Solve {ccc}{-2}&{1}&{-1}{-2}&{1}&{-1}{-2}&{1}&{5}

$$\left| \begin{array} { c c c } { - 2 } & { 1 } & { - 1 } \\ { - 2 } & { 1 } & { - 1 } \\ { - 2 } & { 1 } & { 5 } \end{array} \right|$$

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&1&-1\\-2&1&-1\\-2&1&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&1&-1&-2&1\\-2&1&-1&-2&1\\-2&1&5&-2&1\end{matrix}\right)$$

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

$$-2\times 5-\left(-2\right)-\left(-2\right)=-6$$

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

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

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

$$-6-\left(-6\right)$$

Subtract -6 from -6.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&1&-1\\-2&1&-1\\-2&1&5\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}1&-1\\1&5\end{matrix}\right))-det(\left(\begin{matrix}-2&-1\\-2&5\end{matrix}\right))-det(\left(\begin{matrix}-2&1\\-2&1\end{matrix}\right))$$

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

$$-2\left(5-\left(-1\right)\right)-\left(-2\times 5-\left(-2\left(-1\right)\right)\right)-\left(-2-\left(-2\right)\right)$$

Simplify.

$$-2\times 6-\left(-12\right)$$

Add the terms to obtain the final result.

$$0$$

Factor

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