Question

Consider the system of linear equations\[ \begin{cases}x + y + t = 4, \\2x - 4t = 7, \\x + y + z = 5, \\x - 3y - z - 10t = \lambda,\end{cases}\]where \( x, y, z, t \) are variables and \( \lambda \) is a constant. Then which one of the following is true?

• A. If \( \lambda = 1 \), then the system has a unique solution.
• B. If \( \lambda = 2 \), then the system has infinitely many solutions.
• C. If \( \lambda = 1 \), then the system has infinitely many solutions.
• D. If \( \lambda = 2 \), then the system has a unique solution.

Answer 1

The Correct Option is C

We need to analyze the system of linear equations given below and determine for which value of \( \lambda \) the system has infinitely many solutions:

\(\begin{cases} x + y + t = 4, \\ 2x - 4t = 7, \\ x + y + z = 5, \\ x - 3y - z - 10t = \lambda, \end{cases}\)

To begin, let's rewrite these equations for clarity:

1. The first equation is: \(x + y + t = 4\).
2. The second equation is: \(2x - 4t = 7\).
3. The third equation is: \(x + y + z = 5\).
4. The fourth equation is: \(x - 3y - z - 10t = \lambda\).

We observe that equations (3) and (4) both contain the variable \( z \), so let's start by eliminating it.

Subtract equation (3) from equation (4):

\((x - 3y - z - 10t) - (x + y + z) = \lambda - 5\)

This results in:

\(-4y - 10t - 5 = \lambda - 5\)

We can simplify this equation as:

\(-4y - 10t = \lambda - 5 + 5 \Rightarrow -4y - 10t = \lambda\)

Continuing further, we observe the system of equations:

1. \(x + y + t = 4\)
2. \(2x - 4t = 7\)
3. \(-4y - 10t = \lambda\)

The crucial part here is to determine if there exist solutions that satisfy all the equations simultaneously. Solving for \( x \) and \( t \) via the second equation:

From (2), express \( x \) as:

\(x = \frac{7}{2} + 2t\)

Substitute this \( x \) into equation (1):

\(\left( \frac{7}{2} + 2t \right) + y + t = 4\)

Simplifying gives us:

\(y + 3t = \frac{1}{2}\)

We also have the simplified equation (3): \(-4y - 10t = \lambda\).

Substitute \( y = \frac{1}{2} - 3t \) into \( -4y - 10t = \lambda \):

\(-4\left(\frac{1}{2} - 3t\right) - 10t = \lambda\)

This simplifies to:

\(-2 + 12t - 10t = \lambda \Rightarrow 2t - 2 = \lambda\)

Rearranging gives us:

\(t = \frac{\lambda + 2}{2}\)

Substitute this \( t \) back to find \( y \):

\(y = \frac{1}{2} - 3\left(\frac{\lambda + 2}{2}\right)\) gives \(y = \frac{1}{2} - \frac{3\lambda}{2} - 3\) which simplifies to \(y = -\frac{3\lambda}{2} - \frac{5}{2}\).

When you try solving for \( x, y, z, \) and \( t \), we see that there are solutions available contingent to \( \lambda \). When \(\lambda = 1\), we realize that this allows for dependencies that result in infinitely many solutions by varying \( t \).

Thus, considering all these calculations and dependencies involved, the option stating 'If \( \lambda = 1 \), then the system has infinitely many solutions.' is indeed accurate.