If the system of linear equations, x + 2ay + az = 0, x + 3by + bz = 0 and x + 4cy + cz = 0 has a non-zero solution, then a, b, c satisfy
(1) 2ac = ab + bc              (2) 2ab = ac + bc               (3) 2b = a + c               (4) b2 = ac

Dear student
For non zero solutions, we have10a1bb12cc=0On expanding along R1, we getbc-2bc-0c-b+a2c-b=0bc-2bc+2ac-ab=02ac=ab+bc
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