Find solution for matrix B

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$$A*B^T*A^-1=beginpmatrix
8_ & 0_ & -1_\
11_ & -6_ & -3_\
-7_ & 11_ & 4_
endpmatrix$$



Find matrix B



After I transfer it to: $$B^T=A^-1beginpmatrix
8_ & 0_ & -1_\
11_ & -6_ & -3_\
-7_ & 11_ & 4_
endpmatrixA$$



Is it valid to transpose both sides of the equation, so I can get (possibly) the value of B?



Is this general direction for solving the problem, legit?




linear-algebra matrices




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  • Hint: $(ABC)^T=C^TB^TA^T$
    – Laars Helenius
    Aug 6 at 19:31










  • Do you have $A$ specified?
    – mvw
    Aug 6 at 19:35










  • This looks correct, and yes, you can apply the transpose to both sides of the equation to solve for $B$, although your answer would be in terms of the matrix, $A$, unless you were given this.
    – Matt.P
    Aug 6 at 19:35











  • ok thanks. Solved it
    – user6394019
    Aug 6 at 19:36














up vote
0
down vote

favorite












$$A*B^T*A^-1=beginpmatrix
8_ & 0_ & -1_\
11_ & -6_ & -3_\
-7_ & 11_ & 4_
endpmatrix$$



Find matrix B



After I transfer it to: $$B^T=A^-1beginpmatrix
8_ & 0_ & -1_\
11_ & -6_ & -3_\
-7_ & 11_ & 4_
endpmatrixA$$



Is it valid to transpose both sides of the equation, so I can get (possibly) the value of B?



Is this general direction for solving the problem, legit?




linear-algebra matrices




share|cite|improve this question



















  • Hint: $(ABC)^T=C^TB^TA^T$
    – Laars Helenius
    Aug 6 at 19:31










  • Do you have $A$ specified?
    – mvw
    Aug 6 at 19:35










  • This looks correct, and yes, you can apply the transpose to both sides of the equation to solve for $B$, although your answer would be in terms of the matrix, $A$, unless you were given this.
    – Matt.P
    Aug 6 at 19:35











  • ok thanks. Solved it
    – user6394019
    Aug 6 at 19:36












up vote
0
down vote

favorite









up vote
0
down vote

favorite











$$A*B^T*A^-1=beginpmatrix
8_ & 0_ & -1_\
11_ & -6_ & -3_\
-7_ & 11_ & 4_
endpmatrix$$



Find matrix B



After I transfer it to: $$B^T=A^-1beginpmatrix
8_ & 0_ & -1_\
11_ & -6_ & -3_\
-7_ & 11_ & 4_
endpmatrixA$$



Is it valid to transpose both sides of the equation, so I can get (possibly) the value of B?



Is this general direction for solving the problem, legit?




linear-algebra matrices




share|cite|improve this question











$$A*B^T*A^-1=beginpmatrix
8_ & 0_ & -1_\
11_ & -6_ & -3_\
-7_ & 11_ & 4_
endpmatrix$$



Find matrix B



After I transfer it to: $$B^T=A^-1beginpmatrix
8_ & 0_ & -1_\
11_ & -6_ & -3_\
-7_ & 11_ & 4_
endpmatrixA$$



Is it valid to transpose both sides of the equation, so I can get (possibly) the value of B?



Is this general direction for solving the problem, legit?





linear-algebra matrices





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 6 at 19:28









user6394019

30311




30311











  • Hint: $(ABC)^T=C^TB^TA^T$
    – Laars Helenius
    Aug 6 at 19:31










  • Do you have $A$ specified?
    – mvw
    Aug 6 at 19:35










  • This looks correct, and yes, you can apply the transpose to both sides of the equation to solve for $B$, although your answer would be in terms of the matrix, $A$, unless you were given this.
    – Matt.P
    Aug 6 at 19:35











  • ok thanks. Solved it
    – user6394019
    Aug 6 at 19:36
















  • Hint: $(ABC)^T=C^TB^TA^T$
    – Laars Helenius
    Aug 6 at 19:31










  • Do you have $A$ specified?
    – mvw
    Aug 6 at 19:35










  • This looks correct, and yes, you can apply the transpose to both sides of the equation to solve for $B$, although your answer would be in terms of the matrix, $A$, unless you were given this.
    – Matt.P
    Aug 6 at 19:35











  • ok thanks. Solved it
    – user6394019
    Aug 6 at 19:36















Hint: $(ABC)^T=C^TB^TA^T$
– Laars Helenius
Aug 6 at 19:31




Hint: $(ABC)^T=C^TB^TA^T$
– Laars Helenius
Aug 6 at 19:31












Do you have $A$ specified?
– mvw
Aug 6 at 19:35




Do you have $A$ specified?
– mvw
Aug 6 at 19:35












This looks correct, and yes, you can apply the transpose to both sides of the equation to solve for $B$, although your answer would be in terms of the matrix, $A$, unless you were given this.
– Matt.P
Aug 6 at 19:35





This looks correct, and yes, you can apply the transpose to both sides of the equation to solve for $B$, although your answer would be in terms of the matrix, $A$, unless you were given this.
– Matt.P
Aug 6 at 19:35













ok thanks. Solved it
– user6394019
Aug 6 at 19:36




ok thanks. Solved it
– user6394019
Aug 6 at 19:36










1 Answer
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Let right hand side of matrix be C. Eigen values of C are : 7.7, 0.42, -2.127. As they are distinct, C matrix is diagonalisable. So, $C = ADA^-1$



D is diagonal matrix with diagonal elements as eigen values. A is some non-singular matrix with columns as eigen vectors






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    Let right hand side of matrix be C. Eigen values of C are : 7.7, 0.42, -2.127. As they are distinct, C matrix is diagonalisable. So, $C = ADA^-1$



    D is diagonal matrix with diagonal elements as eigen values. A is some non-singular matrix with columns as eigen vectors






    share|cite|improve this answer

























      up vote
      0
      down vote













      Let right hand side of matrix be C. Eigen values of C are : 7.7, 0.42, -2.127. As they are distinct, C matrix is diagonalisable. So, $C = ADA^-1$



      D is diagonal matrix with diagonal elements as eigen values. A is some non-singular matrix with columns as eigen vectors






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Let right hand side of matrix be C. Eigen values of C are : 7.7, 0.42, -2.127. As they are distinct, C matrix is diagonalisable. So, $C = ADA^-1$



        D is diagonal matrix with diagonal elements as eigen values. A is some non-singular matrix with columns as eigen vectors






        share|cite|improve this answer













        Let right hand side of matrix be C. Eigen values of C are : 7.7, 0.42, -2.127. As they are distinct, C matrix is diagonalisable. So, $C = ADA^-1$



        D is diagonal matrix with diagonal elements as eigen values. A is some non-singular matrix with columns as eigen vectors







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 6 at 19:40









        Magneto

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