Solve {lll}{3}&{1}&{4}{2}&{5}&{7}{8}&{3}&{1}

$$\left( \begin{array} { l l l } { 3 } & { 1 } & { 4 } \\ { 2 } & { 5 } & { 7 } \\ { 8 } & { 3 } & { 1 } \end{array} \right)$$

Calculate Determinant

$$-130$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&1&4\\2&5&7\\8&3&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&1&4&3&1\\2&5&7&2&5\\8&3&1&8&3\end{matrix}\right)$$

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

$$3\times 5+7\times 8+4\times 2\times 3=95$$

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

$$8\times 5\times 4+3\times 7\times 3+2=225$$

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

$$95-225$$

Subtract 225 from 95.

$$-130$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&1&4\\2&5&7\\8&3&1\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}5&7\\3&1\end{matrix}\right))-det(\left(\begin{matrix}2&7\\8&1\end{matrix}\right))+4det(\left(\begin{matrix}2&5\\8&3\end{matrix}\right))$$

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

$$3\left(5-3\times 7\right)-\left(2-8\times 7\right)+4\left(2\times 3-8\times 5\right)$$

Simplify.

$$3\left(-16\right)-\left(-54\right)+4\left(-34\right)$$

Add the terms to obtain the final result.

$$-130$$

Evaluate

$$\left(\begin{matrix}3&1&4\\2&5&7\\8&3&1\end{matrix}\right)$$