Solve {ccc}{1}&{1}&{1}{1}&{1}&{2}{5}&{6}&{10}

$$\left( \begin{array} { c c c } { 1 } & { 1 } & { 1 } \\ { 1 } & { 1 } & { 2 } \\ { 5 } & { 6 } & { 10 } \end{array} \right)$$

Calculate Determinant

$$-1$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&1\\1&1&2\\5&6&10\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\1&1&2&1&1\\5&6&10&5&6\end{matrix}\right)$$

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

$$10+2\times 5+6=26$$

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

$$5+6\times 2+10=27$$

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

$$26-27$$

Subtract 27 from 26.

$$-1$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&1\\1&1&2\\5&6&10\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}1&2\\6&10\end{matrix}\right))-det(\left(\begin{matrix}1&2\\5&10\end{matrix}\right))+det(\left(\begin{matrix}1&1\\5&6\end{matrix}\right))$$

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

$$10-6\times 2-\left(10-5\times 2\right)+6-5$$

Simplify.

$$-2+1$$

Add the terms to obtain the final result.

$$-1$$

Evaluate

$$\left(\begin{matrix}1&1&1\\1&1&2\\5&6&10\end{matrix}\right)$$