Solveeit Logo
Login

Question

Question: If the system of equations \[x{\text{ }} = {\text{ }}cy{\text{ }} + {\text{ }}bz,{\text{ }}y{\text{ ...

If the system of equations x = cy + bz, y = az + cxx{\text{ }} = {\text{ }}cy{\text{ }} + {\text{ }}bz,{\text{ }}y{\text{ }} = {\text{ }}az{\text{ }} + {\text{ }}cx and z = bx + ayz{\text{ }} = {\text{ }}bx{\text{ }} + {\text{ }}ay has a non-zero solution and at least one of a, b, ca,{\text{ }}b,{\text{ }}c is a proper fraction, a2+b2+c2{a^2} + {b^2} + {c^2}is

A. >2 > 2

B. >3 > 3

C. <3 < 3

D. <2 < 2

Explanation

Solution

Hint- In this question, provided that system is a homogeneous equation in 33 variables, its determinant will be zero because it has non-zero value which implies infinite solutions, hence the determinant of the matrix coefficient is zero after we get the correct answer by simplifying the equation.

Complete step-by-step answer:
Provided the equations of the system

x = cy + bz y = az + cx z=bx+ay  x{\text{ }} = {\text{ }}cy{\text{ }} + {\text{ }}bz \\\ y{\text{ }} = {\text{ }}az{\text{ }} + {\text{ }}cx \\\ z = bx + ay \\\

Since we know that if the system of homogeneous equations in 33 variables has non-zero value or infinite solution then matrix determinant is zero
Now we are going to write above 33 equation in the form of a determinant matrix and make it equal to zero, so that we get
\left| {\begin{array}{*{20}{c}} 1&{ - c}&{ - b} \\\ { - c}&1&{ - a} \\\ { - b}&{ - a}&1 \end{array}} \right| = 0

Now we're going to solve it with the help of first raw so we're getting

⇒1(1−a2)+c(−c−ab)−b(ac+b)=0 ⇒1−2abc−a2−b2−c2=0  \Rightarrow 1(1 - {a^2}) + c( - c - ab) - b(ac + b) = 0 \\\ \Rightarrow 1 - 2abc - {a^2} - {b^2} - {c^2} = 0 \\\

Further simplifying above equation we get
a2+b2+c2+2abc=1{a^2} + {b^2} + {c^2} + 2abc = 1
By adding b2c2{b^2}{c^2}in both the sides of equation

a2+b2c2+2ab=1−b2−c2+b2c2=(1−b2)(1−c2) ⇒(a+bc)2=(1−b2)(1−c2)  {a^2} + {b^2}{c^2} + 2ab = 1 - {b^2} - {c^2} + {b^2}{c^2} = (1 - {b^2})(1 - {c^2}) \\\ \Rightarrow {(a + bc)^2} = (1 - {b^2})(1 - {c^2}) \\\

Similarly,

(b+ac)2=(1−a2)(1−b2) (c+ab)2=(1−a2)(1−b2)  {(b + ac)^2} = (1 - {a^2})(1 - {b^2}) \\\ {(c + ab)^2} = (1 - {a^2})(1 - {b^2}) \\\

Hence, (1 − a2), (1 − b2)\left( {1{\text{ }} - {\text{ }}{a^2}} \right),{\text{ }}\left( {1{\text{ }} - {\text{ }}{b^2}} \right)and (1 − c2)\left( {1{\text{ }} - {\text{ }}{c^2}} \right)all have same sign. Since at least one of them is a proper fraction, it implies all of them are positive.
So 1 − a2 > 0, 1 − b2 > 0, 1 − c2 > 01{\text{ }} - {\text{ }}{a^2}{\text{ }} > {\text{ }}0,{\text{ }}1{\text{ }} - {\text{ }}{b^2}{\text{ }} > {\text{ }}0,{\text{ }}1{\text{ }} - {\text{ }}{c^2}{\text{ }} > {\text{ }}0
By adding above 33 equation we get
\Rightarrow {a^2} + {b^2} + {c^2} < 3$$$$ \Rightarrow {a^2} + {b^2} + {c^2} < 3
Hence the correct answer is option C.

Note-When matrix has non zero solution this means that the lines are either equivalent (the number of solutions is endless) or parallel (there are no solutions). The slopes must be different if the determinant is non-zero, and the lines must intersect in exactly one point.