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

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

Evaluate

$$j+7k+5i$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-i\left(-1\right)+j\left(-2\right)+k\times 3\times 2=i+6k-2j$$

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

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

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

$$i+6k-2j-\left(-4i-k-3j\right)$$

Subtract -k-4i-3j from i-2j+6k.

$$j+7k+5i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$5i-j\left(-1\right)+k\times 7$$

Add the terms to obtain the final result.

$$j+7k+5i$$