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Consider the system of equations x + y + z = 1, 2x + 3y + 2z = 1, 2x + 3y + (a2 - 1)z = a + 1 then
- a)System has a unique solution for |a| = √3
- b)System is inconsistence for |a| = √3
- c)System is inconsistence for a = 4
- d)System is inconsistence for a = 3
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Consider the system of equations x + y + z = 1, 2x + 3y + 2z = 1, 2x +...
Given system of linear equations:
x + y + z = 1 ….(1)
2x + 3y + 2z = 1 ….(2)
2x + 3y + (a2 - 1)z = a + 1 …..(3)
Consider a2 - 1 = 2
then LHS of (2) and (3) are same but RHS are not.
Hence a2 = 3 => |a| = √3
For |a| = √3, system is inconsistence.
So option (b) is correct.
x + y + z = 1 ….(1)
2x + 3y + 2z = 1 ….(2)
2x + 3y + (a2 - 1)z = a + 1 …..(3)
Consider a2 - 1 = 2
then LHS of (2) and (3) are same but RHS are not.
Hence a2 = 3 => |a| = √3
For |a| = √3, system is inconsistence.
So option (b) is correct.
Most Upvoted Answer
Consider the system of equations x + y + z = 1, 2x + 3y + 2z = 1, 2x +...
Given system of linear equations:
x + y + z = 1 ….(1)
2x + 3y + 2z = 1 ….(2)
2x + 3y + (a2 – 1)z = a + 1 …..(3)
Consider a2 – 1 = 2
then LHS of (2) and (3) are same but RHS are not.
Hence a2 = 3 => |a| = √3
For |a| = √3, system is inconsistence.
So option (b) is correct.
x + y + z = 1 ….(1)
2x + 3y + 2z = 1 ….(2)
2x + 3y + (a2 – 1)z = a + 1 …..(3)
Consider a2 – 1 = 2
then LHS of (2) and (3) are same but RHS are not.
Hence a2 = 3 => |a| = √3
For |a| = √3, system is inconsistence.
So option (b) is correct.
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Community Answer
Consider the system of equations x + y + z = 1, 2x + 3y + 2z = 1, 2x +...
To solve the system of equations, we can use the method of elimination.
First, let's multiply the second equation by 2 to get rid of the coefficient of x:
2(2x) + 2(3y) + 2(2z) = 2(1)
4x + 6y + 4z = 2
Now, we have the system of equations:
x + y + z = 1
4x + 6y + 4z = 2
We can subtract the first equation from the second equation:
(4x + 6y + 4z) - (x + y + z) = 2 - 1
3x + 5y + 3z = 1
Now, we have the system of equations:
x + y + z = 1
3x + 5y + 3z = 1
We can subtract 3 times the first equation from the second equation:
(3x + 5y + 3z) - 3(x + y + z) = 1 - 3(1)
3x + 5y + 3z - 3x - 3y - 3z = 1 - 3
2y = -2
y = -1
Substituting y = -1 into the first equation:
x + (-1) + z = 1
x + z = 2
Let's call a^2 = z. Then we have:
x + a^2 = 2
x = 2 - a^2
So, the solution to the system of equations is:
x = 2 - a^2
y = -1
z = a^2
First, let's multiply the second equation by 2 to get rid of the coefficient of x:
2(2x) + 2(3y) + 2(2z) = 2(1)
4x + 6y + 4z = 2
Now, we have the system of equations:
x + y + z = 1
4x + 6y + 4z = 2
We can subtract the first equation from the second equation:
(4x + 6y + 4z) - (x + y + z) = 2 - 1
3x + 5y + 3z = 1
Now, we have the system of equations:
x + y + z = 1
3x + 5y + 3z = 1
We can subtract 3 times the first equation from the second equation:
(3x + 5y + 3z) - 3(x + y + z) = 1 - 3(1)
3x + 5y + 3z - 3x - 3y - 3z = 1 - 3
2y = -2
y = -1
Substituting y = -1 into the first equation:
x + (-1) + z = 1
x + z = 2
Let's call a^2 = z. Then we have:
x + a^2 = 2
x = 2 - a^2
So, the solution to the system of equations is:
x = 2 - a^2
y = -1
z = a^2
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Consider the system of equations x + y + z = 1, 2x + 3y + 2z = 1, 2x + 3y + (a2 - 1)z = a + 1 thena)System has a unique solution for |a| = √3b)System is inconsistence for |a| = √3c)System is inconsistence for a = 4d)System is inconsistence for a = 3Correct answer is option 'B'. Can you explain this answer?
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