Solve {lll}{1}&{3}&{7}{2}&{8}&{26}{3}&{15}&{63}

$$\left| \begin{array} { l l l } { 1 } & { 3 } & { 7 } \\ { 2 } & { 8 } & { 26 } \\ { 3 } & { 15 } & { 63 } \end{array} \right|$$

Evaluate

$$12$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&7\\2&8&26\\3&15&63\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&7&1&3\\2&8&26&2&8\\3&15&63&3&15\end{matrix}\right)$$

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

$$8\times 63+3\times 26\times 3+7\times 2\times 15=948$$

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

$$3\times 8\times 7+15\times 26+63\times 2\times 3=936$$

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

$$948-936$$

Subtract 936 from 948.

$$12$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&7\\2&8&26\\3&15&63\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}8&26\\15&63\end{matrix}\right))-3det(\left(\begin{matrix}2&26\\3&63\end{matrix}\right))+7det(\left(\begin{matrix}2&8\\3&15\end{matrix}\right))$$

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

$$8\times 63-15\times 26-3\left(2\times 63-3\times 26\right)+7\left(2\times 15-3\times 8\right)$$

Simplify.

$$114-3\times 48+7\times 6$$

Add the terms to obtain the final result.

$$12$$

Factor

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