Solve {ccc}{0}&{1}&{5}{3.5}&{0}&{1}{12}&{13}&{14}

$$\left| \begin{array} { c c c } { 0 } & { 1 } & { 5 } \\ { 3.5 } & { 0 } & { 1 } \\ { 12 } & { 13 } & { 14 } \end{array} \right|$$

Evaluate

$$190.5$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0&1&5\\3.5&0&1\\12&13&14\end{matrix}\right))$$

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

$$\left(\begin{matrix}0&1&5&0&1\\3.5&0&1&3.5&0\\12&13&14&12&13\end{matrix}\right)$$

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

$$12+5\times 3.5\times 13=239.5$$

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

$$14\times 3.5=49$$

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

$$239.5-49$$

Subtract 49 from 239.5.

$$190.5$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0&1&5\\3.5&0&1\\12&13&14\end{matrix}\right))$$

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.

$$-det(\left(\begin{matrix}3.5&1\\12&14\end{matrix}\right))+5det(\left(\begin{matrix}3.5&0\\12&13\end{matrix}\right))$$

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

$$-\left(3.5\times 14-12\right)+5\times 3.5\times 13$$

Simplify.

$$-37+5\times 45.5$$

Add the terms to obtain the final result.

$$190.5$$

Factor

$$\frac{3 \cdot 127}{2} = 190\frac{1}{2} = 190.5$$