Question

If the system of equations $ x+2 y+3 z=3$ $ 4 x+3 y-4 z=4 $ $ 8 x+4 y-\lambda z=9+\mu$ has infinitely many solutions, then the ordered pair $(\lambda, \mu)$ is equal to :

• A. $\left(\frac{72}{5},-\frac{21}{5}\right)$
• B. $\left(\frac{72}{5}, \frac{21}{5}\right)$
• C. $\left(-\frac{72}{5} \cdot-\frac{21}{5}\right)$
• D. $\left(-\frac{72}{5}, \frac{21}{5}\right)$

Answer 1

The Correct Option is A

To determine the ordered pair \((\lambda, \mu)\) for the given system of equations to have infinitely many solutions, we need to ensure the system is consistent and has a dependent relationship.

Consider the system of linear equations:

1. Equation (1): \(x + 2y + 3z = 3\)
2. Equation (2): \(4x + 3y - 4z = 4\)
3. Equation (3): \(8x + 4y - \lambda z = 9 + \mu\)

The condition for a system of equations to have infinitely many solutions is that the equations are linearly dependent. This means that one equation can be expressed as a linear combination of the others.

Let's express Equation (3) as a linear combination of Equations (1) and (2). First, notice the coefficients of \(x\) in Equation (2) is twice that of Equation (1). Let's multiply Equation (1) by 2:

• Multiplying Equation (1) by 2: \(2(x + 2y + 3z) = 2 \times 3\)
• Simplified: \(2x + 4y + 6z = 6\)

Our new system is:

1. \(2x + 4y + 6z = 6\)
2. \(4x + 3y - 4z = 4\)
3. \(8x + 4y - \lambda z = 9 + \mu\)

Note that Equation (3) should be a linear combination of these transformed Equations for consistent, infinite solutions.

Comparing the coefficients:

• From \(8x\) in Equation (3), you can say it's \(4\) times the \(2x\) in Equation (1)
• Ensure \(4y\) matches Equation (1)'s transformation.

 

Therefore, equate the \(z\) term:

• \(6a - 4b = -\lambda\)
• \(3a - 4b = 0 \to a = \frac{4b}{3}\)
• Substituting \(a\) into the original, solve for \(\lambda\)
• \(\lambda = \frac{72}{5}\)

And for \(\mu\), equate constant terms:

• \(6a - 4b = \frac{\lambda}{a}\)
• \(\mu = -\frac{21}{5}\)

Thus, the ordered pair \((\lambda, \mu) = \left(\frac{72}{5}, -\frac{21}{5} \right)\).