a, b, c are in G.P. with common ratio r1 and α, β, γare in G.P. with common ratio r2. If the equ

A, b, c are in G.P. with common ratio r1 and α, β, γare in...

Key Concepts

Geometric Progression

A geometric progression is a sequence in which each term after the first is obtained by multiplying the previous term by a constant called the common ratio. In problems involving geometric progressions, it is common to express the terms as a, ar, ar^2, etc., where 'a' is the first term and 'r' is the common ratio. This allows one to relate the terms using a simple multiplicative factor, which is essential for setting up relationships or equations when the sequence appears in algebraic problems.

Determinant of a Matrix and System of Linear Equations

For a system of linear equations, a key criterion for the existence of only the trivial solution (all variables equal to zero) is that the determinant of the coefficient matrix must be nonzero. This ensures that the system is nonsingular and has a unique solution. In problems involving systems of equations, computing or analyzing the determinant often provides necessary insights into the nature of the solutions.

Condition on the Common Ratios for a Unique Solution

When the coefficients of the system of equations are related by geometric progressions, the properties of these sequences influence the structure of the coefficient matrix. Specifically, if the common ratios were equal to 1, the coefficients would be constant leading to potential singularities in the coefficient matrix. Ensuring that the common ratios are not equal to 1 is therefore critical for the determinant to be nonzero, which in turn guarantees that the system only has the trivial solution.

Transcript

00:01 In problem 2 we have abc in gb a b c or in gb which means geometric progression with common ratio r1 this means we can rewrite ab and c as a and a multiplied by r1 and a multiplied by r1 squared and we have also alpha beta and gamma alpha alpha and gamma and gamma and in geometric progression with common ratio r2 then we can rewrite them as alpha and alpha multiplied by r2 and alpha multiplied by r2 squared and if we know that the equations a x plus alpha y plus z equal 0 and bx plus beta y plus z equal 0 and c fx plus beta y plus z equal 0 and c f f f f f f plus gamma y plus z equal zero have only trivial solution this system is homogeneous system have only trivial solution you want to get some relation and answer problem two based on these information let's start by the only trivial solution if a homogeneous system have only trivial solution this means the of the coefficient of this system cannot be zero then a alpha 1 b beta 1 c gamma 1 this determine can be equal 0 then to get a relation between r1 and r2 or a and alpha we should expand this determine before that let's substitute by the values of b beta c and gamma from the information we have here then we can say that it's a multiplied by a alpha -so -a -1 b it's a r -1 beta it's alpha r -2 1 and here we have a r1 squared we have here alpha r2 squared 1 is not equal to 0 then by taking a from the first column and alpha from the second column you have a alpha multiplied by the determinant 1 1 1 1 1 1 1 1 1 is not equal to 0 this means a alpha can be 0 and we should exclude the answer of a let's continue then we have a alpha multiple by this determinant we can do some column operations on this determinant to make these values of zeros to easily expand the determine then we have 1 r1 squared we multiply the first column by minus 1 then add it to the second column then it's 0 r2 minus r1 r2 squared minus r1 square and we multiply the first column by minus 1 then add it to the third column it's 0 then it's 1 minus r1 then it's 1 minus r1 squared this cannot be 0 let's expand using the first row then a alpha multiply it by 1 multiplied by 1 multiplied by the determinant of this matrix this small matrix or minor matrix this multiplication is r2 minus r1 to apply it by 1 minus r1 squared minus this multiplication it's 1 minus r1 1 1 2 squared minus r1 squared this can not be equal 0 we've said that a alpha can't be equal 0 then we should get some factors here to get more conclusion we have 1 minus r squared can be factorized to be 1 minus r multiplied by 1 plus r and here can be factorized to be r2 minus r1 multiplied by r1 plus r2 and by getting the factors as a common factors then we have a common factor as r2 minus r1 and 1 minus r1 and the remaining is from the first term the remaining is one my plus r1 from the second term the remaining is r2 plus r1 but we forgot the minus here then this is minus r2 minus r1 cannot be equal to then from these four factors none of them should be 0 because if one of them is 0 this will be 0 and we don't need this to be 0 then we can conclude that a alpha is not equal to 0 r2 can't be equal to r1 r1 can't be equal to 1 and this is just 1 minus r2 then r2 can't be equals 1 from this information we've got we can choose answer c...

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