edited by
856 views

2 Answers

Best answer
10 votes
10 votes
Rearranging the system of equations, we get: x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0. 
Put in Determinant and expand.

\begin{vmatrix} 1 &-c &-b \\ c&-1 &a \\ b& a &-1 \end{vmatrix}=0 So a^2+b^2+c^2+2abc=1 Ans= B
selected by
Answer:

Related questions

3 votes
3 votes
2 answers
1
Ruturaj Mohanty asked Dec 27, 2018
732 views
If $a/b=c/d$, then which of the following does not hold good?$(a+b)/b=(c+d)/d$$(a+c)/(b+d)=(a-c)/(b-d)$$(a+b)/(a-b)=(c+d)/(c-d)$$(a+c)/(b-d)=(a-c)(b+d)$
4 votes
4 votes
1 answer
3
Ruturaj Mohanty asked Dec 27, 2018
4,607 views
$N$ is the smallest number that has $5$ factors. How many factors does $N-1$ have?$4$$6$$5$$3$