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iegen values and eigen vectors

Finding of eigenvalues and eigenvectors

This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial.

Matrix A:(

4

1

7

5

8

2

3

7

8

)

()

CellsClean+−

Find

More:Diagonal matrixJordan decompositionMatrix exponential

Display decimals, number of significant digits:

Clean

  1. Insert in A

    Insert in B

    Clean

     

    1. Find eigenvalues from the characteristic polynomial:

      |4−λ1758−λ2378−λ|=−λ3+20⁢λ2−88⁢λ+243=(?)−(λ+(−15.3))⋅(λ+(−2.36+3.22⋅ⅈ))⋅(λ+(−2.36−3.22⋅ⅈ))

      Details (Triangle's rule)

      Details (Rule of Sarrus)

      Details (Montante's method (Bareiss algorithm))

      Details (Gaussian elimination)

      1. λ1≈15.3
      2. λ2≈2.36−3.22⋅ⅈ
      3. λ3≈2.36+3.22⋅ⅈ
    2. For every λ we find its own vector(s):

      1. λ1≈15.3

        A−λ1⁢I≈(−11.3175−7.28237−7.28)

        A⁢v=λ⁢v *

        (A−λ⁢I)⋅v=0

        So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:

        (−11.31705−7.282037−7.280)

        ⋅(−0.0886)

        ~R1∕(−11.3)→R1−11.3≠0?Divide row 1 by −11.3:a1,1=−11.3⋅(1−11.3)=1a1,2=1⋅(1−11.3)≈−0.0886a1,3=7⋅(1−11.3)≈−0.620a1,4=0⋅(1−11.3)=0(1−0.0886−0.62005−7.282037−7.280)

        ⋅(−5)

        ~R2−5⋅R1→R2?Subtract 5 × row 1 from row 2:a2,1=5−51⋅1=0a2,2=−7.28−51⋅(−0.0886)≈−6.84a2,3=2−51⋅(−0.620)≈5.10a2,4=0−51⋅0=0(1−0.0886−0.62000−6.845.10037−7.280)

        ⋅(−3)

        ~R3−3⋅R1→R3?Subtract 3 × row 1 from row 3:a3,1=3−31⋅1=0a3,2=7−31⋅(−0.0886)≈7.27a3,3=−7.28−31⋅(−0.620)≈−5.42a3,4=0−31⋅0=0(1−0.0886−0.62000−6.845.10007.27−5.420)

        ⋅(−0.146)

        ~R2∕(−6.84)→R2−6.84≠0?Divide row 2 by −6.84:a2,1=0⋅(1−6.84)=0a2,2=−6.84⋅(1−6.84)=1a2,3=5.10⋅(1−6.84)≈−0.746a2,4=0⋅(1−6.84)=0(1−0.0886−0.620001−0.746007.27−5.420)

        ⋅(−7.27)

        ~R3−7.27⋅R2→R3?Subtract 7.27 × row 2 from row 3:a3,1=0−7.271⋅0=0a3,2=7.27−7.271⋅1=0a3,3=−5.42−7.271⋅(−0.746)=0a3,4=0−7.271⋅0=0(1−0.0886−0.620001−0.74600000)

        ⋅(0.0886)

        ~R1−−0.0886⋅R2→R1?Subtract −0.0886 × row 2 from row 1:a1,1=1−−0.08861⋅0=1a1,2=−0.0886−−0.08861⋅1=0a1,3=−0.620−−0.08861⋅(−0.746)≈−0.687a1,4=0−−0.08861⋅0=0(10−0.687001−0.74600000)
        {x1+−0.687⋅x3=0x2+−0.746⋅x3=0 (1)
        • Find the variable x2 from the equation 2 of the system (1):

          x2=0.746⁢x3

        • Find the variable x1 from the equation 1 of the system (1):

          x1=0.687⁢x3

        Answer:

        • x1=0.687⁢x3
        • x2=0.746⁢x3
        • x3=x3
        (−11.3≠0,−6.84≠0)

        General Solution: X=(0.687⁢x30.746⁢x3x3)

        The solution set: {x3⋅(0.6870.7461)}

        Let x3=1, v1≈(0.6870.7461)

      2. λ2≈2.36−3.22⋅ⅈ

        A−λ2⁢I≈(1.64+3.22⋅ⅈ1755.64+3.22⋅ⅈ2375.64+3.22⋅ⅈ)

        A⁢v=λ⁢v *

        (A−λ⁢I)⋅v=0

        So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:

        (1.64+3.22⋅ⅈ17055.64+3.22⋅ⅈ20375.64+3.22⋅ⅈ0)

        ⋅(0.126−0.247⋅ⅈ)

        ~R1∕(1.64+3.22⋅ⅈ)→R11.64+3.22⋅ⅈ≠0?Divide row 1 by 1.64+3.22⋅ⅈ:a1,1=1.64+3.22⋅ⅈ⋅(11.64+3.22⋅ⅈ)=1a1,2=1⋅(11.64+3.22⋅ⅈ)≈0.126−0.247⋅ⅈa1,3=7⋅(11.64+3.22⋅ⅈ)≈0.882−1.73⋅ⅈa1,4=0⋅(11.64+3.22⋅ⅈ)=0(10.126−0.247⋅ⅈ0.882−1.73⋅ⅈ055.64+3.22⋅ⅈ20375.64+3.22⋅ⅈ0)

        ⋅(−5)

        ~R2−5⋅R1→R2?Subtract 5 × row 1 from row 2:a2,1=5−51⋅1=0a2,2=5.64+3.22⋅ⅈ−51⋅(0.126−0.247⋅ⅈ)≈5.01+4.45⋅ⅈa2,3=2−51⋅(0.882−1.73⋅ⅈ)≈−2.41+8.64⋅ⅈa2,4=0−51⋅0=0(10.126−0.247⋅ⅈ0.882−1.73⋅ⅈ005.01+4.45⋅ⅈ−2.41+8.64⋅ⅈ0375.64+3.22⋅ⅈ0)

        ⋅(−3)

        ~R3−3⋅R1→R3?Subtract 3 × row 1 from row 3:a3,1=3−31⋅1=0a3,2=7−31⋅(0.126−0.247⋅ⅈ)≈6.62+0.740⋅ⅈa3,3=5.64+3.22⋅ⅈ−31⋅(0.882−1.73⋅ⅈ)≈3.00+8.40⋅ⅈa3,4=0−31⋅0=0(10.126−0.247⋅ⅈ0.882−1.73⋅ⅈ005.01+4.45⋅ⅈ−2.41+8.64⋅ⅈ006.62+0.740⋅ⅈ3.00+8.40⋅ⅈ0)

        ⋅(0.112−0.0991⋅ⅈ)

        ~R2∕(5.01+4.45⋅ⅈ)→R25.01+4.45⋅ⅈ≠0?Divide row 2 by 5.01+4.45⋅ⅈ:a2,1=0⋅(15.01+4.45⋅ⅈ)=0a2,2=5.01+4.45⋅ⅈ⋅(15.01+4.45⋅ⅈ)=1a2,3=−2.41+8.64⋅ⅈ⋅(15.01+4.45⋅ⅈ)≈0.587+1.20⋅ⅈa2,4=0⋅(15.01+4.45⋅ⅈ)=0(10.126−0.247⋅ⅈ0.882−1.73⋅ⅈ0010.587+1.20⋅ⅈ006.62+0.740⋅ⅈ3.00+8.40⋅ⅈ0)

        ⋅(−6.62−0.740⋅ⅈ)

        ~R3−6.62+0.740⋅ⅈ⋅R2→R3?Subtract 6.62+0.740⋅ⅈ × row 2 from row 3:a3,1=0−6.62+0.740⋅ⅈ1⋅0=0a3,2=6.62+0.740⋅ⅈ−6.62+0.740⋅ⅈ1⋅1=0a3,3=3.00+8.40⋅ⅈ−6.62+0.740⋅ⅈ1⋅(0.587+1.20⋅ⅈ)=0a3,4=0−6.62+0.740⋅ⅈ1⋅0=0(10.126−0.247⋅ⅈ0.882−1.73⋅ⅈ0010.587+1.20⋅ⅈ00000)

        ⋅(−0.126+0.247⋅ⅈ)

        ~R1−0.126−0.247⋅ⅈ⋅R2→R1?Subtract 0.126−0.247⋅ⅈ × row 2 from row 1:a1,1=1−0.126−0.247⋅ⅈ1⋅0=1a1,2=0.126−0.247⋅ⅈ−0.126−0.247⋅ⅈ1⋅1=0a1,3=0.882−1.73⋅ⅈ−0.126−0.247⋅ⅈ1⋅(0.587+1.20⋅ⅈ)≈0.511−1.73⋅ⅈa1,4=0−0.126−0.247⋅ⅈ1⋅0=0(100.511−1.73⋅ⅈ0010.587+1.20⋅ⅈ00000)
        {x1+0.511−1.73⋅ⅈ⋅x3=0x2+0.587+1.20⋅ⅈ⋅x3=0 (1)
        • Find the variable x2 from the equation 2 of the system (1):

          x2=−0.587−1.20⋅ⅈ⁢x3

        • Find the variable x1 from the equation 1 of the system (1):

          x1=−0.511+1.73⋅ⅈ⁢x3

        Answer:

        • x1=−0.511+1.73⋅ⅈ⁢x3
        • x2=−0.587−1.20⋅ⅈ⁢x3
        • x3=x3
        (1.64+3.22⋅ⅈ≠0,5.01+4.45⋅ⅈ≠0)

        General Solution: X=(−0.511+1.73⋅ⅈ⁢x3−0.587−1.20⋅ⅈ⁢x3x3)

        The solution set: {x3⋅(−0.511+1.73⋅ⅈ−0.587−1.20⋅ⅈ1)}

        Let x3=1, v2≈(−0.511+1.73⋅ⅈ−0.587−1.20⋅ⅈ1)

      3. λ3≈2.36+3.22⋅ⅈ

        A−λ3⁢I≈(1.64−3.22⋅ⅈ1755.64−3.22⋅ⅈ2375.64−3.22⋅ⅈ)

        A⁢v=λ⁢v *

        (A−λ⁢I)⋅v=0

        So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:

        (1.64−3.22⋅ⅈ17055.64−3.22⋅ⅈ20375.64−3.22⋅ⅈ0)

        ⋅(0.126+0.247⋅ⅈ)

        ~R1∕(1.64−3.22⋅ⅈ)→R11.64−3.22⋅ⅈ≠0?Divide row 1 by 1.64−3.22⋅ⅈ:a1,1=1.64−3.22⋅ⅈ⋅(11.64−3.22⋅ⅈ)=1a1,2=1⋅(11.64−3.22⋅ⅈ)≈0.126+0.247⋅ⅈa1,3=7⋅(11.64−3.22⋅ⅈ)≈0.882+1.73⋅ⅈa1,4=0⋅(11.64−3.22⋅ⅈ)=0(10.126+0.247⋅ⅈ0.882+1.73⋅ⅈ055.64−3.22⋅ⅈ20375.64−3.22⋅ⅈ0)

        ⋅(−5)

        ~R2−5⋅R1→R2?Subtract 5 × row 1 from row 2:a2,1=5−51⋅1=0a2,2=5.64−3.22⋅ⅈ−51⋅(0.126+0.247⋅ⅈ)≈5.01−4.45⋅ⅈa2,3=2−51⋅(0.882+1.73⋅ⅈ)≈−2.41−8.64⋅ⅈa2,4=0−51⋅0=0(10.126+0.247⋅ⅈ0.882+1.73⋅ⅈ005.01−4.45⋅ⅈ−2.41−8.64⋅ⅈ0375.64−3.22⋅ⅈ0)

        ⋅(−3)

        ~R3−3⋅R1→R3?Subtract 3 × row 1 from row 3:a3,1=3−31⋅1=0a3,2=7−31⋅(0.126+0.247⋅ⅈ)≈6.62−0.740⋅ⅈa3,3=5.64−3.22⋅ⅈ−31⋅(0.882+1.73⋅ⅈ)≈3.00−8.40⋅ⅈa3,4=0−31⋅0=0(10.126+0.247⋅ⅈ0.882+1.73⋅ⅈ005.01−4.45⋅ⅈ−2.41−8.64⋅ⅈ006.62−0.740⋅ⅈ3.00−8.40⋅ⅈ0)

        ⋅(0.112+0.0991⋅ⅈ)

        ~R2∕(5.01−4.45⋅ⅈ)→R25.01−4.45⋅ⅈ≠0?Divide row 2 by 5.01−4.45⋅ⅈ:a2,1=0⋅(15.01−4.45⋅ⅈ)=0a2,2=5.01−4.45⋅ⅈ⋅(15.01−4.45⋅ⅈ)=1a2,3=−2.41−8.64⋅ⅈ⋅(15.01−4.45⋅ⅈ)≈0.587−1.20⋅ⅈa2,4=0⋅(15.01−4.45⋅ⅈ)=0(10.126+0.247⋅ⅈ0.882+1.73⋅ⅈ0010.587−1.20⋅ⅈ006.62−0.740⋅ⅈ3.00−8.40⋅ⅈ0)

        ⋅(−6.62+0.740⋅ⅈ)

        ~R3−6.62−0.740⋅ⅈ⋅R2→R3?Subtract 6.62−0.740⋅ⅈ × row 2 from row 3:a3,1=0−6.62−0.740⋅ⅈ1⋅0=0a3,2=6.62−0.740⋅ⅈ−6.62−0.740⋅ⅈ1⋅1=0a3,3=3.00−8.40⋅ⅈ−6.62−0.740⋅ⅈ1⋅(0.587−1.20⋅ⅈ)=0a3,4=0−6.62−0.740⋅ⅈ1⋅0=0(10.126+0.247⋅ⅈ0.882+1.73⋅ⅈ0010.587−1.20⋅ⅈ00000)

        ⋅(−0.126−0.247⋅ⅈ)

        ~R1−0.126+0.247⋅ⅈ⋅R2→R1?Subtract 0.126+0.247⋅ⅈ × row 2 from row 1:a1,1=1−0.126+0.247⋅ⅈ1⋅0=1a1,2=0.126+0.247⋅ⅈ−0.126+0.247⋅ⅈ1⋅1=0a1,3=0.882+1.73⋅ⅈ−0.126+0.247⋅ⅈ1⋅(0.587−1.20⋅ⅈ)≈0.511+1.73⋅ⅈa1,4=0−0.126+0.247⋅ⅈ1⋅0=0(100.511+1.73⋅ⅈ0010.587−1.20⋅ⅈ00000)
        {x1+0.511+1.73⋅ⅈ⋅x3=0x2+0.587−1.20⋅ⅈ⋅x3=0 (1)
        • Find the variable x2 from the equation 2 of the system (1):

          x2=−0.587+1.20⋅ⅈ⁢x3

        • Find the variable x1 from the equation 1 of the system (1):

          x1=−0.511−1.73⋅ⅈ⁢x3

        Answer:

        • x1=−0.511−1.73⋅ⅈ⁢x3
        • x2=−0.587+1.20⋅ⅈ⁢x3
        • x3=x3
        (1.64−3.22⋅ⅈ≠0,5.01−4.45⋅ⅈ≠0)

        General Solution: X=(−0.511−1.73⋅ⅈ⁢x3−0.587+1.20⋅ⅈ⁢x3x3)

        The solution set: {x3⋅(−0.511−1.73⋅ⅈ−0.587+1.20⋅ⅈ1)}

        Let x3=1, v3≈(−0.511−1.73⋅ⅈ−0.587+1.20⋅ⅈ1)

  2. Insert in A

    Insert in B

    Clean

     

    (417582378)≈(0.687−0.511+1.734⋅ⅈ−0.511−1.734⋅ⅈ0.746−0.587−1.202⋅ⅈ−0.587+1.202⋅ⅈ111)⋅(15.2820002.359−3.215⋅ⅈ0002.359+3.215⋅ⅈ)⋅(0.3210.4620.435−0.160−0.178⋅ⅈ−0.231+0.160⋅ⅈ0.282+0.003⋅ⅈ−0.160+0.178⋅ⅈ−0.231−0.160⋅ⅈ0.282−0.003⋅ⅈ)

    Details

  3. Insert in A

    Insert in B

    Clean

     

    1. Find eigenvalues from the characteristic polynomial:

      |4−λ1758−λ2378−λ|=−λ3+20⁢λ2−88⁢λ+243=(?)−(λ+(−15.282))⋅(λ+(−2.359+3.215⋅ⅈ))⋅(λ+(−2.359−3.215⋅ⅈ))

      Details (Triangle's rule)

      Details (Rule of Sarrus)

      Details (Montante's method (Bareiss algorithm))

      Details (Gaussian elimination)

      1. λ1≈15.282
      2. λ2≈2.359−3.215⋅ⅈ
      3. λ3≈2.359+3.215⋅ⅈ
    2. For every λ we find its own vector(s):

      1. λ1≈15.282

        A−λ1⁢I≈(−11.282175−7.282237−7.282)

        A⁢v=λ⁢v *

        (A−λ⁢I)⋅v=0

        So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:

        (−11.2821705−7.2822037−7.2820)

        ⋅(−0.089)

        ~R1∕(−11.282)→R1−11.282≠0?Divide row 1 by −11.282:a1,1=−11.282⋅(1−11.282)=1a1,2=1⋅(1−11.282)≈−0.089a1,3=7⋅(1−11.282)≈−0.620a1,4=0⋅(1−11.282)=0(1−0.089−0.62005−7.2822037−7.2820)

        ⋅(−5)

        ~R2−5⋅R1→R2?Subtract 5 × row 1 from row 2:a2,1=5−51⋅1=0a2,2=−7.282−51⋅(−0.089)≈−6.839a2,3=2−51⋅(−0.620)≈5.102a2,4=0−51⋅0=0(1−0.089−0.62000−6.8395.102037−7.2820)

        ⋅(−3)

        ~R3−3⋅R1→R3?Subtract 3 × row 1 from row 3:a3,1=3−31⋅1=0a3,2=7−31⋅(−0.089)≈7.266a3,3=−7.282−31⋅(−0.620)≈−5.421a3,4=0−31⋅0=0(1−0.089−0.62000−6.8395.102007.266−5.4210)

        ⋅(−0.146)

        ~R2∕(−6.839)→R2−6.839≠0?Divide row 2 by −6.839:a2,1=0⋅(1−6.839)=0a2,2=−6.839⋅(1−6.839)=1a2,3=5.102⋅(1−6.839)≈−0.746a2,4=0⋅(1−6.839)=0(1−0.089−0.620001−0.746007.266−5.4210)

        ⋅(−7.266)

        ~R3−7.266⋅R2→R3?Subtract 7.266 × row 2 from row 3:a3,1=0−7.2661⋅0=0a3,2=7.266−7.2661⋅1=0a3,3=−5.421−7.2661⋅(−0.746)=0a3,4=0−7.2661⋅0=0(1−0.089−0.620001−0.74600000)

        ⋅(0.089)

        ~R1−−0.089⋅R2→R1?Subtract −0.089 × row 2 from row 1:a1,1=1−−0.0891⋅0=1a1,2=−0.089−−0.0891⋅1=0a1,3=−0.620−−0.0891⋅(−0.746)≈−0.687a1,4=0−−0.0891⋅0=0(10−0.687001−0.74600000)
        {x1+−0.687⋅x3=0x2+−0.746⋅x3=0 (1)
        • Find the variable x2 from the equation 2 of the system (1):

          x2=0.746⁢x3

        • Find the variable x1 from the equation 1 of the system (1):

          x1=0.687⁢x3

        Answer:

        • x1=0.687⁢x3
        • x2=0.746⁢x3
        • x3=x3
        (−11.282≠0,−6.839≠0)

        General Solution: X=(0.687⁢x30.746⁢x3x3)

        The solution set: {x3⋅(0.6870.7461)}

        Let x3=1, v1≈(0.6870.7461)

      2. λ2≈2.359−3.215⋅ⅈ

        A−λ2⁢I≈(1.641+3.215⋅ⅈ1755.641+3.215⋅ⅈ2375.641+3.215⋅ⅈ)

        A⁢v=λ⁢v *

        (A−λ⁢I)⋅v=0

        So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:

        (1.641+3.215⋅ⅈ17055.641+3.215⋅ⅈ20375.641+3.215⋅ⅈ0)

        ⋅(0.126−0.247⋅ⅈ)

        ~R1∕(1.641+3.215⋅ⅈ)→R11.641+3.215⋅ⅈ≠0?Divide row 1 by 1.641+3.215⋅ⅈ:a1,1=1.641+3.215⋅ⅈ⋅(11.641+3.215⋅ⅈ)=1a1,2=1⋅(11.641+3.215⋅ⅈ)≈0.126−0.247⋅ⅈa1,3=7⋅(11.641+3.215⋅ⅈ)≈0.882−1.727⋅ⅈa1,4=0⋅(11.641+3.215⋅ⅈ)=0(10.126−0.247⋅ⅈ0.882−1.727⋅ⅈ055.641+3.215⋅ⅈ20375.641+3.215⋅ⅈ0)

        ⋅(−5)

        ~R2−5⋅R1→R2?Subtract 5 × row 1 from row 2:a2,1=5−51⋅1=0a2,2=5.641+3.215⋅ⅈ−51⋅(0.126−0.247⋅ⅈ)≈5.011+4.449⋅ⅈa2,3=2−51⋅(0.882−1.727⋅ⅈ)≈−2.408+8.636⋅ⅈa2,4=0−51⋅0=0(10.126−0.247⋅ⅈ0.882−1.727⋅ⅈ005.011+4.449⋅ⅈ−2.408+8.636⋅ⅈ0375.641+3.215⋅ⅈ0)

        ⋅(−3)

        ~R3−3⋅R1→R3?Subtract 3 × row 1 from row 3:a3,1=3−31⋅1=0a3,2=7−31⋅(0.126−0.247⋅ⅈ)≈6.622+0.740⋅ⅈa3,3=5.641+3.215⋅ⅈ−31⋅(0.882−1.727⋅ⅈ)≈2.996+8.397⋅ⅈa3,4=0−31⋅0=0(10.126−0.247⋅ⅈ0.882−1.727⋅ⅈ005.011+4.449⋅ⅈ−2.408+8.636⋅ⅈ006.622+0.740⋅ⅈ2.996+8.397⋅ⅈ0)

        ⋅(0.112−0.099⋅ⅈ)

        ~R2∕(5.011+4.449⋅ⅈ)→R25.011+4.449⋅ⅈ≠0?Divide row 2 by 5.011+4.449⋅ⅈ:a2,1=0⋅(15.011+4.449⋅ⅈ)=0a2,2=5.011+4.449⋅ⅈ⋅(15.011+4.449⋅ⅈ)=1a2,3=−2.408+8.636⋅ⅈ⋅(15.011+4.449⋅ⅈ)≈0.587+1.202⋅ⅈa2,4=0⋅(15.011+4.449⋅ⅈ)=0(10.126−0.247⋅ⅈ0.882−1.727⋅ⅈ0010.587+1.202⋅ⅈ006.622+0.740⋅ⅈ2.996+8.397⋅ⅈ0)

        ⋅(−6.622−0.740⋅ⅈ)

        ~R3−6.622+0.740⋅ⅈ⋅R2→R3?Subtract 6.622+0.740⋅ⅈ × row 2 from row 3:a3,1=0−6.622+0.740⋅ⅈ1⋅0=0a3,2=6.622+0.740⋅ⅈ−6.622+0.740⋅ⅈ1⋅1=0a3,3=2.996+8.397⋅ⅈ−6.622+0.740⋅ⅈ1⋅(0.587+1.202⋅ⅈ)=0a3,4=0−6.622+0.740⋅ⅈ1⋅0=0(10.126−0.247⋅ⅈ0.882−1.727⋅ⅈ0010.587+1.202⋅ⅈ00000)

        ⋅(−0.126+0.247⋅ⅈ)

        ~R1−0.126−0.247⋅ⅈ⋅R2→R1?Subtract 0.126−0.247⋅ⅈ × row 2 from row 1:a1,1=1−0.126−0.247⋅ⅈ1⋅0=1a1,2=0.126−0.247⋅ⅈ−0.126−0.247⋅ⅈ1⋅1=0a1,3=0.882−1.727⋅ⅈ−0.126−0.247⋅ⅈ1⋅(0.587+1.202⋅ⅈ)≈0.511−1.734⋅ⅈa1,4=0−0.126−0.247⋅ⅈ1⋅0=0(100.511−1.734⋅ⅈ0010.587+1.202⋅ⅈ00000)
        {x1+0.511−1.734⋅ⅈ⋅x3=0x2+0.587+1.202⋅ⅈ⋅x3=0 (1)
        • Find the variable x2 from the equation 2 of the system (1):

          x2=−0.587−1.202⋅ⅈ⁢x3

        • Find the variable x1 from the equation 1 of the system (1):

          x1=−0.511+1.734⋅ⅈ⁢x3

        Answer:

        • x1=−0.511+1.734⋅ⅈ⁢x3
        • x2=−0.587−1.202⋅ⅈ⁢x3
        • x3=x3
        (1.641+3.215⋅ⅈ≠0,5.011+4.449⋅ⅈ≠0)

        General Solution: X=(−0.511+1.734⋅ⅈ⁢x3−0.587−1.202⋅ⅈ⁢x3x3)

        The solution set: {x3⋅(−0.511+1.734⋅ⅈ−0.587−1.202⋅ⅈ1)}

        Let x3=1, v2≈(−0.511+1.734⋅ⅈ−0.587−1.202⋅ⅈ1)

      3. λ3≈2.359+3.215⋅ⅈ

        A−λ3⁢I≈(1.641−3.215⋅ⅈ1755.641−3.215⋅ⅈ2375.641−3.215⋅ⅈ)

        A⁢v=λ⁢v *

        (A−λ⁢I)⋅v=0

        So we have a homogeneous system of linear equations, we solve it by Gaussian Elimination:

        (1.641−3.215⋅ⅈ17055.641−3.215⋅ⅈ20375.641−3.215⋅ⅈ0)

        ⋅(0.126+0.247⋅ⅈ)

        ~R1∕(1.641−3.215⋅ⅈ)→R11.641−3.215⋅ⅈ≠0?Divide row 1 by 1.641−3.215⋅ⅈ:a1,1=1.641−3.215⋅ⅈ⋅(11.641−3.215⋅ⅈ)=1a1,2=1⋅(11.641−3.215⋅ⅈ)≈0.126+0.247⋅ⅈa1,3=7⋅(11.641−3.215⋅ⅈ)≈0.882+1.727⋅ⅈa1,4=0⋅(11.641−3.215⋅ⅈ)=0(10.126+0.247⋅ⅈ0.882+1.727⋅ⅈ055.641−3.215⋅ⅈ20375.641−3.215⋅ⅈ0)

        ⋅(−5)

        ~R2−5⋅R1→R2?Subtract 5 × row 1 from row 2:a2,1=5−51⋅1=0a2,2=5.641−3.215⋅ⅈ−51⋅(0.126+0.247⋅ⅈ)≈5.011−4.449⋅ⅈa2,3=2−51⋅(0.882+1.727⋅ⅈ)≈−2.408−8.636⋅ⅈa2,4=0−51⋅0=0(10.126+0.247⋅ⅈ0.882+1.727⋅ⅈ005.011−4.449⋅ⅈ−2.408−8.636⋅ⅈ0375.641−3.215⋅ⅈ0)

        ⋅(−3)

        ~R3−3⋅R1→R3?Subtract 3 × row 1 from row 3:a3,1=3−31⋅1=0a3,2=7−31⋅(0.126+0.247⋅ⅈ)≈6.622−0.740⋅ⅈa3,3=5.641−3.215⋅ⅈ−31⋅(0.882+1.727⋅ⅈ)≈2.996−8.397⋅ⅈa3,4=0−31⋅0=0(10.126+0.247⋅ⅈ0.882+1.727⋅ⅈ005.011−4.449⋅ⅈ−2.408−8.636⋅ⅈ006.622−0.740⋅ⅈ2.996−8.397⋅ⅈ0)

        ⋅(0.112+0.099⋅ⅈ)

        ~R2∕(5.011−4.449⋅ⅈ)→R25.011−4.449⋅ⅈ≠0?Divide row 2 by 5.011−4.449⋅ⅈ:a2,1=0⋅(15.011−4.449⋅ⅈ)=0a2,2=5.011−4.449⋅ⅈ⋅(15.011−4.449⋅ⅈ)=1a2,3=−2.408−8.636⋅ⅈ⋅(15.011−4.449⋅ⅈ)≈0.587−1.202⋅ⅈa2,4=0⋅(15.011−4.449⋅ⅈ)=0(10.126+0.247⋅ⅈ0.882+1.727⋅ⅈ0010.587−1.202⋅ⅈ006.622−0.740⋅ⅈ2.996−8.397⋅ⅈ0)

        ⋅(−6.622+0.740⋅ⅈ)

        ~R3−6.622−0.740⋅ⅈ⋅R2→R3?Subtract 6.622−0.740⋅ⅈ × row 2 from row 3:a3,1=0−6.622−0.740⋅ⅈ1⋅0=0a3,2=6.622−0.740⋅ⅈ−6.622−0.740⋅ⅈ1⋅1=0a3,3=2.996−8.397⋅ⅈ−6.622−0.740⋅ⅈ1⋅(0.587−1.202⋅ⅈ)=0a3,4=0−6.622−0.740⋅ⅈ1⋅0=0(10.126+0.247⋅ⅈ0.882+1.727⋅ⅈ0010.587−1.202⋅ⅈ00000)

        ⋅(−0.126−0.247⋅ⅈ)

        ~R1−0.126+0.247⋅ⅈ⋅R2→R1?Subtract 0.126+0.247⋅ⅈ × row 2 from row 1:a1,1=1−0.126+0.247⋅ⅈ1⋅0=1a1,2=0.126+0.247⋅ⅈ−0.126+0.247⋅ⅈ1⋅1=0a1,3=0.882+1.727⋅ⅈ−0.126+0.247⋅ⅈ1⋅(0.587−1.202⋅ⅈ)≈0.511+1.734⋅ⅈa1,4=0−0.126+0.247⋅ⅈ1⋅0=0(100.511+1.734⋅ⅈ0010.587−1.202⋅ⅈ00000)
        {x1+0.511+1.734⋅ⅈ⋅x3=0x2+0.587−1.202⋅ⅈ⋅x3=0 (1)
        • Find the variable x2 from the equation 2 of the system (1):

          x2=−0.587+1.202⋅ⅈ⁢x3

        • Find the variable x1 from the equation 1 of the system (1):

          x1=−0.511−1.734⋅ⅈ⁢x3

        Answer:

        • x1=−0.511−1.734⋅ⅈ⁢x3
        • x2=−0.587+1.202⋅ⅈ⁢x3
        • x3=x3
        (1.641−3.215⋅ⅈ≠0,5.011−4.449⋅ⅈ≠0)

        General Solution: X=(−0.511−1.734⋅ⅈ⁢x3−0.587+1.202⋅ⅈ⁢x3x3)

        The solution set: {x3⋅(−0.511−1.734⋅ⅈ−0.587+1.202⋅ⅈ1)}

        Let x3=1, v3≈(−0.511−1.734⋅ⅈ−0.587+1.202⋅ⅈ1)

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