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Given a system of equations: x 2y 2z = b1 5 x y 3z = b2 which one of the following is true regarding its solution? the system has a unique solution for any given b1 and b2 the system will have infinitely many solutions for any given b1 and b2 whether or not a solution exists depends on the given b1 and b2 the system would have no solution for any values b1 and b2?
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Given a system of equations: x 2y 2z = b1 5 x y 3z = b2 which ...
System of Equations
The given system of equations is:
x + 2y + 2z = b1
5x + y + 3z = b2
Possible Solutions
To determine the nature of the solution for any given values of b1 and b2, we need to analyze the system of equations.
Unique Solution
A system of equations has a unique solution if and only if the number of equations is equal to the number of variables, and the determinant of the coefficient matrix is non-zero.
In this case, the number of variables (x, y, z) is equal to the number of equations (2). However, we need to check the determinant of the coefficient matrix.
The coefficient matrix is:
| 1 2 2 |
| 5 1 3 |
Calculating the determinant of this matrix, we get:
Det = (1*1*3) + (2*3*5) + (2*5*1) - (2*1*2) - (1*5*2) - (3*2*1)
Det = 3 + 30 + 10 - 4 - 10 - 6
Det = 23
Since the determinant is non-zero (Det ≠ 0), we can conclude that the system of equations has a unique solution for any given values of b1 and b2.
Therefore, the statement "the system has a unique solution for any given b1 and b2" is true.
Infinitely Many Solutions
A system of equations has infinitely many solutions if and only if the number of equations is less than the number of variables, and the determinant of the coefficient matrix is zero.
In this case, the number of variables (x, y, z) is greater than the number of equations (2). Additionally, the determinant is non-zero.
Therefore, the statement "the system will have infinitely many solutions for any given b1 and b2" is not true.
No Solution
A system of equations has no solution if and only if the number of equations is equal to the number of variables, and the determinant of the coefficient matrix is zero.
In this case, the number of variables (x, y, z) is equal to the number of equations (2). However, the determinant is non-zero.
Therefore, the statement "the system would have no solution for any values b1 and b2" is not true.
Conclusion
Based on the analysis of the system of equations, we can conclude that the system has a unique solution for any given values of b1 and b2.
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Given a system of equations: x 2y 2z = b1 5 x y 3z = b2 which one of the following is true regarding its solution? the system has a unique solution for any given b1 and b2 the system will have infinitely many solutions for any given b1 and b2 whether or not a solution exists depends on the given b1 and b2 the system would have no solution for any values b1 and b2?
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Given a system of equations: x 2y 2z = b1 5 x y 3z = b2 which one of the following is true regarding its solution? the system has a unique solution for any given b1 and b2 the system will have infinitely many solutions for any given b1 and b2 whether or not a solution exists depends on the given b1 and b2 the system would have no solution for any values b1 and b2? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Given a system of equations: x 2y 2z = b1 5 x y 3z = b2 which one of the following is true regarding its solution? the system has a unique solution for any given b1 and b2 the system will have infinitely many solutions for any given b1 and b2 whether or not a solution exists depends on the given b1 and b2 the system would have no solution for any values b1 and b2? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given a system of equations: x 2y 2z = b1 5 x y 3z = b2 which one of the following is true regarding its solution? the system has a unique solution for any given b1 and b2 the system will have infinitely many solutions for any given b1 and b2 whether or not a solution exists depends on the given b1 and b2 the system would have no solution for any values b1 and b2?.
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