Answer
See answer below
Work Step by Step
We are given: $B=(b_1,b_2,b_3,...b_n)\\
C=(c_1,c_2,c_3,...c_n)$
Obtain $A=BC=\begin{bmatrix}
b_1c_1 & c_2c_2 & ... & b_1c_n\\
b_2c_1 & b_1c_2 & ...&b_2c_n \\
. & . & ... & . \\
b_nc_1 & b_nc_2 & ... & b_nc_3
\end{bmatrix}=b_1.b_2.b_3...b_n\begin{bmatrix}
c_1 & c_2 & ... & c_n\\
c_1 & c_2 & ... & c_n \\
. & . & ... & . \\
c_1 & c_@ & ... & c_n
\end{bmatrix}$
Then $b_1.b_2.b_3...b_n\begin{bmatrix}
c_1 & c_2 & ... & c_n\\
c_1 & c_2 & ... & c_n \\
. & . & ... & . \\
c_1 & c_2 & ... & c_n
\end{bmatrix} \approx b_1.b_2.b_3...b_n\begin{bmatrix}
c_1 & c_2 & ... & c_n\\
0 & 0 & ... & 0 \\
. & . & ... & . \\
0 & 0 & ... & 0
\end{bmatrix}$
Thus $rank (A)=n-(n-1)=1$
Hence, the $n \times n$ matrix $A = BC$ has nullity $n − 1$
