Solve {c|cc}{-1}&{-2}&{-1}{0}&{-4}&{-2}{0}&{0}&{4}

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

Calculate Determinant

$$16$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-\left(-4\right)\times 4=16$$

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

$$\text{true}$$

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

$$16$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-\left(-4\right)\times 4$$

Simplify.

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

Add the terms to obtain the final result.

$$16$$

Evaluate

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