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Decomposition of vector in basis calculator
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 1. `(1,2),(3,1),(8,1)`
2. `(1,2),(1,3),(2,4)`
3. `(3,7),(-1,2),(3,20)`
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Example (with steps)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 methodTotal Equations are `2` `x_1+3x_2=8 -> (1)` `2x_1+x_2=1 -> (2)`
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)` |
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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