Question

Let two matrices $A=\left[\begin{array}{ccc}1& 2\omega & {\omega }^2\\ 3& -1& 5\\ 2& i& \omega \end{array}\right]$ and $B=\left[\begin{array}{cc}2& 3\\ 1& i\\ 2& \omega \end{array}\right],$ where $i^2=-1$ and $\omega$ is a cube root of unity. Then which of the following statement(s) is(are) correct ?

• A. $\sum _{r=1}^3{a}_{1r}\cdot b_{r1}=2$
• B. $\sum _{r=1}^3{a}_{1r}\cdot b_{r1}=0$
• C. $\sum _{r=1}^3{b}_{r2}\cdot a_{3r}=4+\omega$
• D. $\sum _{r=1}^3{b}_{r2}\cdot a_{3r}=5+{\omega }^2$

Answer 1

Answer: (B), (D)

$\sum _{r=1}^3{b}_{r2}\cdot a_{3r}=5+{\omega }^2$

Using property of matrix multiplication:
$\sum _{r=1}^3{a}_{1r}\cdot b_{r1}=(AB{)}_{11}$
$\Rightarrow (AB{)}_{11}=1\cdot 2+2\omega \cdot 1+{\omega }^2\cdot 2=2(1+\omega +{\omega }^2)=0$
$(\because 1+\omega +{\omega }^2=0,$ as $\omega$ is cube root of unity)
Similarly, $\sum _{r=1}^3{b}_{r2}\cdot a_{3r}=(AB{)}_{32}$
$\Rightarrow A{B}_{32}=2\cdot 3+i^2+\omega \cdot \omega$
$\Rightarrow A{B}_{32}=5+{\omega }^2=4-\omega$