Solve {ccc}{i}&{j}&{K}{1}&{-1}&{1}{3}&{2}&{-12}

$$\left| \begin{array} { c c c } { i } & { j } & { K } \\ { 1 } & { - 1 } & { 1 } \\ { 3 } & { 2 } & { - 12 } \end{array} \right|$$

Evaluate

$$5\left(K+3j+2i\right)$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&j&K\\1&-1&1\\3&2&-12\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&j&K&i&j\\1&-1&1&1&-1\\3&2&-12&3&2\end{matrix}\right)$$

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

$$-i\left(-12\right)+j\times 3+K\times 2=2K+3j+12i$$

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

$$3\left(-1\right)K+2i-12j=2i-12j-3K$$

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

$$2K+3j+12i-\left(2i-12j-3K\right)$$

Subtract -3K+2i-12j from 12i+3j+2K.

$$5K+15j+10i$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&j&K\\1&-1&1\\3&2&-12\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}-1&1\\2&-12\end{matrix}\right))-jdet(\left(\begin{matrix}1&1\\3&-12\end{matrix}\right))+Kdet(\left(\begin{matrix}1&-1\\3&2\end{matrix}\right))$$

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

$$i\left(-\left(-12\right)-2\right)-j\left(-12-3\right)+K\left(2-3\left(-1\right)\right)$$

Simplify.

$$10i-j\left(-15\right)+K\times 5$$

Add the terms to obtain the final result.

$$5K+15j+10i$$