Solve {rrr}{1}&{7}&{28}{0}&{-5}&{34}{-1}&{-7}&{3}

$$\left| \begin{array} { r r r } { 1 } & { 7 } & { 28 } \\ { 0 } & { - 5 } & { 34 } \\ { - 1 } & { - 7 } & { 3 } \end{array} \right|$$

Evaluate

$$-155$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&7&28\\0&-5&34\\-1&-7&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&7&28&1&7\\0&-5&34&0&-5\\-1&-7&3&-1&-7\end{matrix}\right)$$

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

$$-5\times 3+7\times 34\left(-1\right)=-253$$

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

$$-\left(-5\right)\times 28-7\times 34=-98$$

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

$$-253-\left(-98\right)$$

Subtract -98 from -253.

$$-155$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&7&28\\0&-5&34\\-1&-7&3\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-5&34\\-7&3\end{matrix}\right))-7det(\left(\begin{matrix}0&34\\-1&3\end{matrix}\right))+28det(\left(\begin{matrix}0&-5\\-1&-7\end{matrix}\right))$$

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

$$-5\times 3-\left(-7\times 34\right)-7\left(-\left(-34\right)\right)+28\left(-\left(-\left(-5\right)\right)\right)$$

Simplify.

$$223-7\times 34+28\left(-5\right)$$

Add the terms to obtain the final result.

$$-155$$

Factor

$$-155$$