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Solution
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Solution provided by AtoZmath.com
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Matrix Diagonalization calculator
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 1. `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
2. `[[6,-2,2],[-2,3,-1],[2,-1,3]]`
3. `[[3,2,4],[2,0,2],[4,2,3]]`
4. `[[1,1,1],[-1,-3,-3],[2,4,4]]`
5. `[[2,3],[4,10]]`
6. `[[5,1],[4,2]]`
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Example1. Find Diagonal Matrix ...
`[[8,-6,2],[-6,7,-4],[2,-4,3]]`Solution:| Here `A` | = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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A can be diagonalized if there exists an invertible matrix P and diagonal matrix D such that `A=PDP^-1`
Find eigenvalues of the matrix `A` `|A-lamdaI|=0` | `(8-lamda)` | `-6` | `2` | | | `-6` | `(7-lamda)` | `-4` | | | `2` | `-4` | `(3-lamda)` | |
| = 0 |
`:.(8-lamda)((7-lamda) × (3-lamda) - (-4) × (-4))-(-6)((-6) × (3-lamda) - (-4) × 2)+2((-6) × (-4) - (7-lamda) × 2)=0` `:.(8-lamda)((21-10lamda+lamda^2)-16)+6((-18+6lamda)-(-8))+2(24-(14-2lamda))=0` `:.(8-lamda)(5-10lamda+lamda^2)+6(-10+6lamda)+2(10+2lamda)=0` `:. (40-85lamda+18lamda^2-lamda^3)+(-60+36lamda)+(20+4lamda)=0` `:.(-lamda^3+18lamda^2-45lamda)=0` `:.-lamda(lamda-3)(lamda-15)=0` `:.` The eigenvalues of the matrix A are given by `lamda=0,3,15`
1. The diagonal matrix D is composed of the eigenvalues | `:.D` | = | | `0` | `0` | `0` | | | `0` | `3` | `0` | | | `0` | `0` | `15` | |
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1. Eigenvectors for `lamda=0` | = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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Now, reduce this matrix `R_1 larr R_1-:8` | = | | `1` `1=8-:8`
`R_1 larr R_1-:8` | `-3/4` `-3/4=-6-:8`
`R_1 larr R_1-:8` | `1/4` `1/4=2-:8`
`R_1 larr R_1-:8` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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`R_2 larr R_2+6xx R_1` | = | | `1` | `-3/4` | `1/4` | | | `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1` | `5/2` `5/2=7+6xx-3/4`
`R_2 larr R_2+6xx R_1` | `-5/2` `-5/2=-4+6xx1/4`
`R_2 larr R_2+6xx R_1` | | | `2` | `-4` | `3` | |
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`R_3 larr R_3-2xx R_1` | = | | `1` | `-3/4` | `1/4` | | | `0` | `5/2` | `-5/2` | | | `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1` | `-5/2` `-5/2=-4-2xx-3/4`
`R_3 larr R_3-2xx R_1` | `5/2` `5/2=3-2xx1/4`
`R_3 larr R_3-2xx R_1` | |
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`R_2 larr R_2xx2/5` | = | | `1` | `-3/4` | `1/4` | | | `0` `0=0xx2/5`
`R_2 larr R_2xx2/5` | `1` `1=5/2xx2/5`
`R_2 larr R_2xx2/5` | `-1` `-1=-5/2xx2/5`
`R_2 larr R_2xx2/5` | | | `0` | `-5/2` | `5/2` | |
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`R_1 larr R_1+3/4xx R_2` | = | | `1` `1=1+3/4xx0`
`R_1 larr R_1+3/4xx R_2` | `0` `0=-3/4+3/4xx1`
`R_1 larr R_1+3/4xx R_2` | `-1/2` `-1/2=1/4+3/4xx-1`
`R_1 larr R_1+3/4xx R_2` | | | `0` | `1` | `-1` | | | `0` | `-5/2` | `5/2` | |
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`R_3 larr R_3+5/2xx R_2` | = | | `1` | `0` | `-1/2` | | | `0` | `1` | `-1` | | | `0` `0=0+5/2xx0`
`R_3 larr R_3+5/2xx R_2` | `0` `0=-5/2+5/2xx1`
`R_3 larr R_3+5/2xx R_2` | `0` `0=5/2+5/2xx-1`
`R_3 larr R_3+5/2xx R_2` | |
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The system associated with the eigenvalue `lamda=0` | `(A-0I)` | | = | | `1` | `0` | `-1/2` | | | `0` | `1` | `-1` | | | `0` | `0` | `0` | |
| | | = | |
`=>x_1-1/2x_3=0,x_2-x_3=0` `=>x_1=1/2x_3,x_2=x_3` `:.` eigenvectors corresponding to the eigenvalue `lamda=0` is Let `x_3=1`
2. Eigenvectors for `lamda=3` | = | | `5` | `-6` | `2` | | | `-6` | `4` | `-4` | | | `2` | `-4` | `0` | |
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Now, reduce this matrix `R_1 larr R_1-:5` | = | | `1` `1=5-:5`
`R_1 larr R_1-:5` | `-6/5` `-6/5=-6-:5`
`R_1 larr R_1-:5` | `2/5` `2/5=2-:5`
`R_1 larr R_1-:5` | | | `-6` | `4` | `-4` | | | `2` | `-4` | `0` | |
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`R_2 larr R_2+6xx R_1` | = | | `1` | `-6/5` | `2/5` | | | `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1` | `-16/5` `-16/5=4+6xx-6/5`
`R_2 larr R_2+6xx R_1` | `-8/5` `-8/5=-4+6xx2/5`
`R_2 larr R_2+6xx R_1` | | | `2` | `-4` | `0` | |
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`R_3 larr R_3-2xx R_1` | = | | `1` | `-6/5` | `2/5` | | | `0` | `-16/5` | `-8/5` | | | `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1` | `-8/5` `-8/5=-4-2xx-6/5`
`R_3 larr R_3-2xx R_1` | `-4/5` `-4/5=0-2xx2/5`
`R_3 larr R_3-2xx R_1` | |
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`R_2 larr R_2xx-5/16` | = | | `1` | `-6/5` | `2/5` | | | `0` `0=0xx-5/16`
`R_2 larr R_2xx-5/16` | `1` `1=-16/5xx-5/16`
`R_2 larr R_2xx-5/16` | `1/2` `1/2=-8/5xx-5/16`
`R_2 larr R_2xx-5/16` | | | `0` | `-8/5` | `-4/5` | |
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`R_1 larr R_1+6/5xx R_2` | = | | `1` `1=1+6/5xx0`
`R_1 larr R_1+6/5xx R_2` | `0` `0=-6/5+6/5xx1`
`R_1 larr R_1+6/5xx R_2` | `1` `1=2/5+6/5xx1/2`
`R_1 larr R_1+6/5xx R_2` | | | `0` | `1` | `1/2` | | | `0` | `-8/5` | `-4/5` | |
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`R_3 larr R_3+8/5xx R_2` | = | | `1` | `0` | `1` | | | `0` | `1` | `1/2` | | | `0` `0=0+8/5xx0`
`R_3 larr R_3+8/5xx R_2` | `0` `0=-8/5+8/5xx1`
`R_3 larr R_3+8/5xx R_2` | `0` `0=-4/5+8/5xx1/2`
`R_3 larr R_3+8/5xx R_2` | |
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The system associated with the eigenvalue `lamda=3` | `(A-3I)` | | = | | `1` | `0` | `1` | | | `0` | `1` | `1/2` | | | `0` | `0` | `0` | |
| | | = | |
`=>x_1+x_3=0,x_2+1/2x_3=0` `=>x_1=-x_3,x_2=-1/2x_3` `:.` eigenvectors corresponding to the eigenvalue `lamda=3` is Let `x_3=1`
3. Eigenvectors for `lamda=15` | = | | `-7` | `-6` | `2` | | | `-6` | `-8` | `-4` | | | `2` | `-4` | `-12` | |
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Now, reduce this matrix `R_1 larr R_1-:-7` | = | | `1` `1=-7-:-7`
`R_1 larr R_1-:-7` | `6/7` `6/7=-6-:-7`
`R_1 larr R_1-:-7` | `-2/7` `-2/7=2-:-7`
`R_1 larr R_1-:-7` | | | `-6` | `-8` | `-4` | | | `2` | `-4` | `-12` | |
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`R_2 larr R_2+6xx R_1` | = | | `1` | `6/7` | `-2/7` | | | `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1` | `-20/7` `-20/7=-8+6xx6/7`
`R_2 larr R_2+6xx R_1` | `-40/7` `-40/7=-4+6xx-2/7`
`R_2 larr R_2+6xx R_1` | | | `2` | `-4` | `-12` | |
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`R_3 larr R_3-2xx R_1` | = | | `1` | `6/7` | `-2/7` | | | `0` | `-20/7` | `-40/7` | | | `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1` | `-40/7` `-40/7=-4-2xx6/7`
`R_3 larr R_3-2xx R_1` | `-80/7` `-80/7=-12-2xx-2/7`
`R_3 larr R_3-2xx R_1` | |
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`R_2 larr R_2xx-7/20` | = | | `1` | `6/7` | `-2/7` | | | `0` `0=0xx-7/20`
`R_2 larr R_2xx-7/20` | `1` `1=-20/7xx-7/20`
`R_2 larr R_2xx-7/20` | `2` `2=-40/7xx-7/20`
`R_2 larr R_2xx-7/20` | | | `0` | `-40/7` | `-80/7` | |
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`R_1 larr R_1-6/7xx R_2` | = | | `1` `1=1-6/7xx0`
`R_1 larr R_1-6/7xx R_2` | `0` `0=6/7-6/7xx1`
`R_1 larr R_1-6/7xx R_2` | `-2` `-2=-2/7-6/7xx2`
`R_1 larr R_1-6/7xx R_2` | | | `0` | `1` | `2` | | | `0` | `-40/7` | `-80/7` | |
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`R_3 larr R_3+40/7xx R_2` | = | | `1` | `0` | `-2` | | | `0` | `1` | `2` | | | `0` `0=0+40/7xx0`
`R_3 larr R_3+40/7xx R_2` | `0` `0=-40/7+40/7xx1`
`R_3 larr R_3+40/7xx R_2` | `0` `0=-80/7+40/7xx2`
`R_3 larr R_3+40/7xx R_2` | |
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The system associated with the eigenvalue `lamda=15` | `(A-15I)` | | = | | `1` | `0` | `-2` | | | `0` | `1` | `2` | | | `0` | `0` | `0` | |
| | | = | |
`=>x_1-2x_3=0,x_2+2x_3=0` `=>x_1=2x_3,x_2=-2x_3` `:.` eigenvectors corresponding to the eigenvalue `lamda=15` is Let `x_3=1`
2. The eigenvectors compose the columns of matrix P | `:.P` | = | | `1/2` | `-1` | `2` | | | `1` | `-1/2` | `-2` | | | `1` | `1` | `1` | |
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3. Now find `P^-1` | `|P|` | = | | `1/2` | `-1` | `2` | | | `1` | `-1/2` | `-2` | | | `1` | `1` | `1` | |
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`=1/2 xx (-1/2 × 1 - (-2) × 1) +1 xx (1 × 1 - (-2) × 1) +2 xx (1 × 1 - (-1/2) × 1)` `=1/2 xx (-1/2 +2) +1 xx (1 +2) +2 xx (1 +1/2)` `=1/2 xx (3/2) - +1 xx (3) +2 xx (3/2)` `= 3/4 +3 +3` `=27/4` | `Adj(P)` | = | | Adj | | `1/2` | `-1` | `2` | | | `1` | `-1/2` | `-2` | | | `1` | `1` | `1` | |
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| = | | `+(-1/2 × 1 - (-2) × 1)` | `-(1 × 1 - (-2) × 1)` | `+(1 × 1 - (-1/2) × 1)` | | | `-(-1 × 1 - 2 × 1)` | `+(1/2 × 1 - 2 × 1)` | `-(1/2 × 1 - (-1) × 1)` | | | `+(-1 × (-2) - 2 × (-1/2))` | `-(1/2 × (-2) - 2 × 1)` | `+(1/2 × (-1/2) - (-1) × 1)` | |
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| = | | `+(-1/2 +2)` | `-(1 +2)` | `+(1 +1/2)` | | | `-(-1 -2)` | `+(1/2 -2)` | `-(1/2 +1)` | | | `+(2 +1)` | `-(-1 -2)` | `+(-1/4 +1)` | |
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| = | | `3/2` | `-3` | `3/2` | | | `3` | `-3/2` | `-3/2` | | | `3` | `3` | `3/4` | |
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| = | | `3/2` | `3` | `3` | | | `-3` | `-3/2` | `3` | | | `3/2` | `-3/2` | `3/4` | |
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`"Now, "P^(-1)=1/|P| × Adj(P)` | = | `1/27/4` × | | `3/2` | `3` | `3` | | | `-3` | `-3/2` | `3` | | | `3/2` | `-3/2` | `3/4` | |
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| = | | `2/9` | `4/9` | `4/9` | | | `-4/9` | `-2/9` | `4/9` | | | `2/9` | `-2/9` | `1/9` | |
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| `:.P^-1` | = | | `2/9` | `4/9` | `4/9` | | | `-4/9` | `-2/9` | `4/9` | | | `2/9` | `-2/9` | `1/9` | |
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4. Now verify that `A=PDP^(-1)` | `P×D` | = | | `1/2` | `-1` | `2` | | | `1` | `-1/2` | `-2` | | | `1` | `1` | `1` | |
| × | | `0` | `0` | `0` | | | `0` | `3` | `0` | | | `0` | `0` | `15` | |
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| = | | `1/2×0-1×0+2×0` | `1/2×0-1×3+2×0` | `1/2×0-1×0+2×15` | | | `1×0-1/2×0-2×0` | `1×0-1/2×3-2×0` | `1×0-1/2×0-2×15` | | | `1×0+1×0+1×0` | `1×0+1×3+1×0` | `1×0+1×0+1×15` | |
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| = | | `0+0+0` | `0-3+0` | `0+0+30` | | | `0+0+0` | `0-3/2+0` | `0+0-30` | | | `0+0+0` | `0+3+0` | `0+0+15` | |
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| = | | `0` | `-3` | `30` | | | `0` | `-3/2` | `-30` | | | `0` | `3` | `15` | |
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| `(P × D)×(P^-1)` | = | | `0` | `-3` | `30` | | | `0` | `-3/2` | `-30` | | | `0` | `3` | `15` | |
| × | | `2/9` | `4/9` | `4/9` | | | `-4/9` | `-2/9` | `4/9` | | | `2/9` | `-2/9` | `1/9` | |
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| = | | `0×2/9-3×-4/9+30×2/9` | `0×4/9-3×-2/9+30×-2/9` | `0×4/9-3×4/9+30×1/9` | | | `0×2/9-3/2×-4/9-30×2/9` | `0×4/9-3/2×-2/9-30×-2/9` | `0×4/9-3/2×4/9-30×1/9` | | | `0×2/9+3×-4/9+15×2/9` | `0×4/9+3×-2/9+15×-2/9` | `0×4/9+3×4/9+15×1/9` | |
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| = | | `0+4/3+20/3` | `0+2/3-20/3` | `0-4/3+10/3` | | | `0+2/3-20/3` | `0+1/3+20/3` | `0-2/3-10/3` | | | `0-4/3+10/3` | `0-2/3-10/3` | `0+4/3+5/3` | |
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| = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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| `:.P*D*P^-1` | = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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| And `A` | = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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