The system of linear equations(4d - 1) x + y + z = 0, - y + z = 0, (4d - 1 )z = 0 has a non-trivial solution, if thevalue of d is ___ .Correct answer is '0.25'. Can you explain this answer?

Question

Answer

The system of homogeneous linear equations has a non-trivial solution, if

Answer

The system of homogeneous linear equations has a non-trivial solution, if

Answer

Given equations:
(4d - 1)x + y + z = 0 ...(1)
-y + z = 0 ...(2)
(4d - 1)z = 0 ...(3)

Non-trivial solution:
A non-trivial solution means that the variables (x, y, z) cannot all be zero. In other words, there exists at least one solution where at least one of the variables is non-zero.

Analysis of equations:
Let's analyze each equation individually to determine the conditions for a non-trivial solution.

Equation (3):
(4d - 1)z = 0

For this equation to hold true, either (4d - 1) = 0 or z = 0.

If (4d - 1) = 0, then d = 1/4.

If z = 0, then equation (2) becomes -y + 0 = 0, which implies y = 0.

So, we have two possible cases:
1) (4d - 1) = 0 (i.e., d = 1/4)
2) z = 0 and y = 0

Equation (2):
-y + z = 0

If y = 0 and z = 0, then this equation is satisfied.

Equation (1):
(4d - 1)x + y + z = 0

Substituting y = 0 and z = 0, we get (4d - 1)x = 0.

If (4d - 1) = 0, then x can take any value.

If x = 0, then (4d - 1)x = 0 regardless of the value of d.

Conclusion:
From the above analysis, we can conclude that for the given system of equations to have a non-trivial solution, the value of d must be 1/4.

This means that when d = 1/4, the system of equations has a solution where at least one of the variables (x, y, z) is non-zero.

Answer

The system of homogeneous linear equations has a non-trivial solution, if

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