Solve {ccc}{i}&{j}&{k}{2}&{-4}&{1}{3}&{-1}&{-2}=

$$\left| \begin{array} { c c c } { i } & { j } & { k } \\ { 2 } & { - 4 } & { 1 } \\ { 3 } & { - 1 } & { - 2 } \end{array} \right| =$$

Evaluate

$$7j+10k+9i$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-4i\left(-2\right)+j\times 3+k\times 2\left(-1\right)=3j-2k+8i$$

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

$$3\left(-4\right)k-i-2\times 2j=-i-12k-4j$$

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

$$3j-2k+8i-\left(-i-12k-4j\right)$$

Subtract -12k-i-4j from 8i+3j-2k.

$$7j+10k+9i$$

Steps Using Expansion by Minors

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

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

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.

$$idet(\left(\begin{matrix}-4&1\\-1&-2\end{matrix}\right))-jdet(\left(\begin{matrix}2&1\\3&-2\end{matrix}\right))+kdet(\left(\begin{matrix}2&-4\\3&-1\end{matrix}\right))$$

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

$$i\left(-4\left(-2\right)-\left(-1\right)\right)-j\left(2\left(-2\right)-3\right)+k\left(2\left(-1\right)-3\left(-4\right)\right)$$

Simplify.

$$9i-j\left(-7\right)+k\times 10$$

Add the terms to obtain the final result.

$$7j+10k+9i$$