Solve {rrr}{3}&{1}&{0}{1}&{-1}&{2}{1}&{1}&{1}

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

Calculate Determinant

$$-8$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\left(-1\right)+2=-1$$

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

$$2\times 3+1=7$$

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

$$-1-7$$

Subtract 7 from -1.

$$-8$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$3\left(-1-2\right)-\left(1-2\right)$$

Simplify.

$$3\left(-3\right)-\left(-1\right)$$

Add the terms to obtain the final result.

$$-8$$

Evaluate

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