Question

The value of the determinant $\left|\begin{array}{c}1+a\\ 1\\ 1\end{array}\begin{array}{c}1\\ 1+a\\ 1\end{array}\begin{array}{c}1\\ 1\\ 1+a\end{array}\right|$ is

• A. $a^3(1-\frac{2}{a})$
• B. $a^3(1+\frac{3}{a})$
• C. $a^3(1-\frac{3}{a})$
• D. $a^3(1+\frac{2}{a})$

Answer 1

Answer: (B)

$a^3(1+\frac{3}{a})$

Let $\Delta =\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+a& 1\\ 1& 1& 1+a\end{array}\right|$

Expand the above determinant as shown below:

$\Delta =(1+a)[(1+a)(1+a)-\left(1\right)\left(1\right)]-1[\left(1\right)(1+a)-\left(1\right)\left(1\right)]+1[\left(1\right)\left(1\right)-\left(1\right)(1+a)]$

$\Delta =(1+a)[{(1+a)}^2-1]-1[1+a-1]+1[1-1-a]$

$\Delta =(1+a)[1+a^2+2a-1]-1\left[a\right]+1[-a]$

$\Delta =(1+a)[a^2+2a]-a-a$

$\Delta =a^2+2a+a^3+2a^2-2a$

$\Delta =a^3+3a^2$

$\Delta =a^3(1+\frac{3}{a})$

Hence, option $B$ is correct.