Solve {rrr}{8}&{0}&{0}{2}&{9}&{-4}{1}&{7}&{-3}

$$\left| \begin{array} { r r r } { 8 } & { 0 } & { 0 } \\ { 2 } & { 9 } & { - 4 } \\ { 1 } & { 7 } & { - 3 } \end{array} \right|$$

Evaluate

$$8$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}8&0&0\\2&9&-4\\1&7&-3\end{matrix}\right))$$

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

$$\left(\begin{matrix}8&0&0&8&0\\2&9&-4&2&9\\1&7&-3&1&7\end{matrix}\right)$$

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

$$8\times 9\left(-3\right)=-216$$

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

$$7\left(-4\right)\times 8=-224$$

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

$$-216-\left(-224\right)$$

Subtract -224 from -216.

$$8$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}8&0&0\\2&9&-4\\1&7&-3\end{matrix}\right))$$

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.

$$8det(\left(\begin{matrix}9&-4\\7&-3\end{matrix}\right))$$

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

$$8\left(9\left(-3\right)-7\left(-4\right)\right)$$

Simplify.

$$8$$

Factor

$$2^{3}$$