Solve {ccc}{35}&{26}&{1}{30}&{22}&{1}{20}&{14}&{1}

$$\left| \begin{array} { c c c } { 35 } & { 26 } & { 1 } \\ { 30 } & { 22 } & { 1 } \\ { 20 } & { 14 } & { 1 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}35&26&1\\30&22&1\\20&14&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}35&26&1&35&26\\30&22&1&30&22\\20&14&1&20&14\end{matrix}\right)$$

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

$$35\times 22+26\times 20+30\times 14=1710$$

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

$$20\times 22+14\times 35+30\times 26=1710$$

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

$$1710-1710$$

Subtract 1710 from 1710.

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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).

$$det(\left(\begin{matrix}35&26&1\\30&22&1\\20&14&1\end{matrix}\right))$$

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.

$$35det(\left(\begin{matrix}22&1\\14&1\end{matrix}\right))-26det(\left(\begin{matrix}30&1\\20&1\end{matrix}\right))+det(\left(\begin{matrix}30&22\\20&14\end{matrix}\right))$$

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

$$35\left(22-14\right)-26\left(30-20\right)+30\times 14-20\times 22$$

Simplify.

$$35\times 8-26\times 10-20$$

Add the terms to obtain the final result.

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Factor

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