Solve {rrr}{1}&{-2}&{0}{4}&{-2}&{-1}{-3}&{1}&{2}

$$\left( \begin{array} { r r r } { 1 } & { - 2 } & { 0 } \\ { 4 } & { - 2 } & { - 1 } \\ { - 3 } & { 1 } & { 2 } \end{array} \right)$$

Calculate Determinant

$$7$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&-2&0\\4&-2&-1\\-3&1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&-2&0&1&-2\\4&-2&-1&4&-2\\-3&1&2&-3&1\end{matrix}\right)$$

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

$$-2\times 2-2\left(-1\right)\left(-3\right)=-10$$

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

$$-1+2\times 4\left(-2\right)=-17$$

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

$$-10-\left(-17\right)$$

Subtract -17 from -10.

$$7$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&-2&0\\4&-2&-1\\-3&1&2\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-2&-1\\1&2\end{matrix}\right))-\left(-2det(\left(\begin{matrix}4&-1\\-3&2\end{matrix}\right))\right)$$

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

$$-2\times 2-\left(-1\right)-\left(-2\left(4\times 2-\left(-3\left(-1\right)\right)\right)\right)$$

Simplify.

$$-3-\left(-2\times 5\right)$$

Add the terms to obtain the final result.

$$7$$

Evaluate

$$\left(\begin{matrix}1&-2&0\\4&-2&-1\\-3&1&2\end{matrix}\right)$$