Solve {lll}{2}&{4}&{1}{2}&{4}&{1}{2}&{4}&{1}

$$\left| \begin{array} { l l l } { 2 } & { 4 } & { 1 } \\ { 2 } & { 4 } & { 1 } \\ { 2 } & { 4 } & { 1 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 4+4\times 2+2\times 4=24$$

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

$$24-24$$

Subtract 24 from 24.

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Steps Using Expansion by Minors

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

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

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.

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

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

$$2\left(4-4\right)-4\left(2-2\right)+2\times 4-2\times 4$$

Simplify.

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Factor

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