Solve {lll}{2}&{0}&{0}{0}&{2}&{0}{3}&{0}&{4}

$$\left| \begin{array} { l l l } { 2 } & { 0 } & { 0 } \\ { 0 } & { 2 } & { 0 } \\ { 3 } & { 0 } & { 4 } \end{array} \right|$$

Evaluate

$$16$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&0&0\\0&2&0\\3&0&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&0&0&2&0\\0&2&0&0&2\\3&0&4&3&0\end{matrix}\right)$$

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

$$2\times 2\times 4=16$$

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

$$\text{true}$$

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

$$16$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&0&0\\0&2&0\\3&0&4\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}2&0\\0&4\end{matrix}\right))$$

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

$$2\times 2\times 4$$

Simplify.

$$2\times 8$$

Add the terms to obtain the final result.

$$16$$

Factor

$$2^{4}$$