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

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

Here, `vec (a_1)=(1,2),vec (a_2)=(3,1),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
`x_1+3x_2=8`

`2x_1+x_2=1`

Solution of equations using Elimination method

Total Equations are `2`

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

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

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Select the equations `(1)` and `(2)`, and eliminate the variable `x_1`.

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

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Now use back substitution method
From (3)
`5x_2=15`

`=>x_2=(15)/(5)=3`

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

`=>x_1+3(3)=8`

`=>x_1+9=8`

`=>x_1=8-9=-1`

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

`x_1=-1,x_2=3`

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

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2. Find Decomposition of vector in basis
`A=(1,2)`, `B=(1,3)`, `C=(2,4)`

Solution:
Here `vec A=(1,2),vec B=(1,3),vec C=(2,4)`

Here, `vec (a_1)=(1,2),vec (a_2)=(1,3),vec b=(2,4)`

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

So system of linear equations are
`x_1+x_2=2`

`2x_1+3x_2=4`

Solution of equations using Elimination method

Total Equations are `2`

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

`2x_1+3x_2=4 -> (2)`

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Select the equations `(1)` and `(2)`, and eliminate the variable `x_1`.

`x_1+x_2=2`	` xx 2->`		``	`2x_1`	`+`	`2x_2`	`=`	`4`	``
		−
`2x_1+3x_2=4`	` xx 1->`		``	`2x_1`	`+`	`3x_2`	`=`	`4`	``
					`-`	`x_2`	`=`	`0`	` -> (3)`

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Now use back substitution method
From (3)
`-x_2=0`

`=>x_2=0`

From (1)
`x_1+x_2=2`

`=>x_1+(0)=2`

`=>x_1=2`

Solution using Elimination method.
`x_1=2,x_2=0`

`x_1=2,x_2=0`

So, `vec b = 2 vec(a_1)`

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