Problem 110
Question
If the system of linear equations \(x+2 a y+a z=0\) \(x+3 b y+b z=0\) \(x+4 c y+c z=0\) has a non-zero solution, then \(a, b, c\) (A) are in A. P. (B) are in G.P. (C) are in H.P. (D) satisfy \(a+2 b+3 c=0\)
Step-by-Step Solution
Verified Answer
(A) are in Arithmetic Progression (A.P.).
1Step 1: Write down the system of equations
The system of linear equations is given as:\[\begin{align*} x + 2ay + az &= 0 \ x + 3by + bz &= 0 \ x + 4cy + cz &= 0 \end{align*}\] We need to find the condition on \(a, b, c\) under which this system has a non-zero solution.
2Step 2: Setup the coefficient matrix
The coefficient matrix from the given system of equations is: \[ \begin{pmatrix} 1 & 2a & a \ 1 & 3b & b \ 1 & 4c & c \end{pmatrix} \] For the system to have a non-zero solution, the determinant of this matrix must be zero.
3Step 3: Compute the determinant of the matrix
Calculate the determinant of the coefficient matrix:\[ \text{det} \begin{pmatrix} 1 & 2a & a \ 1 & 3b & b \ 1 & 4c & c \end{pmatrix} = 1 \cdot (3bc - 4cb) - 2a \cdot (b - c) + a \cdot (b - c) = 0 \] Simplify the determinant to:\[ c(3b - a) - 2ab + ac = 0 \] Further simplify it to find conditions on \(a, b, c\).
4Step 4: Simplify the determinant equation
The simplified determinant condition is:\[ c(b - a) + ab - 2ac = 0 \] Rearrange to identify the condition: \[ ab + c(b-a) - 2ac = 0 \] Factor and simplify further: \[ a(b - 2c) = -c(b - a) \] If \(a, b, c\) are not zero, this simplies to checking a recognizable form.
5Step 5: Recognize the pattern
If \(c = b - a\), this would mean that adding step differences forms a series. An Arithmetic Progression (A.P) means equal difference in sequential terms. Let's calculate and check the form to validate A.P.:\[ c = b - a \Rightarrow a, b, c \text{ are in Arithmetic Progression if } a - b + c = 0 \]
6Step 6: Conclusion: Confirm the pattern
From the simplification of our determinant condition and recognizing the arithmetic relation, we see that, indeed, \(a, b, c\) meet the requirements to be in Arithmetic Progression.
Key Concepts
DeterminantNon-zero SolutionArithmetic Progression
Determinant
In a system of linear equations, the determinant of the coefficient matrix plays a crucial role in determining the type of solutions the system will have.
For a system to have a non-zero solution, its determinant must be zero. This is because a zero determinant implies that the matrix is singular, which leads to an infinite number of solutions for the linear system, including non-zero solutions.
In our given system of equations:\[\begin{pmatrix} 1 & 2a & a \ 1 & 3b & b \ 1 & 4c & c \end{pmatrix}\]setting up the determinant involves cross-multiplying and finding the sum and difference of those products.
The process involves calculating:\[1 \cdot (3bc - 4cb) - 2a \cdot (b - c) + a \cdot (b - c)\]Successfully simplifying this to zero gives us the necessary conditions on the constants involved, enabling the system to have a non-zero solution.
For a system to have a non-zero solution, its determinant must be zero. This is because a zero determinant implies that the matrix is singular, which leads to an infinite number of solutions for the linear system, including non-zero solutions.
In our given system of equations:\[\begin{pmatrix} 1 & 2a & a \ 1 & 3b & b \ 1 & 4c & c \end{pmatrix}\]setting up the determinant involves cross-multiplying and finding the sum and difference of those products.
The process involves calculating:\[1 \cdot (3bc - 4cb) - 2a \cdot (b - c) + a \cdot (b - c)\]Successfully simplifying this to zero gives us the necessary conditions on the constants involved, enabling the system to have a non-zero solution.
Non-zero Solution
A non-zero solution in the context of a linear system refers to a scenario where not all variables are equal to zero.
It's significant because it demonstrates that the variables can take non-zero values to satisfy all equations simultaneously.
If our system's determinant is zero, it means that instead of a unique solution where all variable values are zero, we may have infinitely many solutions with at least one non-zero variable.
This condition sets our system apart from having no solutions or a singular trivial solution, extending the possibility of finding varied solutions within the set of linear equations.
Therefore, calculating and assigning value to each variable in its non-zero state remains essential in exploring solution spaces in linear systems.
It's significant because it demonstrates that the variables can take non-zero values to satisfy all equations simultaneously.
If our system's determinant is zero, it means that instead of a unique solution where all variable values are zero, we may have infinitely many solutions with at least one non-zero variable.
This condition sets our system apart from having no solutions or a singular trivial solution, extending the possibility of finding varied solutions within the set of linear equations.
Therefore, calculating and assigning value to each variable in its non-zero state remains essential in exploring solution spaces in linear systems.
Arithmetic Progression
Arithmetic progression (A.P) is a sequence of numbers in which each term after the first is obtained by adding a constant difference to the preceding term.
Applied to our linear equation conditions, we need to prove that constants \(a, b, c\) forming the coefficients must be in A.P.In A.P., if \(c = b - a\), it implies that the differences between consecutive terms are equal, confirming the sequence as an arithmetic one.
Thus, the equation simplifies to:\[a - b + c = 0\]This results in evenly spaced values along the sequence, showing the characteristic of an arithmetic progression.
The identification of this pattern in the determinants and coefficients not only confirms the solution conditions but also satisfies the requirement for the desired property of the coefficients within the equations given.
Applied to our linear equation conditions, we need to prove that constants \(a, b, c\) forming the coefficients must be in A.P.In A.P., if \(c = b - a\), it implies that the differences between consecutive terms are equal, confirming the sequence as an arithmetic one.
Thus, the equation simplifies to:\[a - b + c = 0\]This results in evenly spaced values along the sequence, showing the characteristic of an arithmetic progression.
The identification of this pattern in the determinants and coefficients not only confirms the solution conditions but also satisfies the requirement for the desired property of the coefficients within the equations given.
Other exercises in this chapter
Problem 108
If \(\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|=x+i y\), then (A) \(x=3, y=1\) (B) \(x=1, y=3\) (C) \(x=0, y=3\) (D)
View solution Problem 109
If \((\omega \neq 1)\) is a cubic root of unity, then \(\left|\begin{array}{ccc}1 & 1+i+\omega^{2} & \omega^{2} \\ 1-i & -1 & \omega^{2}-1 \\ -i & -1+\omega-i &
View solution Problem 111
If \(1, \omega, \omega^{2}\) are the cube roots of unity, then \(\left|\begin{array}{lll}1 & \omega^{n} & \omega^{2 \pi} \\ \omega^{n} & \omega^{2 n} & 1 \\ \om
View solution Problem 112
If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\) are in G.P., then the value of the determinant \(\quad\) \(\left|\begin{array}{lll}\log a_{n} & \log a_{n+1} &
View solution