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

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

Evaluate

$$-3-2i+\left(3-i\right)k-7j$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$ik+2i+3\times 3j=9j+ik+2i$$

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

$$2j+3k-3-\left(9j+ik+2i\right)$$

Subtract ik+2i+9j from -3+2j+3k.

$$-3-2i+\left(3-i\right)k-7j$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$i\left(-2+3i\right)-j\times 7+k\left(3-i\right)$$

Add the terms to obtain the final result.

$$-3-2i+\left(3-i\right)k-7j$$