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Home > Matrix & Vector > Matrix operation > SVD - Singular Value Decomposition example
17. SVD - Singular Value Decomposition example ( Enter your problem )
( Enter your problem )
  1. Example `[[4,0],[3,-5]]`
  2. Example `[[1,0,1,0],[0,1,0,1]]`
  3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
  4. Example `[[2,3],[4,10]]`

3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`


Find SVD - Singular Value Decomposition ...
`[[1,1,1],[-1,-3,-3],[2,4,4]]`

Solution:
`A = `
`1``1``1`
`-1``-3``-3`
`2``4``4`


Here we are trying to find out two solutions using `A*A'` and `A'*A`

`1^"st"` Solution using `A*A'` for normalized vectors `u_i`

`A * A'`
`A^T` = 
`1``1``1`
`-1``-3``-3`
`2``4``4`
T
 = 
`1``-1``2`
`1``-3``4`
`1``-3``4`


`A×(A^T)`=
`1``1``1`
`-1``-3``-3`
`2``4``4`
×
`1``-1``2`
`1``-3``4`
`1``-3``4`


=
`1×1+1×1+1×1``1×(-1)+1×(-3)+1×(-3)``1×2+1×4+1×4`
`-1×1+(-3)×1+(-3)×1``-1×(-1)+(-3)×(-3)+(-3)×(-3)``-1×2+(-3)×4+(-3)×4`
`2×1+4×1+4×1``2×(-1)+4×(-3)+4×(-3)``2×2+4×4+4×4`


=
`1+1+1``-1+(-3)+(-3)``2+4+4`
`-1+(-3)+(-3)``1+9+9``-2+(-12)+(-12)`
`2+4+4``-2+(-12)+(-12)``4+16+16`


=
`3``-7``10`
`-7``19``-26`
`10``-26``36`
`A * A' = `
`3``-7``10`
`-7``19``-26`
`10``-26``36`


Find Eigen vector for `A * A'`

`|A * A'-lamdaI|=0`

 `(3-lamda)`  `-7`  `10` 
 `-7`  `(19-lamda)`  `-26` 
 `10`  `-26`  `(36-lamda)` 
 = 0


`:.(3-lamda)((19-lamda) × (36-lamda) - (-26) × (-26))-(-7)((-7) × (36-lamda) - (-26) × 10)+10((-7) × (-26) - (19-lamda) × 10)=0`

`:.(3-lamda)((684-55lamda+lamda^2)-676)+7((-252+7lamda)-(-260))+10(182-(190-10lamda))=0`

`:.(3-lamda)(8-55lamda+lamda^2)+7(8+7lamda)+10(-8+10lamda)=0`

`:. (24-173lamda+58lamda^2-lamda^3)+(56+49lamda)+(-80+100lamda)=0`

`:.(-lamda^3+58lamda^2-24lamda)=0`

`:.-lamda(lamda-0.4168)(lamda-57.5832)=0`

`:.lamda=0 or (lamda-0.4168)=0 or (lamda-57.5832)=0`

`:.lamda=0 or lamda=0.4168 or lamda=57.5832`

`:.` The eigenvalues of the matrix `A * A'` are given by `lamda=0,0.4168,57.5832`

1. Eigenvectors for `lamda=57.5832`



1. Eigenvectors for `lamda=57.5832`

`A * A'-lamdaI = `
3-710
-719-26
10-2636
 - `57.5832` 
100
010
001


 = 
3-710
-719-26
10-2636
 - 
57.583200
057.58320
0057.5832

 = 
`-54.5832``-7``10`
`-7``-38.5832``-26`
`10``-26``-21.5832`


Now, reduce this matrix
`R_1 larr R_1-:(-54.5832)`

 = 
`1``0.1282``-0.1832`
`-7``-38.5832``-26`
`10``-26``-21.5832`


`R_2 larr R_2+7xx R_1`

 = 
`1``0.1282``-0.1832`
`0``-37.6855``-27.2824`
`10``-26``-21.5832`


`R_3 larr R_3-10xx R_1`

 = 
`1``0.1282``-0.1832`
`0``-37.6855``-27.2824`
`0``-27.2824``-19.7511`


`R_2 larr R_2-:(-37.6855)`

 = 
`1``0.1282``-0.1832`
`0``1``0.724`
`0``-27.2824``-19.7511`


`R_1 larr R_1-0.1282xx R_2`

 = 
`1``0``-0.276`
`0``1``0.724`
`0``-27.2824``-19.7511`


`R_3 larr R_3+27.2824xx R_2`

 = 
`1``0``-0.276`
`0``1``0.724`
`0``0``0`


The system associated with the eigenvalue `lamda=57.5832`

`(A * A'-57.5832I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-0.276`
`0``1``0.724`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-0.276x_3=0,x_2+0.724x_3=0`

`=>x_1=0.276x_3,x_2=-0.724x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=57.5832` is

`v=`
`0.276x_3`
`-0.724x_3`
`x_3`


Let `x_3=1`

`v_1=`
`0.276`
`-0.724`
`1`
`v_1=`
`0.276`
`-0.724`
`1`


2. Eigenvectors for `lamda=0.4168`



2. Eigenvectors for `lamda=0.4168`

`A * A'-lamdaI = `
3-710
-719-26
10-2636
 - `0.4168` 
100
010
001


 = 
3-710
-719-26
10-2636
 - 
0.416800
00.41680
000.4168

 = 
`2.5832``-7``10`
`-7``18.5832``-26`
`10``-26``35.5832`


Now, reduce this matrix
`R_1 larr R_1-:2.5832`

 = 
`1``-2.7098``3.8711`
`-7``18.5832``-26`
`10``-26``35.5832`


`R_2 larr R_2+7xx R_1`

 = 
`1``-2.7098``3.8711`
`0``-0.3854``1.098`
`10``-26``35.5832`


`R_3 larr R_3-10xx R_1`

 = 
`1``-2.7098``3.8711`
`0``-0.3854``1.098`
`0``1.098``-3.1283`


`R_2 larr R_2-:(-0.3854)`

 = 
`1``-2.7098``3.8711`
`0``1``-2.849`
`0``1.098``-3.1283`


`R_1 larr R_1+2.7098xx R_2`

 = 
`1``0``-3.849`
`0``1``-2.849`
`0``1.098``-3.1283`


`R_3 larr R_3-1.098xx R_2`

 = 
`1``0``-3.849`
`0``1``-2.849`
`0``0``0`


The system associated with the eigenvalue `lamda=0.4168`

`(A * A'-0.4168I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-3.849`
`0``1``-2.849`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-3.849x_3=0,x_2-2.849x_3=0`

`=>x_1=3.849x_3,x_2=2.849x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0.4168` is

`v=`
`3.849x_3`
`2.849x_3`
`x_3`


Let `x_3=1`

`v_2=`
`3.849`
`2.849`
`1`
`v_2=`
`3.849`
`2.849`
`1`


3. Eigenvectors for `lamda=0`



3. Eigenvectors for `lamda=0`

`A * A'-lamdaI = `
3-710
-719-26
10-2636
 - `0` 
100
010
001


 = 
`3``-7``10`
`-7``19``-26`
`10``-26``36`


Now, reduce this matrix
`R_1 larr R_1-:3`

 = 
`1``-2.3333``3.3333`
`-7``19``-26`
`10``-26``36`


`R_2 larr R_2+7xx R_1`

 = 
`1``-2.3333``3.3333`
`0``2.6667``-2.6667`
`10``-26``36`


`R_3 larr R_3-10xx R_1`

 = 
`1``-2.3333``3.3333`
`0``2.6667``-2.6667`
`0``-2.6667``2.6667`


`R_2 larr R_2-:2.6667`

 = 
`1``-2.3333``3.3333`
`0``1``-1`
`0``-2.6667``2.6667`


`R_1 larr R_1+2.3333xx R_2`

 = 
`1``0``1`
`0``1``-1`
`0``-2.6667``2.6667`


`R_3 larr R_3+2.6667xx R_2`

 = 
`1``0``1`
`0``1``-1`
`0``0``0`


The system associated with the eigenvalue `lamda=0`

`(A * A'-0I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``1`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1+x_3=0,x_2-x_3=0`

`=>x_1=-x_3,x_2=x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0` is

`v=`
`-x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_3=`
`-1`
`1`
`1`
`v_3=`
`-1`
`1`
`1`


For Eigenvector-1 `(0.276,-0.724,1)`, Length L = `sqrt(|0.27605|^2+|-0.72395|^2+|1|^2)=1.265`

So, normalizing gives `u_1=((0.276)/(1.265),(-0.724)/(1.265),(1)/(1.265))=(0.2182,-0.5723,0.7905)`

For Eigenvector-2 `(3.849,2.849,1)`, Length L = `sqrt(|3.84895|^2+|2.84895|^2+|1|^2)=4.8919`

So, normalizing gives `u_2=((3.849)/(4.8919),(2.849)/(4.8919),(1)/(4.8919))=(0.7868,0.5824,0.2044)`

For Eigenvector-3 `(-1,1,1)`, Length L = `sqrt(|-1|^2+|1|^2+|1|^2)=1.7321`

So, normalizing gives `u_3=((-1)/(1.7321),(1)/(1.7321),(1)/(1.7321))=(-0.5774,0.5774,0.5774)`

`2^"nd"` Solution using `A'*A` for normalized vectors `v_i`

`A' * A`
`A^T` = 
`1``1``1`
`-1``-3``-3`
`2``4``4`
T
 = 
`1``-1``2`
`1``-3``4`
`1``-3``4`


`(A^T)×A`=
`1``-1``2`
`1``-3``4`
`1``-3``4`
×
`1``1``1`
`-1``-3``-3`
`2``4``4`


=
`1×1+(-1)×(-1)+2×2``1×1+(-1)×(-3)+2×4``1×1+(-1)×(-3)+2×4`
`1×1+(-3)×(-1)+4×2``1×1+(-3)×(-3)+4×4``1×1+(-3)×(-3)+4×4`
`1×1+(-3)×(-1)+4×2``1×1+(-3)×(-3)+4×4``1×1+(-3)×(-3)+4×4`


=
`1+1+4``1+3+8``1+3+8`
`1+3+8``1+9+16``1+9+16`
`1+3+8``1+9+16``1+9+16`


=
`6``12``12`
`12``26``26`
`12``26``26`
`A' * A = `
`6``12``12`
`12``26``26`
`12``26``26`


Find Eigen vector for `A' * A`

`|A' * A-lamdaI|=0`

 `(6-lamda)`  `12`  `12` 
 `12`  `(26-lamda)`  `26` 
 `12`  `26`  `(26-lamda)` 
 = 0


`:.(6-lamda)((26-lamda) × (26-lamda) - 26 × 26)-12(12 × (26-lamda) - 26 × 12)+12(12 × 26 - (26-lamda) × 12)=0`

`:.(6-lamda)((676-52lamda+lamda^2)-676)-12((312-12lamda)-312)+12(312-(312-12lamda))=0`

`:.(6-lamda)(-52lamda+lamda^2)-12(-12lamda)+12(12lamda)=0`

`:. (-312lamda+58lamda^2-lamda^3)-(-144lamda)+(144lamda)=0`

`:.(-lamda^3+58lamda^2-24lamda)=0`

`:.-lamda(lamda-0.4168)(lamda-57.5832)=0`

`:.lamda=0 or (lamda-0.4168)=0 or (lamda-57.5832)=0`

`:.lamda=0 or lamda=0.4168 or lamda=57.5832`

`:.` The eigenvalues of the matrix `A' * A` are given by `lamda=0,0.4168,57.5832`

1. Eigenvectors for `lamda=57.5832`



1. Eigenvectors for `lamda=57.5832`

`A' * A-lamdaI = `
61212
122626
122626
 - `57.5832` 
100
010
001


 = 
61212
122626
122626
 - 
57.583200
057.58320
0057.5832

 = 
`-51.5832``12``12`
`12``-31.5832``26`
`12``26``-31.5832`


Now, reduce this matrix
`R_1 larr R_1-:(-51.5832)`

 = 
`1``-0.2326``-0.2326`
`12``-31.5832``26`
`12``26``-31.5832`


`R_2 larr R_2-12xx R_1`

 = 
`1``-0.2326``-0.2326`
`0``-28.7916``28.7916`
`12``26``-31.5832`


`R_3 larr R_3-12xx R_1`

 = 
`1``-0.2326``-0.2326`
`0``-28.7916``28.7916`
`0``28.7916``-28.7916`


`R_2 larr R_2-:(-28.7916)`

 = 
`1``-0.2326``-0.2326`
`0``1``-1`
`0``28.7916``-28.7916`


`R_1 larr R_1+0.2326xx R_2`

 = 
`1``0``-0.4653`
`0``1``-1`
`0``28.7916``-28.7916`


`R_3 larr R_3-28.7916xx R_2`

 = 
`1``0``-0.4653`
`0``1``-1`
`0``0``0`


The system associated with the eigenvalue `lamda=57.5832`

`(A' * A-57.5832I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-0.4653`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-0.4653x_3=0,x_2-x_3=0`

`=>x_1=0.4653x_3,x_2=x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=57.5832` is

`v=`
`0.4653x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_1=`
`0.4653`
`1`
`1`
`v_1=`
`0.4653`
`1`
`1`


2. Eigenvectors for `lamda=0.4168`



2. Eigenvectors for `lamda=0.4168`

`A' * A-lamdaI = `
61212
122626
122626
 - `0.4168` 
100
010
001


 = 
61212
122626
122626
 - 
0.416800
00.41680
000.4168

 = 
`5.5832``12``12`
`12``25.5832``26`
`12``26``25.5832`


Now, reduce this matrix
`R_1 larr R_1-:5.5832`

 = 
`1``2.1493``2.1493`
`12``25.5832``26`
`12``26``25.5832`


`R_2 larr R_2-12xx R_1`

 = 
`1``2.1493``2.1493`
`0``-0.2084``0.2084`
`12``26``25.5832`


`R_3 larr R_3-12xx R_1`

 = 
`1``2.1493``2.1493`
`0``-0.2084``0.2084`
`0``0.2084``-0.2084`


`R_2 larr R_2-:(-0.2084)`

 = 
`1``2.1493``2.1493`
`0``1``-1`
`0``0.2084``-0.2084`


`R_1 larr R_1-2.1493xx R_2`

 = 
`1``0``4.2986`
`0``1``-1`
`0``0.2084``-0.2084`


`R_3 larr R_3-0.2084xx R_2`

 = 
`1``0``4.2986`
`0``1``-1`
`0``0``0`


The system associated with the eigenvalue `lamda=0.4168`

`(A' * A-0.4168I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``4.2986`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1+4.2986x_3=0,x_2-x_3=0`

`=>x_1=-4.2986x_3,x_2=x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0.4168` is

`v=`
`-4.2986x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_2=`
`-4.2986`
`1`
`1`
`v_2=`
`-4.2986`
`1`
`1`


3. Eigenvectors for `lamda=0`



3. Eigenvectors for `lamda=0`

`A' * A-lamdaI = `
61212
122626
122626
 - `0` 
100
010
001


 = 
`6``12``12`
`12``26``26`
`12``26``26`


Now, reduce this matrix
`R_1 larr R_1-:6`

 = 
`1``2``2`
`12``26``26`
`12``26``26`


`R_2 larr R_2-12xx R_1`

 = 
`1``2``2`
`0``2``2`
`12``26``26`


`R_3 larr R_3-12xx R_1`

 = 
`1``2``2`
`0``2``2`
`0``2``2`


`R_2 larr R_2-:2`

 = 
`1``2``2`
`0``1``1`
`0``2``2`


`R_1 larr R_1-2xx R_2`

 = 
`1``0``0`
`0``1``1`
`0``2``2`


`R_3 larr R_3-2xx R_2`

 = 
`1``0``0`
`0``1``1`
`0``0``0`


The system associated with the eigenvalue `lamda=0`

`(A' * A-0I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``0`
`0``1``1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1=0,x_2+x_3=0`

`=>x_1=0,x_2=-x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0` is

`v=`
`0`
`-x_3`
`x_3`


Let `x_3=1`

`v_3=`
`0`
`-1`
`1`
`v_3=`
`0`
`-1`
`1`


For Eigenvector-1 `(0.4653,1,1)`, Length L = `sqrt(|0.46527|^2+|1|^2+|1|^2)=1.4888`

So, normalizing gives `v_1=((0.4653)/(1.4888),(1)/(1.4888),(1)/(1.4888))=(0.3125,0.6717,0.6717)`

For Eigenvector-2 `(-4.2986,1,1)`, Length L = `sqrt(|-4.2986|^2+|1|^2+|1|^2)=4.5253`

So, normalizing gives `v_2=((-4.2986)/(4.5253),(1)/(4.5253),(1)/(4.5253))=(-0.9499,0.221,0.221)`

For Eigenvector-3 `(0,-1,1)`, Length L = `sqrt(|0|^2+|-1|^2+|1|^2)=1.4142`

So, normalizing gives `v_3=((0)/(1.4142),(-1)/(1.4142),(1)/(1.4142))=(0,-0.7071,0.7071)`

`1^"st"` SVD Solution using `A*A'`

`:. U = ``[u_1,u_2,u_3]``=`
`0.2182``0.7868``-0.5774`
`-0.5723``0.5824``0.5774`
`0.7905``0.2044``0.5774`


`:. Sigma = `
`sqrt(57.5832)``0``0`
`0``sqrt(0.4168)``0`
`0``0``sqrt(0)`
`=`
`7.5884``0``0`
`0``0.6456``0`
`0``0``0`


`V` is found using formula `v_i=1/sigma_i A^T*u_i`

`:. V = `
`0.3125``0.9498``0`
`0.6717``-0.2212``-0.7071`
`0.6717``-0.2212``0.7071`



`2^"nd"` SVD Solution using `A'*A`

`U` is found using formula `u_i=1/sigma_i A*v_i`

`:. U = `
`0.2182``-0.7867``-0.5774`
`-0.5723``-0.5826``0.5774`
`0.7905``-0.2042``0.5774`


`:. Sigma = `
`sqrt(57.5832)``0``0`
`0``sqrt(0.4168)``0`
`0``0``sqrt(0)`
`=`
`7.5884``0``0`
`0``0.6456``0`
`0``0``0`


`:. V = ``[v_1,v_2,v_3]``=`
`0.3125``-0.9499``0`
`0.6717``0.221``-0.7071`
`0.6717``0.221``0.7071`



Verify `1^"st"` Solution `A = U Sigma V^T`


`U×Sigma`=
`0.21822``0.7868``-0.57735`
`-0.57228``0.58238``0.57735`
`0.79049``0.20442``0.57735`
×
`7.58836``0``0`
`0``0.64559``0`
`0``0``0`


=
`0.21822×7.58836+0.7868×0+(-0.57735)×0``0.21822×0+0.7868×0.64559+(-0.57735)×0``0.21822×0+0.7868×0+(-0.57735)×0`
`-0.57228×7.58836+0.58238×0+0.57735×0``-0.57228×0+0.58238×0.64559+0.57735×0``-0.57228×0+0.58238×0+0.57735×0`
`0.79049×7.58836+0.20442×0+0.57735×0``0.79049×0+0.20442×0.64559+0.57735×0``0.79049×0+0.20442×0+0.57735×0`


=
`1.65593+0+0``0+0.50795+0``0+0+0`
`-4.34267+0+0``0+0.37598+0``0+0+0`
`5.99852+0+0``0+0.13197+0``0+0+0`


=
`1.65593``0.50795``0`
`-4.34267``0.37598``0`
`5.99852``0.13197``0`


`(U × Sigma)×(V^T)`=
`1.65593``0.50795``0`
`-4.34267``0.37598``0`
`5.99852``0.13197``0`
×
`0.31252``0.6717``0.6717`
`0.94983``-0.22119``-0.22119`
`0``-0.70711``0.70711`


=
`1.65593×0.31252+0.50795×0.94983+0×0``1.65593×0.6717+0.50795×(-0.22119)+0×(-0.70711)``1.65593×0.6717+0.50795×(-0.22119)+0×0.70711`
`-4.34267×0.31252+0.37598×0.94983+0×0``-4.34267×0.6717+0.37598×(-0.22119)+0×(-0.70711)``-4.34267×0.6717+0.37598×(-0.22119)+0×0.70711`
`5.99852×0.31252+0.13197×0.94983+0×0``5.99852×0.6717+0.13197×(-0.22119)+0×(-0.70711)``5.99852×0.6717+0.13197×(-0.22119)+0×0.70711`


=
`0.51751+0.48247+0``1.11229+(-0.11235)+0``1.11229+(-0.11235)+0`
`-1.35717+0.35712+0``-2.91697+(-0.08316)+0``-2.91697+(-0.08316)+0`
`1.87466+0.12535+0``4.02921+(-0.02919)+0``4.02921+(-0.02919)+0`


=
`0.99998``0.99994``0.99994`
`-1.00005``-3.00013``-3.00013`
`2.00001``4.00002``4.00002`


`1^"st"` Solution is possible.

`1^"st"` Solution is possible.


Verify `2^"nd"` Solution `A = U Sigma V^T`


`U×Sigma`=
`0.21822``-0.78672``-0.57735`
`-0.57228``-0.58257``0.57735`
`0.7905``-0.20415``0.57735`
×
`7.58836``0``0`
`0``0.64559``0`
`0``0``0`


=
`0.21822×7.58836+(-0.78672)×0+(-0.57735)×0``0.21822×0+(-0.78672)×0.64559+(-0.57735)×0``0.21822×0+(-0.78672)×0+(-0.57735)×0`
`-0.57228×7.58836+(-0.58257)×0+0.57735×0``-0.57228×0+(-0.58257)×0.64559+0.57735×0``-0.57228×0+(-0.58257)×0+0.57735×0`
`0.7905×7.58836+(-0.20415)×0+0.57735×0``0.7905×0+(-0.20415)×0.64559+0.57735×0``0.7905×0+(-0.20415)×0+0.57735×0`


=
`1.65593+0+0``0+(-0.5079)+0``0+0+0`
`-4.34267+0+0``0+(-0.3761)+0``0+0+0`
`5.9986+0+0``0+(-0.1318)+0``0+0+0`


=
`1.65593``-0.5079``0`
`-4.34267``-0.3761``0`
`5.9986``-0.1318``0`


`(U × Sigma)×(V^T)`=
`1.65593``-0.5079``0`
`-4.34267``-0.3761``0`
`5.9986``-0.1318``0`
×
`0.31252``0.67169``0.67169`
`-0.94991``0.22098``0.22098`
`0``-0.70711``0.70711`


=
`1.65593×0.31252+(-0.5079)×(-0.94991)+0×0``1.65593×0.67169+(-0.5079)×0.22098+0×(-0.70711)``1.65593×0.67169+(-0.5079)×0.22098+0×0.70711`
`-4.34267×0.31252+(-0.3761)×(-0.94991)+0×0``-4.34267×0.67169+(-0.3761)×0.22098+0×(-0.70711)``-4.34267×0.67169+(-0.3761)×0.22098+0×0.70711`
`5.9986×0.31252+(-0.1318)×(-0.94991)+0×0``5.9986×0.67169+(-0.1318)×0.22098+0×(-0.70711)``5.9986×0.67169+(-0.1318)×0.22098+0×0.70711`


=
`0.51751+0.48246+0``1.11227+(-0.11224)+0``1.11227+(-0.11224)+0`
`-1.35717+0.35726+0``-2.91693+(-0.08311)+0``-2.91693+(-0.08311)+0`
`1.87468+0.1252+0``4.0292+(-0.02912)+0``4.0292+(-0.02912)+0`


=
`0.99997``1.00004``1.00004`
`-0.99991``-3.00004``-3.00004`
`1.99988``4.00007``4.00007`


`2^"nd"` Solution is possible.

`2^"nd"` Solution is possible.



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