Question

The value of ' \(\lambda\) ' so that the system of equations $$ \begin{aligned} &x+2 y+3=0 \\ &3 x+4 y+4=0 \\ &7 x-10 y+\lambda=0 \end{aligned} $$ is consistent, is (1) \(-39\) (2) 20 (3) \(-20\) (4) none of these

Short answer

The value of \(\lambda\) is \(-39\).

Step-by-step solution

Form the augmented matrix

Rewrite the given system of equations as an augmented matrix: \[ \begin{bmatrix} 1 & 2 & | & -3 \ 3 & 4 & | & -4 \ 7 & -10 & | & -\lambda \end{bmatrix} \]

Row reduce the matrix

Use row operations to transform the matrix to row echelon form:Subtract 3 times the first row from the second row: \[ R_2 = R_2 - 3R_1 \] The matrix becomes: \[ \begin{bmatrix} 1 & 2 & | & -3 \ 0 & -2 & | & 5 \ 7 & -10 & | & -\lambda \end{bmatrix} \]Next, subtract 7 times the first row from the third row: \[ R_3 = R_3 - 7R_1 \]The matrix becomes: \[ \begin{bmatrix} 1 & 2 & | & -3 \ 0 & -2 & | & 5 \ 0 & -24 & | & 21 - \lambda \end{bmatrix} \]

Determine the consistency condition

To make the system consistent, the third row should be a multiple of the second row. We want: \[ R_3 = k R_2 \] Comparing the coefficients of the second and third rows: \[ \frac{-24}{-2} = 12 \ k = 12 \]Now equate the constants from the second and third rows: \[ 12 \times 5 = 21 - \lambda \] Solve for \(\lambda\): \[ 60 = 21 - \lambda \] \[ \lambda = 21 - 60 \] \[ \lambda = -39 \]

Key concepts

Consistency of Equations

For a system of equations to be consistent, it must have at least one solution. Consistency is all about finding out if the given set of equations have any points in common.
There are typically three scenarios you might encounter:

• The system has a unique solution (consistent and independent).
• The system has infinitely many solutions (consistent and dependent).
• The system has no solution (inconsistent).

In the context of the IIT JEE Main problem mentioned, we want the system of equations to be consistent. This means that the third row of our augmented matrix should be a multiple of the second row to ensure there is some overlap in solutions.
When we compare the third row with a transformation of the second row, we need to solve for the parameter \(\backslashlambda\) that makes this condition hold true.

Row Operations

Row operations are key to solving systems of linear equations using matrices. There are three types of row operations you can perform:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting the multiple of one row to/from another row.

These operations help in transforming the matrix into a row echelon form or even into a reduced row echelon form.
In our problem, we used row operations such as subtracting multiples of one row from another to simplify the augmented matrix. This helped in comparing rows to determine the condition under which the system is consistent.
By performing these operations, we can achieve a format where we can easily identify dependencies and therefore understand the relationship between variables in the system.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It includes the coefficients of the variables and the constants from the right side of the equations.
The general form of an augmented matrix for a system of linear equations is:
\begin{bmatrix} a_1 & b_1 & | & c_1 \ a_2 & b_2 & | & c_2 \ ... & ... & | & ... \ a_n & b_n & | & c_n \end{bmatrix}
In the provided problem, the augmented matrix for the system of equations is:
\begin{bmatrix} 1 & 2 & | & -3 \ 3 & 4 & | & -4 \ 7 & -10 & | & -\lambda \end{bmatrix}
This format helps in applying row operations systematically. The goal is to simplify the augmented matrix in a way that makes it easy to determine if the system has solutions, and what those solutions might be.
For our problem, converting the equations into an augmented matrix allowed us to use row operations effectively to find the value of \(\backslashlambda\) that makes the system consistent. By analyzing the rows and ensuring that the transformations are in line with consistency requirements, we arrived at the solution for \(\backslashlambda\).