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The simultaneous equations
2x + ay + z = 20 
x + 3y + 4z = b
x + 2y + 3z = c
has unique solution then what is the value of a, b and c respectively?
  • a)
    a ≠ ± 1, b and c can be any value
  • b)
    a ≠ ± 1, b ≠ 0, c ≠ 1
  • c)
    a ≠ -1, b and c can be any value
  • d)
    a, b and c can be any value
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The simultaneous equations2x + ay + z = 20x + 3y + 4z = bx + 2y + 3z =...
For given system of equations to have unique solution, the determinant of the coefficient matrix should be not equal to 0

 
2 (9−8) − a(3−4) + (2−3) ≠ 0
2 + a − 1 ≠ 0
a ≠ −1
 
Since Δ of coefficient matrix is not equal to 0 then b and c can take any value
Note:
If coefficient matrix’s Δ ≠ 0 then rank of coefficient matrix is always equal rank of augmented matrix so right-hand side of given equation can take any value
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Most Upvoted Answer
The simultaneous equations2x + ay + z = 20x + 3y + 4z = bx + 2y + 3z =...
The given system of equations is:

2x + ay + z = 20 (equation 1)
x + 3y + 4z = b (equation 2)
x + 2y + 3z = c (equation 3)

To determine the value of a, b, and c, we need to use the fact that the system has a unique solution.

One way to solve the system is by using the method of elimination. We can start by eliminating x from equations 2 and 3:

Multiply equation 2 by 2 and equation 3 by -1:

2(x + 3y + 4z) = 2b
-1(x + 2y + 3z) = -c

Simplifying:

2x + 6y + 8z = 2b (equation 4)
-x - 2y - 3z = -c (equation 5)

Now, add equation 1 and equation 5:

2x + ay + z + (-x - 2y - 3z) = 20 + (-c)

Simplifying:

x + (ay - 2y) + (z - 3z) = 20 - c

Combining like terms:

x + (ay - 2y) - 2z = 20 - c (equation 6)

Now, we have the following system of equations:

2x + 6y + 8z = 2b (equation 4)
x + (ay - 2y) - 2z = 20 - c (equation 6)

To have a unique solution, the determinant of the coefficient matrix must be non-zero.

The determinant of the coefficient matrix is given by:

|2 6 8|
|1 a-2 -2|
|0 1 -3|

Expanding along the first row, we have:

2[(a-2)(-3) - (-2)(1)] - 6[(-2)(-3) - 1(0)] + 8[1(1) - (a-2)(0)]

Simplifying:

6a - 12 + 12 - 6 + 8 = 6a - 6

For a unique solution, the determinant must be non-zero:

6a - 6 ≠ 0

Solving for a:

6a ≠ 6
a ≠ 1

Therefore, for the system to have a unique solution, a cannot be 1.

So, the value of a is not determined.

The values of b and c are also not determined from the given information.
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The simultaneous equations2x + ay + z = 20x + 3y + 4z = bx + 2y + 3z = chas unique solution then what is the value of a, b and c respectively?a)a ≠ ± 1, b and c can be any valueb)a ≠ ± 1, b ≠ 0, c ≠ 1c)a ≠ -1, b and c can be any valued)a, b and c can be any valueCorrect answer is option 'C'. Can you explain this answer?
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