Solve {lr}{12}&{-1}{0}&{2}&{3}{4}&{1}&{5}

$$\left| \begin{array} { l r } { 12 } & { - 1 } \\ { 0 } & { 2 } & { 3 } \\ { 4 } & { 1 } & { 5 } \end{array} \right|$$

Evaluate

$$72$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}12&-1&0\\0&2&3\\4&1&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}12&-1&0&12&-1\\0&2&3&0&2\\4&1&5&4&1\end{matrix}\right)$$

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

$$12\times 2\times 5-3\times 4=108$$

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

$$3\times 12=36$$

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

$$108-36$$

Subtract 36 from 108.

$$72$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}12&-1&0\\0&2&3\\4&1&5\end{matrix}\right))$$

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.

$$12det(\left(\begin{matrix}2&3\\1&5\end{matrix}\right))-\left(-det(\left(\begin{matrix}0&3\\4&5\end{matrix}\right))\right)$$

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

$$12\left(2\times 5-3\right)-\left(-\left(-4\times 3\right)\right)$$

Simplify.

$$12\times 7-\left(-\left(-12\right)\right)$$

Add the terms to obtain the final result.

$$72$$

Factor

$$2^{3}\times 3^{2}$$