Question

$\left|\begin{array}{ccc}a+b+nc& (n-1)a& (n-1)b\\ (n-1)c& b+c+na& (n-1)b\\ (n-1)c& (n-1)a& (c+a+nb)\end{array}\right|$ is equal to

• A. $n(a+b+c{)}^2$
• B. $n(a+b+c{)}^3$
• C. $n^3(a+b+c)$
• D. $n^2(a+b+c{)}^2$

Answer 1

Answer: (B)

$n(a+b+c{)}^3$

$\left|\begin{array}{ccc}a+b+nc& (n-1)a& (n-1)b\\ (n-1)c& b+c+na& (n-1)b\\ (n-1)c& (n-1)a& (c+a+nb)\end{array}\right|$
$C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}n(a+b+c)& (n-1)a& (n-1)b\\ n(a+b+c)& b+c+na& (n-1)b\\ n(a+b+c)& (n-1)a& (c+a+nb)\end{array}\right|$
$=n(a+b+c)\left|\begin{array}{ccc}1& (n-1)a& (n-1)b\\ 1& b+c+na& (n-1)b\\ 1& (n-1)a& (c+a+nb)\end{array}\right|$
$R_2\to R_2-R_1,R_3\to R_3-R_1$
$=n(a+b+c)\left|\begin{array}{ccc}1& (n-1)a& (n-1)b\\ 0& b+c+a& 0\\ 0& 0& c+a+b\end{array}\right|$
$=n{(a+b+c)}^3$