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The system of equations, x + y + z = 1, 3 x + 6 y + z = 8, αx + 2 y + 3z = 1 has a unique solution for
  • a)
    α not equal to 0
  • b)
    all integral α
  • c)
    all rational α
  • d)
    all real α
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The system of equations, x + y + z = 1, 3 x + 6 y + z = 8,αx + 2...
The given system of equations has unique solution , if 


⇒1(18−2)−1(9−α) ⇒13−5α ≠ 0 ⇒ α ≠ 13/5 + 1(6−6α) ≠ 0
Therefore , unique solution exists for all integral values of α.
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Most Upvoted Answer
The system of equations, x + y + z = 1, 3 x + 6 y + z = 8,αx + 2...
To solve this system of equations, we can use the method of substitution.

From the first equation, we have:
x + y + z = 1 ---> Equation (1)

From the second equation, we have:
3x + 6y + z = 8 ---> Equation (2)

Let's solve Equation (1) for x:
x = 1 - y - z

Now substitute this value of x in Equation (2):
3(1 - y - z) + 6y + z = 8

Expand and simplify:
3 - 3y - 3z + 6y + z = 8
3y - 2z = 5 ---> Equation (3)

Now we have two equations:
Equation (1): x + y + z = 1
Equation (3): 3y - 2z = 5

We can solve this system of equations using any method we prefer: substitution, elimination, or matrices. Let's use the substitution method:

From Equation (1), we can solve for x:
x = 1 - y - z

Substitute this value of x in Equation (3):
3y - 2z = 5

Now we have two equations:
Equation (3): 3y - 2z = 5
Equation (4): x = 1 - y - z

Now we can solve these two equations simultaneously to find the values of x, y, and z.
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The system of equations, x + y + z = 1, 3 x + 6 y + z = 8,αx + 2...
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