Question

If the system of linear equations \( 2x + y - z = 7 \), \( x - 3y + 2z = 1 \), \( x + 4y + \delta z = k \) (where \(\delta, k \in \mathbb{R}\)) has infinitely many solutions, then \(\delta + k\) is equal to:

• A. -3
• B. 3
• C. 6
• D. 9

Hint:

Before calculating bulky determinants, check if one equation is simply a sum or difference of the others. It can save you minutes of calculation!

Answer 1

The Correct Option is B

To determine when the given system of equations has infinitely many solutions, we need to check the conditions under which the system of equations is consistent and dependent. The given system of linear equations is: \(2x + y - z = 7\), \(x - 3y + 2z = 1\), \(x + 4y + \delta z = k\).

For the system to have infinitely many solutions, the three equations must be linearly dependent, which means one equation should be a linear combination of the others. Let's write the system in matrix form:

The augmented matrix is:

Coefficient Matrix			Constants
2	1	-1	7
1	-3	2	1
1	4	\(\delta\)	k

For the third equation \(x + 4y + \delta z = k\) to be a linear combination of the first two equations, we compare coefficients to find constants such that:

- \(a(2x + y - z) + b(x - 3y + 2z) = x + 4y + \delta z\)

On solving, we have:

• \(2a + b = 1\) (Coefficient of \(x\))
• \(a - 3b = 4\) (Coefficient of \(y\))
• \(-a + 2b = \delta\) (Coefficient of \(z\))
• \(7a + b = k\) (Constant terms)

Solving the first two equations:

• From \(2a + b = 1 \implies b = 1 - 2a\).
• Substituting in \(a - 3b = 4\):
• \(a - 3(1 - 2a) = 4 \implies a - 3 + 6a = 4 \implies 7a = 7 \implies a = 1\).
• Therefore, \(b = 1 - 2(1) = -1\).
• Now calculate \(\delta\): using \(-a + 2b = \delta\), we get:
• \(-1 + 2(-1) = \delta \implies \delta = -3\).
• Finally, calculate \(k\): using \(7a + b = k\), we get:
• \(7(1) + (-1) = k \implies k = 6\).

Thus, the values are: \(\delta = -3\) and \(k = 6\).

Hence, \(\delta + k = -3 + 6 = 3\), which is the answer.