Solve {ccc}{1}&{0}&{1}{-7}&{2}&{1}{2}&{-1}&{-6}

$$\left( \begin{array} { c c c } { 1 } & { 0 } & { 1 } \\ { - 7 } & { 2 } & { 1 } \\ { 2 } & { - 1 } & { - 6 } \end{array} \right)$$

Calculate Determinant

$$-8$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&0&1\\-7&2&1\\2&-1&-6\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&0&1&1&0\\-7&2&1&-7&2\\2&-1&-6&2&-1\end{matrix}\right)$$

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

$$2\left(-6\right)-7\left(-1\right)=-5$$

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

$$2\times 2-1=3$$

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

$$-5-3$$

Subtract 3 from -5.

$$-8$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&0&1\\-7&2&1\\2&-1&-6\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}2&1\\-1&-6\end{matrix}\right))+det(\left(\begin{matrix}-7&2\\2&-1\end{matrix}\right))$$

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

$$2\left(-6\right)-\left(-1\right)+-7\left(-1\right)-2\times 2$$

Simplify.

$$-11+3$$

Add the terms to obtain the final result.

$$-8$$

Evaluate

$$\left(\begin{matrix}1&0&1\\-7&2&1\\2&-1&-6\end{matrix}\right)$$