Solve {ccc}{1}&{0}&{-3}{0}&{2}&{3}{0}&{7}&{0}

$$\left| \begin{array} { c c c } { 1 } & { 0 } & { - 3 } \\ { 0 } & { 2 } & { 3 } \\ { 0 } & { 7 } & { 0 } \end{array} \right|$$

Evaluate

$$-21$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&0&-3\\0&2&3\\0&7&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&0&-3&1&0\\0&2&3&0&2\\0&7&0&0&7\end{matrix}\right)$$

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

$$\text{true}$$

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

$$7\times 3=21$$

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

$$-21$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&0&-3\\0&2&3\\0&7&0\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}2&3\\7&0\end{matrix}\right))-3det(\left(\begin{matrix}0&2\\0&7\end{matrix}\right))$$

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

$$-7\times 3$$

Simplify.

$$-21$$

Factor

$$-21$$