Solve {lll}{a}&{1}&{1}{1}&{a}&{1}{1}&{1}&{9}

$$\left| \begin{array} { l l l } { a } & { 1 } & { 1 } \\ { 1 } & { a } & { 1 } \\ { 1 } & { 1 } & { 9 } \end{array} \right|$$

Evaluate

$$\left(a-1\right)\left(9a+7\right)$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}a&1&1\\1&a&1\\1&1&9\end{matrix}\right))$$

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

$$\left(\begin{matrix}a&1&1&a&1\\1&a&1&1&a\\1&1&9&1&1\end{matrix}\right)$$

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

$$aa\times 9+1+1=9a^{2}+2$$

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

$$a+a+9=2a+9$$

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

$$9a^{2}+2-\left(2a+9\right)$$

Subtract 2a+9 from 9a^{2}+2.

$$\left(a-1\right)\left(9a+7\right)$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}a&1&1\\1&a&1\\1&1&9\end{matrix}\right))$$

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.

$$adet(\left(\begin{matrix}a&1\\1&9\end{matrix}\right))-det(\left(\begin{matrix}1&1\\1&9\end{matrix}\right))+det(\left(\begin{matrix}1&a\\1&1\end{matrix}\right))$$

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

$$a\left(a\times 9-1\right)-\left(9-1\right)+1-a$$

Simplify.

$$a\left(9a-1\right)-8+1-a$$

Add the terms to obtain the final result.

$$\left(a-1\right)\left(9a+7\right)$$

Integrate w.r.t. a

$$С+3a^{3}-a^{2}-7a$$