Find #a, b# so that the system has unique solution: ? #x+y+z=6# #x+2y+3z=10# #x+2y+az=b#

Question

#a != 3#, #b# can be any value

#[(1, 1, 1 | 6),(1, 2, 3 | 10), (1, 2, a | b)]#
Row #2 - 1# and #3 -1#
#=[(1, 1, 1 | 6),(0, 1, 2 | 4), (0, 1, (a-1) | (b-6))]#
Row #1 -2# and #3-2#
#=[(1, 0, -1 | 6),(0, 1, 2 | 4), (0, 0, (a-3) | (b-10))]#
To find unique solution #(b-10)/(a-3)# must be defined, so, #a != 3#

How can I find #a, b#?

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Answer 1 · 2017-06-30T10:26:39

See below

Given

#M=((1,1,1),(1,2,3),(1,2,a))# we have

#det(M)=a-3# so for #a ne 3#, #M# has inverse.

Concluding, the system has unique solution for #a ne 3#.

This result does not depends on #b# values.