Solve {ccc}{3}&{5}&{7}{4}&{11}&{pi}{0}&{5}&{12}

$$\left( \begin{array} { c c c } { 3 } & { 5 } & { 7 } \\ { 4 } & { 11 } & { \pi } \\ { 0 } & { 5 } & { 12 } \end{array} \right)$$

Calculate Determinant

$$296-15\pi$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&5&7\\4&11&\pi \\0&5&12\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&5&7&3&5\\4&11&\pi &4&11\\0&5&12&0&5\end{matrix}\right)$$

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

$$3\times 11\times 12+7\times 4\times 5=536$$

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

$$5\pi \times 3+12\times 4\times 5=15\pi +240$$

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

$$536-\left(15\pi +240\right)$$

Subtract 15\pi +240 from 536.

$$296-15\pi$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&5&7\\4&11&\pi \\0&5&12\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}11&\pi \\5&12\end{matrix}\right))-5det(\left(\begin{matrix}4&\pi \\0&12\end{matrix}\right))+7det(\left(\begin{matrix}4&11\\0&5\end{matrix}\right))$$

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

$$3\left(11\times 12-5\pi \right)-5\times 4\times 12+7\times 4\times 5$$

Simplify.

$$3\left(132-5\pi \right)-5\times 48+7\times 20$$

Add the terms to obtain the final result.

$$296-15\pi$$

Evaluate

$$\left(\begin{matrix}3&5&7\\4&11&\pi \\0&5&12\end{matrix}\right)$$