3 votes 3 votes If $x=cy+bz, \: y=az+cx, \: z=bx+ay,$ where $x,y,z$ are not all zero, then $a^2+b^2+c^2=$ $1+2abc$ $1-2abc$ $1+abc$ $abc-1$ Set Theory & Algebra go-mockgate-1 set-theory&algebra algebra quantitative-aptitude + – Ruturaj Mohanty asked Dec 27, 2018 edited Sep 10, 2020 by ajaysoni1924 Ruturaj Mohanty 856 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
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. So a^2+b^2+c^2+2abc=1 Ans= B Sukhbir Singh answered Dec 29, 2018 selected Jan 5, 2019 by Ruturaj Mohanty Sukhbir Singh comment Share Follow See all 4 Comments See all 4 4 Comments reply gmrishikumar commented Jan 30, 2019 reply Follow Share Why do we take the determinant value and why is it equal to zero? 0 votes 0 votes Dipanshu Rana commented Nov 23, 2019 reply Follow Share Why determinant taken equal to 0 @Sukhbir Singh @Ruturaj Mohanty 0 votes 0 votes `JEET commented Nov 23, 2019 reply Follow Share Because you are finding the solution of the determinant in terms of the given values. Think of this as compared with the quadratic equation. How do you find the roots? Simply by equating it to $\mathbf 0$, right? So, the same is true here. 1 votes 1 votes pritishc commented Feb 7, 2020 reply Follow Share Write the equations in matrix form and use the theorem - "A square matrix M from the equation Mx = 0 is invertible if and only if this homoegenous system of equations has no non-trivial solution". In this case, we want a non-trivial solution (as indicated by "x, y and z are not all zero"), so we set the determinant to zero. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes . MANSI_SOMANI answered Sep 13, 2022 MANSI_SOMANI comment Share Follow See all 0 reply Please log in or register to add a comment.