Solve {lll}{7}&{2}&{1}{1}&{0}&{3}{2}&{4}&{3}=

$$\left| \begin{array} { l l l } { 7 } & { 2 } & { 1 } \\ { 1 } & { 0 } & { 3 } \\ { 2 } & { 4 } & { 3 } \end{array} \right| =$$

Evaluate

$$-74$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}7&2&1\\1&0&3\\2&4&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}7&2&1&7&2\\1&0&3&1&0\\2&4&3&2&4\end{matrix}\right)$$

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

$$2\times 3\times 2+4=16$$

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

$$4\times 3\times 7+3\times 2=90$$

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

$$16-90$$

Subtract 90 from 16.

$$-74$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}7&2&1\\1&0&3\\2&4&3\end{matrix}\right))$$

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.

$$7det(\left(\begin{matrix}0&3\\4&3\end{matrix}\right))-2det(\left(\begin{matrix}1&3\\2&3\end{matrix}\right))+det(\left(\begin{matrix}1&0\\2&4\end{matrix}\right))$$

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

$$7\left(-4\times 3\right)-2\left(3-2\times 3\right)+4$$

Simplify.

$$7\left(-12\right)-2\left(-3\right)+4$$

Add the terms to obtain the final result.

$$-74$$

Factor

$$-74$$