Solve {lll}{2}&{3}&{7}{1}&{7}&{5}{1}&{0}&{0}

$$\left| \begin{array} { l l l } { 2 } & { 3 } & { 7 } \\ { 1 } & { 7 } & { 5 } \\ { 1 } & { 0 } & { 0 } \end{array} \right|$$

Evaluate

$$-34$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&7\\1&7&5\\1&0&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&7&2&3\\1&7&5&1&7\\1&0&0&1&0\end{matrix}\right)$$

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

$$3\times 5=15$$

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

$$7\times 7=49$$

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

$$15-49$$

Subtract 49 from 15.

$$-34$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&7\\1&7&5\\1&0&0\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}7&5\\0&0\end{matrix}\right))-3det(\left(\begin{matrix}1&5\\1&0\end{matrix}\right))+7det(\left(\begin{matrix}1&7\\1&0\end{matrix}\right))$$

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

$$-3\left(-5\right)+7\left(-7\right)$$

Add the terms to obtain the final result.

$$-34$$

Factor

$$-34$$