Chapter 4 - Vector Spaces - 4.7 Change of Basis - Problems - Page 318: 5

$(4,6,-1)$

We know that $a(1,-6,3)+b(0,5,-1)+c(3,-1,-1)=(1,7,7)$. Thus $a+3c=1 \\ -6a+5b-c=7\\ 3a-b-c=7$. Thus $-6a+5b-c=7\\ 15a-5b-5c=35$. Thus $2a+6c=2 \\9a-6c=42 $ Thus $11a=44 \rightarrow a=4$ Substitute: $4+3c=1$ $c=-1$ then $3.4-b-(-1)=7 \rightarrow b=6$ Thus the vector is $(4,6,-1)$.