Solve {rrr}{-3}&{-5}&{3}{0}&{-1}&{0}{7}&{7}&{2}

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

Calculate Determinant

$$27$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-3\left(-1\right)\times 2=6$$

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

$$7\left(-1\right)\times 3=-21$$

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

$$6-\left(-21\right)$$

Subtract -21 from 6.

$$27$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-3\left(-1\right)\times 2+3\left(-7\left(-1\right)\right)$$

Simplify.

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

Add the terms to obtain the final result.

$$27$$

Evaluate

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