Solve {rrr}{3}&{2}&{1}{-2}&{1}&{-7}{3}&{-1}&{8}

$$\left( \begin{array} { r r r } { 3 } & { 2 } & { 1 } \\ { - 2 } & { 1 } & { - 7 } \\ { 3 } & { - 1 } & { 8 } \end{array} \right)$$

Calculate Determinant

$$-8$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&2&1\\-2&1&-7\\3&-1&8\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&2&1&3&2\\-2&1&-7&-2&1\\3&-1&8&3&-1\end{matrix}\right)$$

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

$$3\times 8+2\left(-7\right)\times 3-2\left(-1\right)=-16$$

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

$$3-\left(-7\times 3\right)+8\left(-2\right)\times 2=-8$$

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

$$-16-\left(-8\right)$$

Subtract -8 from -16.

$$-8$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&2&1\\-2&1&-7\\3&-1&8\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}1&-7\\-1&8\end{matrix}\right))-2det(\left(\begin{matrix}-2&-7\\3&8\end{matrix}\right))+det(\left(\begin{matrix}-2&1\\3&-1\end{matrix}\right))$$

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

$$3\left(8-\left(-\left(-7\right)\right)\right)-2\left(-2\times 8-3\left(-7\right)\right)-2\left(-1\right)-3$$

Simplify.

$$3-2\times 5-1$$

Add the terms to obtain the final result.

$$-8$$

Evaluate

$$\left(\begin{matrix}3&2&1\\-2&1&-7\\3&-1&8\end{matrix}\right)$$