Solve {ccc}{0}&{2}&{0}{z}&{3i}&{i}{-i}&{0}&{1+i}=

$$\left| \begin{array} { c c c } { 0 } & { 2 } & { 0 } \\ { z } & { 3 i } & { i } \\ { - i } & { 0 } & { 1 + i } \end{array} \right| =$$

Evaluate

$$\left(-2-2i\right)z+2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0&2&0\\z&3i&i\\-i&0&1+i\end{matrix}\right))$$

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

$$\left(\begin{matrix}0&2&0&0&2\\z&3i&i&z&3i\\-i&0&1+i&-i&0\end{matrix}\right)$$

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

$$2i\left(-i\right)=2$$

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

$$\left(1+i\right)z\times 2=\left(2+2i\right)z$$

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

$$2-\left(2+2i\right)z$$

Subtract \left(2+2i\right)z from 2.

$$\left(-2-2i\right)z+2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0&2&0\\z&3i&i\\-i&0&1+i\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}z&i\\-i&1+i\end{matrix}\right))$$

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

$$-2\left(z\left(1+i\right)-\left(-ii\right)\right)$$

Simplify.

$$-2\left(\left(1+i\right)z-1\right)$$

Add the terms to obtain the final result.

$$\left(-2-2i\right)z+2$$