Solve {ccc}{4}&{20}&{3}{16}&{80}&{6}{28}&{140}&{9}

$$\left| \begin{array} { c c c } { 4 } & { 20 } & { 3 } \\ { 16 } & { 80 } & { 6 } \\ { 28 } & { 140 } & { 9 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&20&3\\16&80&6\\28&140&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&20&3&4&20\\16&80&6&16&80\\28&140&9&28&140\end{matrix}\right)$$

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

$$4\times 80\times 9+20\times 6\times 28+3\times 16\times 140=12960$$

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

$$28\times 80\times 3+140\times 6\times 4+9\times 16\times 20=12960$$

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

$$12960-12960$$

Subtract 12960 from 12960.

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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).

$$det(\left(\begin{matrix}4&20&3\\16&80&6\\28&140&9\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}80&6\\140&9\end{matrix}\right))-20det(\left(\begin{matrix}16&6\\28&9\end{matrix}\right))+3det(\left(\begin{matrix}16&80\\28&140\end{matrix}\right))$$

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

$$4\left(80\times 9-140\times 6\right)-20\left(16\times 9-28\times 6\right)+3\left(16\times 140-28\times 80\right)$$

Simplify.

$$4\left(-120\right)-20\left(-24\right)$$

Add the terms to obtain the final result.

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Factor

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