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

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

Evaluate

$$2ks-3js-j-k+5i$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$k+2\left(-i\right)+3sj=3js+k-2i$$

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

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

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

$$2ks-3js-j-k+5i$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$5i-j\left(3s+1\right)+k\left(2s-1\right)$$

Add the terms to obtain the final result.

$$-j\left(3s+1\right)+k\left(2s-1\right)+5i$$