Solve {lll}{10}&{0}{0}&{1}&{2}{1}&{10}&{1}

$$\left( \begin{array} { l l l } { 10 } & { 0 } \\ { 0 } & { 1 } & { 2 } \\ { 1 } & { 10 } & { 1 } \end{array} \right)$$

Calculate Determinant

$$-190$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}10&0&0\\0&1&2\\1&10&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}10&0&0&10&0\\0&1&2&0&1\\1&10&1&1&10\end{matrix}\right)$$

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

$$10=10$$

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

$$10\times 2\times 10=200$$

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

$$10-200$$

Subtract 200 from 10.

$$-190$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}10&0&0\\0&1&2\\1&10&1\end{matrix}\right))$$

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.

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

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

$$10\left(1-10\times 2\right)$$

Simplify.

$$10\left(-19\right)$$

Add the terms to obtain the final result.

$$-190$$

Evaluate

$$\left(\begin{matrix}10&0&0\\0&1&2\\1&10&1\end{matrix}\right)$$