Problem 282 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Using matrix method, find the values of \(\lambda\) and \(\mu\) so that the system of equation \(2 x-3 y+5 z=12\) \(3 x+y+\lambda z=\mu\) \(x-7 y+8 z=17\) has i. Unique solution \\{Ans. \(\lambda \neq 2\\}\) ii. Infinite solutions \\{Ans. \(\lambda=2, \mu=7\\}\) iii. No solution \\{Ans. \(\lambda=2, \mu \neq 7\\}\)

Step-by-Step Solution

Answer

For a unique solution, \(\lambda \neq 17\). For infinite solutions, \(\lambda = 17, \mu = -19\), and for no solutions, \(\lambda = 17, \mu \neq -19\).

Step 1: Formulate the Augmented Matrix

The system of equations can be written in the form of an augmented matrix as follows: \[\begin{bmatrix} 2 & -3 & 5 & 12 \ 3 & 1 & \lambda & \mu \ 1 & -7 & 8 & 17 \end{bmatrix}\].

Step 2: Formulate the Coefficients Matrix

Then, form the matrix A of coefficients: \[\begin{bmatrix} 2 & -3 & 5 \ 3 & 1 & \lambda \ 1 & -7 & 8 \end{bmatrix}\].

Step 3: Find the determinant of the Coefficients matrix

Next, find the determinant of the matrix A. The determinant \(\Delta\) of a 3x3 matrix \[\begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}\] is \(a(ei−fh)−b(di−fg)+c(dh−ge)\). So here, the determinant \(\Delta\) of our matrix A is \(\Delta = 2*(1*8-(-7)*\lambda) - (-3)*(3*8-1*\lambda)+5*(3*(-7) - 1*1) = 16\lambda + 56\lambda - 33 - 35 = 72\lambda -68 = 4\lambda - 4*(17)\).

Step 4: Condition for Unique Solution

For the system to have a unique solution, determinant \(\Delta\) must not be equal to zero. That means, \(4\lambda - 4*(17) \neq 0\), which simplifies to \(\lambda \neq 17\).

Step 5: Condition for Infinite Solutions

For the system to have infinite solutions, determinant \(\Delta\) must be equal to zero (\(4\lambda - 4*(17) = 0\)), and the relationships between constants after replacing z in first two equations must hold. It means, \(\lambda = 17\) and \(\mu = 12- 5*(17) = -95*.2 = -19\).

Step 6: Condition for No Solution

For the system to have no solution, determinant \(\Delta\) must be equal to zero (\(4\lambda - 4*(17) = 0\)), but the relationships between constants after replacing z in first two equations must not hold. This gives \(\lambda = 17\) and \(\mu \neq -19\).

Key Concepts

Determinant of a matrix

When working with a system of equations, the matrix method involves the use of the determinant of the matrix to determine the nature of the solutions. The determinant provides us with critical information about the matrix. In general terms, the determinant of a 3x3 matrix is calculated using the formula: \[\text{det}(A) = a(ei−fh)−b(di−fg)+c(dh−ge)\].

For a matrix to be invertible, which is necessary to have a unique solution to the system, its determinant must be non-zero. In our specific case, when calculating the determinant of the coefficients matrix, it's important to ensure that you perform the arithmetic correctly and double-check your work to avoid errors that could lead to a misinterpretation of the system's solution.

Conditions for unique solution

Achieving a single, unique solution for a system of equations hinges on the determinant of its coefficients matrix not equating to zero. This is because a non-zero determinant indicates that our matrix is invertible, allowing us to find exact values for the unknown variables. In the provided exercise, we look for the condition under which \(\text{det}(A) eq 0\), explicitly finding the values that \(\text{lambda}\) must not take to avoid a determinant of zero.

To clarify this condition, simplifying the determinant into the form \(\text{lambda} eq k\) where \(\text{k}\) is a specific value or condition, can be of great help to students. It's beneficial to highlight that these specific restrictions are crucial for ensuring the coefficients matrix has the power to dictate a unique solution.

Conditions for infinite solutions

For a system of equations to have infinite solutions, a couple of conditions have to be met. Firstly, the determinant of the coefficients matrix must be zero, symbolizing that the matrix isn't invertible and that there isn't a unique transformation from variable space to solution space. Following the determinant being zero, the actual equations must also align perfectly, implying that the system is consistent and that all equations are essentially multiples of each other.

In the context of our exercise, the infinite solutions occur at the point where \(\text{lambda}\) and \(\text{mu}\) are set in a way that keeps the determinant at zero and maintains the equations' consistency. Explicitly illustrating the calculations and the logic behind setting certain values can greatly aid students in conceptualizing why infinite solutions exist for these specific parameter configurations.

Conditions for no solution

A system of linear equations can have no solution when it's inconsistent, which can happen if the determinant of the coefficients matrix is zero—alike to the infinite solutions scenario—but the equations do not align. This misalignment means that there is at least one equation that contradicts others, leading to a situation where no single solution satisfies all the equations simultaneously.

In the provided exercise, to identify the conditions for no solution, we look for cases where not only does the determinant equal zero but also the constants after substituting the variables do not adhere to the expected proportional relationships. Explaining this with concrete examples, such as contrasting a no solution case against an infinite solutions case with the same determinant of zero, can significantly deepen student understanding of this concept.