Solve {lll}{1}&{5}&{1}{2}&{3}&{0}{4}&{2}&{1}

$$\left( \begin{array} { l l l } { 1 } & { 5 } & { 1 } \\ { 2 } & { 3 } & { 0 } \\ { 4 } & { 2 } & { 1 } \end{array} \right)$$

Calculate Determinant

$$-15$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3+2\times 2=7$$

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

$$4\times 3+2\times 5=22$$

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

$$7-22$$

Subtract 22 from 7.

$$-15$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$3-5\times 2+2\times 2-4\times 3$$

Simplify.

$$3-5\times 2-8$$

Add the terms to obtain the final result.

$$-15$$

Evaluate

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