } & { \lambda } \\ { 0 } & { 5 } & { 0 } \end{array} \right)$$

Calculate Determinant

$$-5\lambda -15$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&-1&3\\-1&2&\lambda \\0&5&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&-1&3&1&-1\\-1&2&\lambda &-1&2\\0&5&0&0&5\end{matrix}\right)$$

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

$$3\left(-1\right)\times 5=-15$$

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

$$5\lambda =5\lambda$$

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

$$-15-5\lambda$$

Subtract 5\lambda from -15.

$$-5\lambda -15$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&-1&3\\-1&2&\lambda \\0&5&0\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}2&\lambda \\5&0\end{matrix}\right))-\left(-det(\left(\begin{matrix}-1&\lambda \\0&0\end{matrix}\right))\right)+3det(\left(\begin{matrix}-1&2\\0&5\end{matrix}\right))$$

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-5\lambda -15$$

Evaluate

$$\left(\begin{matrix}1&-1&3\\-1&2&\lambda \\0&5&0\end{matrix}\right)$$