Solve {ccc}{i}&{J}&{mu}{3}&{-4}&{1}{1}&{1}&{2}

$$\left| \begin{array} { c c c } { i } & { J } & { \mu } \\ { 3 } & { - 4 } & { 1 } \\ { 1 } & { 1 } & { 2 } \end{array} \right|$$

Evaluate

$$-9i+7\mu -5J$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}i&J&\mu \\3&-4&1\\1&1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}i&J&\mu &i&J\\3&-4&1&3&-4\\1&1&2&1&1\end{matrix}\right)$$

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

$$-4i\times 2+J+\mu \times 3=J+3\mu -8i$$

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

$$-4\mu +i+2\times 3J=6J-4\mu +i$$

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

$$J+3\mu -8i-\left(6J-4\mu +i\right)$$

Subtract -4\mu +i+6J from -8i+J+3\mu .

$$-9i+7\mu -5J$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}i&J&\mu \\3&-4&1\\1&1&2\end{matrix}\right))$$

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.

$$idet(\left(\begin{matrix}-4&1\\1&2\end{matrix}\right))-Jdet(\left(\begin{matrix}3&1\\1&2\end{matrix}\right))+\mu det(\left(\begin{matrix}3&-4\\1&1\end{matrix}\right))$$

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

$$i\left(-4\times 2-1\right)-J\left(3\times 2-1\right)+\mu \left(3-\left(-4\right)\right)$$

Simplify.

$$-9i-J\times 5+\mu \times 7$$

Add the terms to obtain the final result.

$$-9i+7\mu -5J$$