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The value of a for which system of equations,
a3x + (a + 1)3 y + (a + 2)3z = 0,
ax + (a + 1) y + (a + 2) z = 0,
x + y + z = 0, has a non–zero solution is
ax + (a + 1) y + (a + 2) z = 0,
x + y + z = 0, has a non–zero solution is
- a)–1
- b)0
- c)1
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The value of a for which system of equations,a3x + (a + 1)3y + (a + 2)...
To find the value of a for which the system of equations has a non-zero solution, we can eliminate the variables x, y, and z by substitution.
First, we can solve the third equation for z: z = -(x + y). Substituting this expression for z into the first and second equations, we get:
a3x + (a + 1)3(-x - y) + (a + 2)3(-x - y) = 0
ax + (a + 1) (-x - y) + (a + 2) (-x - y) = 0
Simplifying these equations, we get:
-2a3x - 2(a + 1)3y - 2(a + 2)3y = 0
-ax - (a + 1)y - (a + 2)y = 0
Then, we can solve the second equation for y: y = -(ax + (a + 2)y)/(a + 1). Substituting this expression for y into the first equation, we get:
-2a3x - 2(a + 1)3(-ax - (a + 2)(-ax - (a + 2)y)/(a + 1)) - 2(a + 2)3(-ax - (a + 2)(-ax - (a + 2)y)/(a + 1)) = 0
Simplifying this equation, we get:
(a + 1)(a + 2)(2a3 + 2a + 4)x = 0
Since we want the value of a for which the system has a non-zero solution, we need the expression on the left-hand side to be equal to 0. This occurs when a = -1 or a = -2/3.
Therefore, the correct answer is (a) -1.
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Community Answer
The value of a for which system of equations,a3x + (a + 1)3y + (a + 2)...
Trivial solution is when a = -2.
To see this, we can start by looking at the third equation, which tells us that at least one of x, y, or z must be zero. Let's assume that z = 0. Then the second equation becomes:
ax (a 1) y = 0
This equation has a non-trivial solution when either a = 0 or y = 0. If a = 0, then the first equation becomes:
0 (-1)3y (-2)3z = 0
This reduces to -27y = 0, which implies that y = 0. But then the second equation becomes:
0 (a 1) 0 = 0
This equation is always true, so there are infinitely many solutions. Therefore, a = 0 is not the value we are looking for.
If y = 0, then the second equation becomes:
ax 0 0 = 0
This equation has a non-trivial solution when x = 0 or a = 0. If x = 0, then the first equation becomes:
0 (a 1)3y (a 2)3z = 0
This reduces to (a+2)3z = 0, which implies that z = 0. But then the third equation becomes:
0 0 0 = 0
This equation is always true, so there are infinitely many solutions. Therefore, x = 0 is not the value we are looking for.
If a = 0, then the first equation becomes:
0 (-1)3y (-2)3z = 0
This reduces to -27y - 8z = 0. If we set y = 1, then we get z = -27/8. Therefore, a = 0 does not give us a non-trivial solution.
Now let's assume that y = 0 and z = 0. Then the second equation becomes:
ax 0 0 = 0
This equation has a non-trivial solution when x = 0 or a = 0. If x = 0, then the third equation becomes:
0 0 0 = 0
This equation is always true, so there are infinitely many solutions. Therefore, x = 0 is not the value we are looking for.
If a = 0, then the first equation becomes:
0 (-1)3y (-2)3z = 0
This reduces to -27y - 8z = 0. If we set y = 1, then we get z = -27/8. Therefore, a = 0 does not give us a non-trivial solution.
Therefore, the only value of a that gives us a non-trivial solution is a = -2. If we set a = -2, then the first equation becomes:
-8x 1y 0z = 0
If we set y = 8, then we get x = -1. Therefore, the system of equations has a non-trivial solution when a = -2 and x = -1, y = 8, z = 0.
To see this, we can start by looking at the third equation, which tells us that at least one of x, y, or z must be zero. Let's assume that z = 0. Then the second equation becomes:
ax (a 1) y = 0
This equation has a non-trivial solution when either a = 0 or y = 0. If a = 0, then the first equation becomes:
0 (-1)3y (-2)3z = 0
This reduces to -27y = 0, which implies that y = 0. But then the second equation becomes:
0 (a 1) 0 = 0
This equation is always true, so there are infinitely many solutions. Therefore, a = 0 is not the value we are looking for.
If y = 0, then the second equation becomes:
ax 0 0 = 0
This equation has a non-trivial solution when x = 0 or a = 0. If x = 0, then the first equation becomes:
0 (a 1)3y (a 2)3z = 0
This reduces to (a+2)3z = 0, which implies that z = 0. But then the third equation becomes:
0 0 0 = 0
This equation is always true, so there are infinitely many solutions. Therefore, x = 0 is not the value we are looking for.
If a = 0, then the first equation becomes:
0 (-1)3y (-2)3z = 0
This reduces to -27y - 8z = 0. If we set y = 1, then we get z = -27/8. Therefore, a = 0 does not give us a non-trivial solution.
Now let's assume that y = 0 and z = 0. Then the second equation becomes:
ax 0 0 = 0
This equation has a non-trivial solution when x = 0 or a = 0. If x = 0, then the third equation becomes:
0 0 0 = 0
This equation is always true, so there are infinitely many solutions. Therefore, x = 0 is not the value we are looking for.
If a = 0, then the first equation becomes:
0 (-1)3y (-2)3z = 0
This reduces to -27y - 8z = 0. If we set y = 1, then we get z = -27/8. Therefore, a = 0 does not give us a non-trivial solution.
Therefore, the only value of a that gives us a non-trivial solution is a = -2. If we set a = -2, then the first equation becomes:
-8x 1y 0z = 0
If we set y = 8, then we get x = -1. Therefore, the system of equations has a non-trivial solution when a = -2 and x = -1, y = 8, z = 0.
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