Solve {rrr}{8}&{2}&{4}{12}&{3}&{-5}{16}&{4}&{3}

$$\left| \begin{array} { r r r } { 8 } & { 2 } & { 4 } \\ { 12 } & { 3 } & { - 5 } \\ { 16 } & { 4 } & { 3 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}8&2&4\\12&3&-5\\16&4&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}8&2&4&8&2\\12&3&-5&12&3\\16&4&3&16&4\end{matrix}\right)$$

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

$$8\times 3\times 3+2\left(-5\right)\times 16+4\times 12\times 4=104$$

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

$$16\times 3\times 4+4\left(-5\right)\times 8+3\times 12\times 2=104$$

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

$$104-104$$

Subtract 104 from 104.

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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).

$$det(\left(\begin{matrix}8&2&4\\12&3&-5\\16&4&3\end{matrix}\right))$$

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.

$$8det(\left(\begin{matrix}3&-5\\4&3\end{matrix}\right))-2det(\left(\begin{matrix}12&-5\\16&3\end{matrix}\right))+4det(\left(\begin{matrix}12&3\\16&4\end{matrix}\right))$$

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

$$8\left(3\times 3-4\left(-5\right)\right)-2\left(12\times 3-16\left(-5\right)\right)+4\left(12\times 4-16\times 3\right)$$

Simplify.

$$8\times 29-2\times 116$$

Add the terms to obtain the final result.

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Factor

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