Let us write the vector ( 1 , − 2 , 5 ) (1, −2, 5) ( 1 , − 2 , 5 ) as a linear combination of the vectors ( 1 , 1 , 1 ) , ( 1 , 2 , 3 ) (1, 1, 1),(1, 2, 3) ( 1 , 1 , 1 ) , ( 1 , 2 , 3 )
and ( 2 , − 1 , 1 ) . (2, −1, 1). ( 2 , − 1 , 1 ) .
Let
( 1 , − 2 , 5 ) = a ( 1 , 1 , 1 ) + b ( 1 , 2 , 3 ) + c ( 2 , − 1 , 1 ) = ( a + b + 2 c , a + 2 b − c , a + 3 b + c ) . (1, −2, 5)=a(1, 1, 1)+b(1, 2, 3)+c(2, −1, 1)=(a+b+2c,a+2b-c,a+3b+c). ( 1 , − 2 , 5 ) = a ( 1 , 1 , 1 ) + b ( 1 , 2 , 3 ) + c ( 2 , − 1 , 1 ) = ( a + b + 2 c , a + 2 b − c , a + 3 b + c ) .
It follows that we get the system
{ a + b + 2 c = 1 a + 2 b − c = − 2 a + 3 b + c = 5 \begin{cases}
a+b+2c=1\\
a+2b-c=-2\\
a+3b+c=5
\end{cases} ⎩ ⎨ ⎧ a + b + 2 c = 1 a + 2 b − c = − 2 a + 3 b + c = 5
which is equivalent after subtracting from the second row the first row, and from the third row the second row to the system
{ a + b + 2 c = 1 b − 3 c = − 3 b + 2 c = 7 \begin{cases}
a+b+2c=1\\
b-3c=-3\\
b+2c=7
\end{cases} ⎩ ⎨ ⎧ a + b + 2 c = 1 b − 3 c = − 3 b + 2 c = 7
and hence the last system is equivalent after subtracting from the third row the second row to the system
{ a + b + 2 c = 1 b − 3 c = − 3 5 c = 10 \begin{cases}
a+b+2c=1\\
b-3c=-3\\
5c=10
\end{cases} ⎩ ⎨ ⎧ a + b + 2 c = 1 b − 3 c = − 3 5 c = 10
It follows that
c = 2 , b = 3 c − 3 = 6 − 3 = 3 , a = 1 − b − 2 c = 1 − 3 − 4 = − 6. c=2,\\ b=3c-3=6-3=3,\\ a=1-b-2c=1-3-4=-6. c = 2 , b = 3 c − 3 = 6 − 3 = 3 , a = 1 − b − 2 c = 1 − 3 − 4 = − 6.
We conclude that ( 1 , − 2 , 5 ) = − 6 ( 1 , 1 , 1 ) + 3 ( 1 , 2 , 3 ) + 2 ( 2 , − 1 , 1 ) . (1, −2, 5)=-6(1, 1, 1)+3(1, 2, 3)+2(2, −1, 1). ( 1 , − 2 , 5 ) = − 6 ( 1 , 1 , 1 ) + 3 ( 1 , 2 , 3 ) + 2 ( 2 , − 1 , 1 ) .
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