Question

If $\omega$ is a root of $x^4=1$, show that $a+b\omega +c{\omega }^2+d{\omega }^3$ is a factor of $\left|\begin{array}{cccc}a& b& c& d\\ b& c& d& a\\ c& d& a& b\\ d& a& b& c\end{array}\right|$ Hence show that the determinant is equal to $-(a+b+c+d)(a-b+c-d)\left\{\right(a-c{)}^2+(b-d{)}^2\}$

• A. $-(a+b+c+d)(a-b+c-d)\left[\right(a-c{)}^2+(b-d{)}^2]$
• B. $(a+b+c+d)(a-b+c-d)\left[\right(a-c{)}^2+(b-d{)}^2]$
• C. $(a+b+c+d)(a+b-c-d)\left[\right(a-b{)}^2+(c-d{)}^2]$
• D. $-(a+b+c+d)(a+b-c-d)\left[\right(a-b{)}^2+(c-d{)}^2]$

Answer 1

The correct option is A $-(a+b+c+d)(a-b+c-d)\left[\right(a-c{)}^2+(b-d{)}^2]$

Roots of $x^4=1$ are $1,\omega ,{\omega }^2,{\omega }^3$ and ${\omega }^4=1;1+\omega +{\omega }^2+{\omega }^3=0$

Let $\Delta =\left|\begin{array}{cccc}a& b& c& d\\ b& c& d& a\\ c& d& a& b\\ d& a& b& c\end{array}\right|$
$\Rightarrow \Delta =\frac{1}{{\omega }^6}\left|\begin{array}{cccc}a& b\omega & c{\omega }^2& d{\omega }^3\\ b& c\omega & d{\omega }^2& a{\omega }^3\\ c& d\omega & a{\omega }^2& b{\omega }^3\\ d& a\omega & b{\omega }^2& c{\omega }^3\end{array}\right|$

Multiplying $R_2,R_3,R_4$ by $\omega ,{\omega }^2,{\omega }^3$ respectively

$\Rightarrow \Delta =\frac{1}{{\omega }^{12}}\left|\begin{array}{cccc}a& b\omega & c{\omega }^2& d{\omega }^2\\ b\omega & c{\omega }^2& d{\omega }^3& a\\ c{\omega }^2& d{\omega }^3& a& b\omega \\ d{\omega }^3& a& b\omega & c{\omega }^2\end{array}\right|.....\{\because {\omega }^4=1\}$

Applying $C_1\to C_1+C_2+C_3+C_4$

$\Rightarrow \Delta =\frac{1}{({\omega }^4{)}^3}\left|\begin{array}{cccc}a+b\omega +c{\omega }^2+d{\omega }^3& b\omega & c{\omega }^2& d{\omega }^3\\ a+b\omega +c{\omega }^2+d{\omega }^3& c{\omega }^2& d{\omega }^3& a\\ a+b\omega +c{\omega }^2+d{\omega }^2& d{\omega }^3& a& b\omega \\ a+b\omega +c{\omega }^2+d{\omega }^3& a& b\omega & c{\omega }^2\end{array}\right|$

$\Rightarrow$ Taking $a+b\omega +c{\omega }^2+d{\omega }^3$ from $C_1$

$\Rightarrow \Delta =(a+b\omega +c{\omega }^2+d{\omega }^3)\left|\begin{array}{cccc}1& b\omega & c{\omega }^2& d{\omega }^3\\ 1& c{\omega }^2& d{\omega }^3& a\\ 1& d{\omega }^3& a& b\omega \\ 1& a& b\omega & c{\omega }^2\end{array}\right|$
Hence $(a+b\omega +c{\omega }^2+d{\omega }^3)$ is a factor of $\Delta$.

Again

$\Delta =(a+bi-c-di)\left|\begin{array}{cccc}1& bi& -c& -di\\ 1& -c& -di& a\\ 1& -di& a& bi\\ 1& a& bi& -c\end{array}\right|$

Multiplying $R_1$ and $R_3$ by $-1$, we get
$\Delta =\left[\right(a-c)+i(b-d\left)\right]\left|\begin{array}{cccc}-1& -bi& c& di\\ 1& -c& -di& a\\ -1& di& -a& -bi\\ 1& a& bi& -c\end{array}\right|$
$R_1\to R_1+R_2+R_3+R_4$
$=\left[\right(a-c)+i(b-d\left)\right]\left|\begin{array}{cccc}0& P& -P& P\\ 1& -c& -di& a\\ -1& di& -a& -bi\\ 1& a& bi& -c\end{array}\right|$
where $P=(a-c)-i(b-d)$
$\Delta =\overline{P}P\left|\begin{array}{cccc}0& 1& -1& 1\\ 1& -c& -di& a\\ -1& di& -a& -bi\\ 1& a& bi& -c\end{array}\right|$
Applying $C_2\to C_2-C_4;C_3\to C_3+C_4$
$=|P{|}^2\left|\begin{array}{cccc}0& 0& 0& 1\\ 1& -c-a& a-di& a\\ -1& di+bi& -a-bi& bi\\ 1& a+c& bi-c& -c\end{array}\right|$

Expanding w.r.t $R_1$
$=-|P{|}^2\left|\begin{array}{ccc}1& -c-a& a-di\\ -1& (b+d)i& -a-bi\\ 1& a+c& bi-c\end{array}\right|$

Applying $R_2\to R_2+R_1;R_3\to R_3-R_1$
$=-|P{|}^2\left|\begin{array}{ccc}1& -c-a& a-di\\ 0& (b+d)i-(c+a)& -i(b+d)\\ 0& 2(a+c)& (b+d)i-(c+a)\end{array}\right|$

Expanding w.r.t $R_1$

$=-|P{|}^2\left|\begin{array}{cc}(b+d)i-(c+a)& -i(b+d)\\ 2(a+c)& (b+d)i-(c+a)\end{array}\right|$

$=-|P{|}^2\left[\right\{(b+d)i-(c+a){\}}^2+2i(a+c)(b+d)]$

$=-|P^2|[-(b+d{)}^2+(c+a{)}^2]$

$=-(a+b+c+d)(a-b+c-d)\left[\right(a-c{)}^2+(b-d{)}^2]$