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The system of simultaneous linear equations
x + y + z = 0
x - y - z = 0 has
x + y + z = 0
x - y - z = 0 has
- a)no solution in R3
- b)a unique solution in R3
- c)infinitely many solution in R3
- d)more than 2 but finitely many solutions in R3
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The system of simultaneous linear equationsx + y + z = 0x - y - z = 0 ...
We are given that the system of simultaneous linear equations,
x + y + z = 0
x - y - z = 0
The coefficient matrix is given by,
rank of coefficient matrix = 2. Here, the rank of matrix < no. of unknowns therefore, the system of equation has infinitely many solution in R3.
x + y + z = 0
x - y - z = 0
The coefficient matrix is given by,
rank of coefficient matrix = 2. Here, the rank of matrix < no. of unknowns therefore, the system of equation has infinitely many solution in R3.
Most Upvoted Answer
The system of simultaneous linear equationsx + y + z = 0x - y - z = 0 ...
To determine the solution to the system of simultaneous linear equations, we can use the method of elimination or substitution. Let's solve the given system of equations using the method of elimination:
Equation 1: x + y + z = 0
Equation 2: x - y - z = 0
1. Elimination method:
To eliminate the variable 'x', we can subtract Equation 2 from Equation 1:
(x + y + z) - (x - y - z) = 0 - 0
x + y + z - x + y + z = 0
2y + 2z = 0
y + z = 0 [Divide by 2]
Now, we have two equations:
Equation 1: x + y + z = 0
Equation 3: y + z = 0
2. Substitution method:
From Equation 3, we can express 'z' in terms of 'y':
z = -y
Now, substitute this value of 'z' in Equation 1:
x + y + (-y) = 0
x = 0
3. Solution in R3:
From the above calculations, we have:
x = 0
y = y
z = -y
This means that for any value of 'y' chosen, we can find the corresponding values of 'x' and 'z'. Therefore, there are infinitely many solutions to the given system of equations in R3.
Hence, the correct answer is option C) infinitely many solutions in R3.
Equation 1: x + y + z = 0
Equation 2: x - y - z = 0
1. Elimination method:
To eliminate the variable 'x', we can subtract Equation 2 from Equation 1:
(x + y + z) - (x - y - z) = 0 - 0
x + y + z - x + y + z = 0
2y + 2z = 0
y + z = 0 [Divide by 2]
Now, we have two equations:
Equation 1: x + y + z = 0
Equation 3: y + z = 0
2. Substitution method:
From Equation 3, we can express 'z' in terms of 'y':
z = -y
Now, substitute this value of 'z' in Equation 1:
x + y + (-y) = 0
x = 0
3. Solution in R3:
From the above calculations, we have:
x = 0
y = y
z = -y
This means that for any value of 'y' chosen, we can find the corresponding values of 'x' and 'z'. Therefore, there are infinitely many solutions to the given system of equations in R3.
Hence, the correct answer is option C) infinitely many solutions in R3.
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The system of simultaneous linear equationsx + y + z = 0x - y - z = 0 hasa)no solution in R3b)a unique solution in R3c)infinitely many solution in R3d)more than 2 but finitely many solutions in R3Correct answer is option 'C'. Can you explain this answer?
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