Solve {lll}{i}&{j}&{k}{1}&{3}&{2}{2}&{-1}&{-10}

$$\left| \begin{array} { l l l } { i } & { j } & { k } \\ { 1 } & { 3 } & { 2 } \\ { 2 } & { - 1 } & { - 10 } \end{array} \right|$$

Evaluate

$$14j-7k-28i$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&j&k\\1&3&2\\2&-1&-10\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&j&k&i&j\\1&3&2&1&3\\2&-1&-10&2&-1\end{matrix}\right)$$

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

$$3i\left(-10\right)+j\times 2\times 2+k\left(-1\right)=4j-k-30i$$

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

$$2\times 3k-2i-10j=-2i+6k-10j$$

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

$$4j-k-30i-\left(-2i+6k-10j\right)$$

Subtract 6k-2i-10j from -30i+4j-k.

$$14j-7k-28i$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&j&k\\1&3&2\\2&-1&-10\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}3&2\\-1&-10\end{matrix}\right))-jdet(\left(\begin{matrix}1&2\\2&-10\end{matrix}\right))+kdet(\left(\begin{matrix}1&3\\2&-1\end{matrix}\right))$$

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

$$i\left(3\left(-10\right)-\left(-2\right)\right)-j\left(-10-2\times 2\right)+k\left(-1-2\times 3\right)$$

Simplify.

$$-28i-j\left(-14\right)+k\left(-7\right)$$

Add the terms to obtain the final result.

$$14j-7k-28i$$