Question

If a, b, c, d denote the determinants of matrices A, B, C, D where
A= $\left[\begin{array}{ccc}0& x-y& x-z\\ y-x& 0& y-z\\ z-x& z-y& 0\end{array}\right]$
B= $\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 4\\ 4& 9& 16\end{array}\right]$
C=$\left[\begin{array}{ccc}0& 0& 2\\ 0& 5& 0\\ 7& 0& 0\end{array}\right]$
D=$\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 8\\ 1& 3& 27\end{array}\right]$
Then the ascending order of a, b, c, d

• A. a b c d
• B. c a d b
• C. c a b d
• D. a c d b

Answer 1

The correct option is C c a b d

A$=\underset{=}{0}$ as A is skew symmetric
c=$\underset{=}{-14}$ (clearly)
for $B{C}_3\to c_3-c_2,c_2\to c_2{c}_1\left|\begin{array}{ccc}1& 0& 0\\ 2& 1& 1\\ 4& 5& 7\end{array}\right|$ expand along $c_1=2$
same for D $\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 6\\ 1& 2& 24\end{array}\right|$ expand along $\underset{=}{12}$
CABD is the order