Solve {ccc}{5}&{2}&{1}{35}&{26}&{1}{15}&{12}&{1}

$$\left| \begin{array} { c c c } { 5 } & { 2 } & { 1 } \\ { 35 } & { 26 } & { 1 } \\ { 15 } & { 12 } & { 1 } \end{array} \right|$$

Evaluate

$$60$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&2&1\\35&26&1\\15&12&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&2&1&5&2\\35&26&1&35&26\\15&12&1&15&12\end{matrix}\right)$$

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

$$5\times 26+2\times 15+35\times 12=580$$

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

$$15\times 26+12\times 5+35\times 2=520$$

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

$$580-520$$

Subtract 520 from 580.

$$60$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&2&1\\35&26&1\\15&12&1\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}26&1\\12&1\end{matrix}\right))-2det(\left(\begin{matrix}35&1\\15&1\end{matrix}\right))+det(\left(\begin{matrix}35&26\\15&12\end{matrix}\right))$$

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

$$5\left(26-12\right)-2\left(35-15\right)+35\times 12-15\times 26$$

Simplify.

$$5\times 14-2\times 20+30$$

Add the terms to obtain the final result.

$$60$$

Factor

$$2^{2}\times 3\times 5$$