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

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

Evaluate

$$4j+3k-i$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-2k+3i-j=3i-2k-j$$

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

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

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

$$4j+3k-i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$4j+3k-i$$