The system of equations
x + y + z = 6
x + 4y + 6z = 20
x + 4y + \[\lambda \]z = \[\mu \]
has NO solution for values of \[\lambda \] and \[\mu \] given by
A. \[\lambda = 6,\,\mu = 20\]
B. \[\lambda = 6,\,\mu \ne 20\]
C. \[\lambda \ne 6,\,\mu = 20\]
D. \[\lambda \ne 6,\,\mu \ne 20\]
Answer: Option B
Related Questions on Linear Algebra
A. 3, 3 + 5j, 6 - j
B. -6 + 5j, 3 + j, 3 - j
C. 3 + j, 3 - j, 5 + j
D. 3, -1 + 3j, -1 - 3j
A. 1024 and -1024
B. 1024√2 and -1024√2
C. 4√2 and -4√2
D. 512√2 and -512√2

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