Answer on Question #56010 – Math – Linear Algebra
Let A=2i−j+k, B=i+3j−2k, C=−2i+j−2k, and D=3i+2j+5k.
Find scalar a, b, c such that D=aA+bB+cC
Solution
First find aA,bB,cC in component form
aA=a(2i−j+k)=2ai−aj+akbB=b(i+3j−2k)=bi+3bj−2bkcC=c(−2i+j−2k)=−2ci+cj−2ck
Now let's sum these equations and compare them with the expression for D in component form
D=aA+bB+cC=2ai−aj+ak+bi+3bj−2bk−2ci+cj−2ck=(2a+b−2c)i+(−a+3b+c)j+(a−2b−2c)kD=3i+2j+5k
Equating expressions that stood near the corresponding unit vectors, we obtain the following system
i:⎩⎨⎧2a+b−2c=3−a+3b+c=2a−2b−2c=5
Express b from the first equation and substitute for b into the second and the third equations of the system:
⎩⎨⎧b=3+2c−2a−a+9+6c−6a+c=2a−6−4c+4a−2c=5⎩⎨⎧b=3+2c−2a7c−7a=−75a−6c=11
Divide the second equation by 7:
⎩⎨⎧b=3+2c−2ac−a=−15a−6c=11
Express c from the second equation and substitute for c into the third equations of the system:
⎩⎨⎧b=3+2c−2ac=−1+a5a−6a+6=11⎩⎨⎧b=3+2c−2ac=−1+a−a=5⎩⎨⎧b=3+2c−2ac=−6a=−5⎩⎨⎧b=1c=−6a=−5
Answer: a=−5, b=1, c=−6.
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