1. Find Decomposition of vector in basis
`A=(3,7)`, `B=(-1,2)`, `C=(3,20)`

Solution:
Here `vec A=(3,7),vec B=(-1,2),vec C=(3,20)`

Here, `vec (a_1)=(3,7),vec (a_2)=(-1,2),vec b=(3,20)`

Form equation from vectors
`vec b = x_1 vec(a_1)+x_2 vec(a_2)`

So system of linear equations are
`3x_1-x_2=3`

`7x_1+2x_2=20`

Solution of equations using Elimination method

Total Equations are `2`

`3x_1-x_2=3 -> (1)`

`7x_1+2x_2=20 -> (2)`

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Select the equations `(1)` and `(2)`, and eliminate the variable `x_2`.

`3x_1-x_2=3`	` xx 2->`		``	`6x_1`	`-`	`2x_2`	`=`	`6`	``
		+
`7x_1+2x_2=20`	` xx 1->`		``	`7x_1`	`+`	`2x_2`	`=`	`20`	``
			``	`13x_1`			`=`	`26`	` -> (3)`

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Now use back substitution method
From (3)
`13x_1=26`

`=>x_1=(26)/(13)=2`

From (1)
`3x_1-x_2=3`

`=>3(2)-x_2=3`

`=>-x_2+6=3`

`=>-x_2=3-6=-3`

`=>x_2=3`

Solution using Elimination method.
`x_1=2,x_2=3`

`x_1=2,x_2=3`

So, `vec b = 2 vec(a_1)+3 vec(a_2)`

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2. Find Decomposition of vector in basis
`A=(3,1)`, `B=(1,2)`, `C=(8,1)`

Solution:
Here `vec A=(3,1),vec B=(1,2),vec C=(8,1)`

Here, `vec (a_1)=(3,1),vec (a_2)=(1,2),vec b=(8,1)`

Form equation from vectors
`vec b = x_1 vec(a_1)+x_2 vec(a_2)`

So system of linear equations are
`3x_1+x_2=8`

`x_1+2x_2=1`

Solution of equations using Elimination method

Total Equations are `2`

`3x_1+x_2=8 -> (1)`

`x_1+2x_2=1 -> (2)`

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Select the equations `(1)` and `(2)`, and eliminate the variable `x_2`.

`3x_1+x_2=8`	` xx 2->`		``	`6x_1`	`+`	`2x_2`	`=`	`16`	``
		−
`x_1+2x_2=1`	` xx 1->`		``	`x_1`	`+`	`2x_2`	`=`	`1`	``
			``	`5x_1`			`=`	`15`	` -> (3)`

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Now use back substitution method
From (3)
`5x_1=15`

`=>x_1=(15)/(5)=3`

From (1)
`3x_1+x_2=8`

`=>3(3)+x_2=8`

`=>x_2+9=8`

`=>x_2=8-9=-1`

Solution using Elimination method.
`x_1=3,x_2=-1`

`x_1=3,x_2=-1`

So, `vec b = 3 vec(a_1)- vec(a_2)`

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