`3x+2y-z+w=0`
`x-y+4z+2w=25`
`-2x+y+2z-w=2`
`x+y+z+w=6`
The above system of equations can be represented by the coefficient matrix A and right hand side vector b as follows:
`A=[[3,2,-1,1],[1,-1,4,2],[-2,1,2,-1],[1,1,1,1]]`
`b=[[0],[25],[2],[6]]`
The augmented matrix can be written as:
`[A b]=[[3,2,-1,1,0],[1,-1,4,2,25],[-2,1,2,-1,2],[1,1,1,1,6]]`
Now let's bring the above matrix in row-echelon form by performing various row operations,
Rewrite the 1st Row`(R_1)` as `(R_1+R_3)`
`[[1,3,1,0,2],[1,-1,4,2,25],[-2,1,2,-1,2],[1,1,1,1,6]]`
Rewrite the 2nd Row`(R_2)` as `(R_2-R_4)`
`[[1,3,1,0,2],[0,-2,3,1,19],[-2,1,2,-1,2],[1,1,1,1,6]]`
Rewrite the 3rd Row`(R_3)` as`(R_3+2R_4)`
`[[1,3,1,0,2],[0,-2,3,1,19],[0,3,4,1,14],[1,1,1,1,6]]`
Rewrite the 4th Row as `(R_4-R_1)`
...
`3x+2y-z+w=0`
`x-y+4z+2w=25`
`-2x+y+2z-w=2`
`x+y+z+w=6`
The above system of equations can be represented by the coefficient matrix A and right hand side vector b as follows:
`A=[[3,2,-1,1],[1,-1,4,2],[-2,1,2,-1],[1,1,1,1]]`
`b=[[0],[25],[2],[6]]`
The augmented matrix can be written as:
`[A b]=[[3,2,-1,1,0],[1,-1,4,2,25],[-2,1,2,-1,2],[1,1,1,1,6]]`
Now let's bring the above matrix in row-echelon form by performing various row operations,
Rewrite the 1st Row`(R_1)` as `(R_1+R_3)`
`[[1,3,1,0,2],[1,-1,4,2,25],[-2,1,2,-1,2],[1,1,1,1,6]]`
Rewrite the 2nd Row`(R_2)` as `(R_2-R_4)`
`[[1,3,1,0,2],[0,-2,3,1,19],[-2,1,2,-1,2],[1,1,1,1,6]]`
Rewrite the 3rd Row`(R_3)` as`(R_3+2R_4)`
`[[1,3,1,0,2],[0,-2,3,1,19],[0,3,4,1,14],[1,1,1,1,6]]`
Rewrite the 4th Row as `(R_4-R_1)`
`[[1,3,1,0,2],[0,-2,3,1,19],[0,3,4,1,14],[0,-2,0,1,4]]`
Rewrite the 2nd Row as `(R_2+R_3)`
`[[1,3,1,0,2],[0,1,7,2,33],[0,3,4,1,14],[0,-2,0,1,4]]`
Rewrite the 3rd Row as `(R_3-3R_2)`
`[[1,3,1,0,2],[0,1,7,2,33],[0,0,-17,-5,-85],[0,-2,0,1,4]]`
Rewrite the 4th Row as `(R_4+2R_2)`
`[[1,3,1,0,2],[0,1,7,2,33],[0,0,-17,-5,-85],[0,0,14,5,70]]`
Rewrite the 3rd Row as `(R_3+R_4)`
`[[1,3,1,0,2],[0,1,7,2,33],[0,0,-3,0,-15],[0,0,14,5,70]]`
Rewrite the 3rd Row by dividing it with -3,
`[[1,3,1,0,2],[0,1,7,2,33],[0,0,1,0,5],[0,0,14,5,70]]`
Rewrite the 4th Row as`(R_4-14R_3)`
`[[1,3,1,0,2],[0,1,7,2,33],[0,0,1,0,5],[0,0,0,5,0]]`
Rewrite the 4th Row by dividing it with 5,
`[[1,3,1,0,2],[0,1,7,2,33],[0,0,1,0,5],[0,0,0,1,0]]`
Now the above matrix is row-echelon form and we can perform back substitution on the corresponding system,
`x+3y+z=2` ----- Eq:1
`y+7z+2w=33` ------ Eq:2
`z=5`
`w=0`
Substitute back the value of z and w in Eq:2,
`y+7(5)+2(0)=33`
`y+35=33`
`y=33-35=-2`
Substitute back the value of y and z in Eq:1,
`x+3(-2)+5=2`
`x-6+5=2`
`x-1=2`
`x=3`
So the solutions are x=3,y=-2,z=5 and w=0
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