Find the inverse of the following matrix: $A$












2












$begingroup$


Find the inverseof the following matrix:



$$A= begin{bmatrix}
0 & 1 & 0 & 0 &0 \
1 & 0 & 0 & 0 &0 \
0 & 0 & 0 & 0&1 \
0 & 0 & 1 & 0 &0 \0&0&0&1&0
end{bmatrix}$$



My attempt : I think the inverse will not exist because $det A=0 $



Is it True ?



Any hints/solution will be appreciated.






linear-algebra







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$endgroup$








  • 1




    $begingroup$
    Columns are independent so the inverse exists.
    $endgroup$
    – Widawensen
    Jan 7 at 15:17
















2












$begingroup$


Find the inverseof the following matrix:



$$A= begin{bmatrix}
0 & 1 & 0 & 0 &0 \
1 & 0 & 0 & 0 &0 \
0 & 0 & 0 & 0&1 \
0 & 0 & 1 & 0 &0 \0&0&0&1&0
end{bmatrix}$$



My attempt : I think the inverse will not exist because $det A=0 $



Is it True ?



Any hints/solution will be appreciated.






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Columns are independent so the inverse exists.
    $endgroup$
    – Widawensen
    Jan 7 at 15:17














2












2








2


0



$begingroup$


Find the inverseof the following matrix:



$$A= begin{bmatrix}
0 & 1 & 0 & 0 &0 \
1 & 0 & 0 & 0 &0 \
0 & 0 & 0 & 0&1 \
0 & 0 & 1 & 0 &0 \0&0&0&1&0
end{bmatrix}$$



My attempt : I think the inverse will not exist because $det A=0 $



Is it True ?



Any hints/solution will be appreciated.






linear-algebra







share|cite|improve this question











$endgroup$




Find the inverseof the following matrix:



$$A= begin{bmatrix}
0 & 1 & 0 & 0 &0 \
1 & 0 & 0 & 0 &0 \
0 & 0 & 0 & 0&1 \
0 & 0 & 1 & 0 &0 \0&0&0&1&0
end{bmatrix}$$



My attempt : I think the inverse will not exist because $det A=0 $



Is it True ?



Any hints/solution will be appreciated.






linear-algebra




linear-algebra






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share|cite|improve this question













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edited Jan 7 at 12:16









user376343

3,3933826




3,3933826










asked Jan 7 at 11:33









jasminejasmine

1,701417




1,701417








  • 1




    $begingroup$
    Columns are independent so the inverse exists.
    $endgroup$
    – Widawensen
    Jan 7 at 15:17














  • 1




    $begingroup$
    Columns are independent so the inverse exists.
    $endgroup$
    – Widawensen
    Jan 7 at 15:17








1




1




$begingroup$
Columns are independent so the inverse exists.
$endgroup$
– Widawensen
Jan 7 at 15:17




$begingroup$
Columns are independent so the inverse exists.
$endgroup$
– Widawensen
Jan 7 at 15:17










5 Answers
5






active

oldest

votes


















1












$begingroup$

Interchange rows three times ($R_1-R_2, R_3-R_4,R_4-R_5$):
$$det(A)= begin{vmatrix}
0 & color{blue}1 & 0 & 0 &0 \
color{red}1 & 0 & 0 & 0 &0 \
0 & 0 & 0 & 0&color{green}1 \
0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
end{vmatrix}=(-1)^3begin{vmatrix}
color{red}1 & 0 & 0 & 0 &0 \
0 & color{blue}1 & 0 & 0 &0 \
0 & 0 & color{brown}1 & 0& 0 \
0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
end{vmatrix}=-1ne 0.$$
Consider the expanded matrix and perform the same row interchanges:
$$begin{vmatrix}
0 & color{blue}1 & 0 & 0 &0 \
color{red}1 & 0 & 0 & 0 &0 \
0 & 0 & 0 & 0&color{green}1 \
0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
end{vmatrix}begin{vmatrix}
1 & 0 & 0 & 0 &0 \
0 & 1 & 0 & 0 &0 \
0 & 0 & 1 & 0&0 \
0 & 0 & 0 & 1 &0 \0&0&0&0&0
end{vmatrix} Rightarrow begin{vmatrix}
color{red}1 & 0 & 0 & 0 &0 \
0 & color{blue}1 & 0 & 0 &0 \
0 & 0 & color{brown}1 & 0& 0 \
0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
end{vmatrix}begin{vmatrix}
0 & 1 & 0 & 0 &0 \
1 & 0 & 0 & 0 &0 \
0 & 0 & 0 & 1&0 \
0 & 0 & 0 & 0 &1 \
0 & 0 & 1 & 0 &0
end{vmatrix} $$
Hence, the inverse matrix is:
$$A^{-1}=begin{vmatrix}
0 & 1 & 0 & 0 &0 \
1 & 0 & 0 & 0 &0 \
0 & 0 & 0 & 1&0 \
0 & 0 & 0 & 0 &1 \
0 & 0 & 1 & 0 &0
end{vmatrix}$$






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$endgroup$





















    3












    $begingroup$

    Note that the given matrix is a so-called permutation matrix, the inverse of it is the permutation matrix associated to the inverse permutation.




    Later edit: The matrix $A$ works on a "generic vector" $x$ with components $x_1,dots, x_5$ as below by mapping it into a vector with components $x_{sigma(1)}, dots,x_{sigma(5)}$ for a particular permutation $sigma$. Explicitly:



    $$
    A
    begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
    =
    begin{bmatrix}
    0 & 1 & 0 & 0 & 0 \
    1 & 0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 & 1 \
    0 & 0 & 1 & 0 & 0 \
    0 & 0 & 0 & 1 & 0
    end{bmatrix}
    begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
    =
    begin{bmatrix}x_2\x_1\x_5\x_3\x_4end{bmatrix}
    $$
    The corresponding permutation $sigma$ (with the inverse $tau$) is of course
    $$
    sigma
    =
    binom{12345}{21534}=(12)(354) ,
    qquadtext{ with }
    tau:=sigma^{-1}=
    binom{12345}{21453}=(12)(345) .
    $$
    Now associate in the natural way the permutation matrix for $tau$...



    Note: It is important to see at a glance, that this matrix is a permutation matrix. (Exactly one $1$ entry on each row and/or column.) This is part of a structure. (One has a mapping / a representation of the symmetric group into the general linear group of the corresponding dimension over $Bbb Z$, or $Bbb Q$ if we prefer a field.) "Other answers" like writing $A$ as a $(2+3)times(2+3)$ block matrix, then doing the work in smaller pieces, may also lead to a good end, but for my taste, if i see a structural answer, and a computational one (Gauss scheme, block split, adjoint matrix, etc...) the structural one wins for me.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      can u elaborate ur answer
      $endgroup$
      – jasmine
      Jan 7 at 11:47



















    2












    $begingroup$

    This matrix is a permutation matrix and also orthogonal. What do you know about inverses of orthogonal matrices?






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    $endgroup$





















      1












      $begingroup$

      The determinant of the matrix is not $0$, the determinant is $1$.



      Hint:



      Take a look at $A^2, A^3,A^4,dots$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        eigenvalue are $0$...i thinks det $A =0$ @Xum
        $endgroup$
        – jasmine
        Jan 7 at 11:35










      • $begingroup$
        @jasmine None of the eigenvalues of the matrix is $0$. And the determinant is $1$.
        $endgroup$
        – 5xum
        Jan 7 at 11:43



















      1












      $begingroup$

      Note that this matrix is obtained from the identity matrix by exchanging columns (or rows), and that columns (and rows) are still linearly independent, so immediately we have that the determinant is not $0$. Even more precisely, we immediately have that the determinant is $1$ or $-1$. (Immediately= without doing any calculation)






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        5 Answers
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        active

        oldest

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        5 Answers
        5






        active

        oldest

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        active

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        active

        oldest

        votes









        1












        $begingroup$

        Interchange rows three times ($R_1-R_2, R_3-R_4,R_4-R_5$):
        $$det(A)= begin{vmatrix}
        0 & color{blue}1 & 0 & 0 &0 \
        color{red}1 & 0 & 0 & 0 &0 \
        0 & 0 & 0 & 0&color{green}1 \
        0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
        end{vmatrix}=(-1)^3begin{vmatrix}
        color{red}1 & 0 & 0 & 0 &0 \
        0 & color{blue}1 & 0 & 0 &0 \
        0 & 0 & color{brown}1 & 0& 0 \
        0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
        end{vmatrix}=-1ne 0.$$
        Consider the expanded matrix and perform the same row interchanges:
        $$begin{vmatrix}
        0 & color{blue}1 & 0 & 0 &0 \
        color{red}1 & 0 & 0 & 0 &0 \
        0 & 0 & 0 & 0&color{green}1 \
        0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
        end{vmatrix}begin{vmatrix}
        1 & 0 & 0 & 0 &0 \
        0 & 1 & 0 & 0 &0 \
        0 & 0 & 1 & 0&0 \
        0 & 0 & 0 & 1 &0 \0&0&0&0&0
        end{vmatrix} Rightarrow begin{vmatrix}
        color{red}1 & 0 & 0 & 0 &0 \
        0 & color{blue}1 & 0 & 0 &0 \
        0 & 0 & color{brown}1 & 0& 0 \
        0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
        end{vmatrix}begin{vmatrix}
        0 & 1 & 0 & 0 &0 \
        1 & 0 & 0 & 0 &0 \
        0 & 0 & 0 & 1&0 \
        0 & 0 & 0 & 0 &1 \
        0 & 0 & 1 & 0 &0
        end{vmatrix} $$
        Hence, the inverse matrix is:
        $$A^{-1}=begin{vmatrix}
        0 & 1 & 0 & 0 &0 \
        1 & 0 & 0 & 0 &0 \
        0 & 0 & 0 & 1&0 \
        0 & 0 & 0 & 0 &1 \
        0 & 0 & 1 & 0 &0
        end{vmatrix}$$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Interchange rows three times ($R_1-R_2, R_3-R_4,R_4-R_5$):
          $$det(A)= begin{vmatrix}
          0 & color{blue}1 & 0 & 0 &0 \
          color{red}1 & 0 & 0 & 0 &0 \
          0 & 0 & 0 & 0&color{green}1 \
          0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
          end{vmatrix}=(-1)^3begin{vmatrix}
          color{red}1 & 0 & 0 & 0 &0 \
          0 & color{blue}1 & 0 & 0 &0 \
          0 & 0 & color{brown}1 & 0& 0 \
          0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
          end{vmatrix}=-1ne 0.$$
          Consider the expanded matrix and perform the same row interchanges:
          $$begin{vmatrix}
          0 & color{blue}1 & 0 & 0 &0 \
          color{red}1 & 0 & 0 & 0 &0 \
          0 & 0 & 0 & 0&color{green}1 \
          0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
          end{vmatrix}begin{vmatrix}
          1 & 0 & 0 & 0 &0 \
          0 & 1 & 0 & 0 &0 \
          0 & 0 & 1 & 0&0 \
          0 & 0 & 0 & 1 &0 \0&0&0&0&0
          end{vmatrix} Rightarrow begin{vmatrix}
          color{red}1 & 0 & 0 & 0 &0 \
          0 & color{blue}1 & 0 & 0 &0 \
          0 & 0 & color{brown}1 & 0& 0 \
          0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
          end{vmatrix}begin{vmatrix}
          0 & 1 & 0 & 0 &0 \
          1 & 0 & 0 & 0 &0 \
          0 & 0 & 0 & 1&0 \
          0 & 0 & 0 & 0 &1 \
          0 & 0 & 1 & 0 &0
          end{vmatrix} $$
          Hence, the inverse matrix is:
          $$A^{-1}=begin{vmatrix}
          0 & 1 & 0 & 0 &0 \
          1 & 0 & 0 & 0 &0 \
          0 & 0 & 0 & 1&0 \
          0 & 0 & 0 & 0 &1 \
          0 & 0 & 1 & 0 &0
          end{vmatrix}$$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Interchange rows three times ($R_1-R_2, R_3-R_4,R_4-R_5$):
            $$det(A)= begin{vmatrix}
            0 & color{blue}1 & 0 & 0 &0 \
            color{red}1 & 0 & 0 & 0 &0 \
            0 & 0 & 0 & 0&color{green}1 \
            0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
            end{vmatrix}=(-1)^3begin{vmatrix}
            color{red}1 & 0 & 0 & 0 &0 \
            0 & color{blue}1 & 0 & 0 &0 \
            0 & 0 & color{brown}1 & 0& 0 \
            0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
            end{vmatrix}=-1ne 0.$$
            Consider the expanded matrix and perform the same row interchanges:
            $$begin{vmatrix}
            0 & color{blue}1 & 0 & 0 &0 \
            color{red}1 & 0 & 0 & 0 &0 \
            0 & 0 & 0 & 0&color{green}1 \
            0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
            end{vmatrix}begin{vmatrix}
            1 & 0 & 0 & 0 &0 \
            0 & 1 & 0 & 0 &0 \
            0 & 0 & 1 & 0&0 \
            0 & 0 & 0 & 1 &0 \0&0&0&0&0
            end{vmatrix} Rightarrow begin{vmatrix}
            color{red}1 & 0 & 0 & 0 &0 \
            0 & color{blue}1 & 0 & 0 &0 \
            0 & 0 & color{brown}1 & 0& 0 \
            0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
            end{vmatrix}begin{vmatrix}
            0 & 1 & 0 & 0 &0 \
            1 & 0 & 0 & 0 &0 \
            0 & 0 & 0 & 1&0 \
            0 & 0 & 0 & 0 &1 \
            0 & 0 & 1 & 0 &0
            end{vmatrix} $$
            Hence, the inverse matrix is:
            $$A^{-1}=begin{vmatrix}
            0 & 1 & 0 & 0 &0 \
            1 & 0 & 0 & 0 &0 \
            0 & 0 & 0 & 1&0 \
            0 & 0 & 0 & 0 &1 \
            0 & 0 & 1 & 0 &0
            end{vmatrix}$$






            share|cite|improve this answer









            $endgroup$



            Interchange rows three times ($R_1-R_2, R_3-R_4,R_4-R_5$):
            $$det(A)= begin{vmatrix}
            0 & color{blue}1 & 0 & 0 &0 \
            color{red}1 & 0 & 0 & 0 &0 \
            0 & 0 & 0 & 0&color{green}1 \
            0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
            end{vmatrix}=(-1)^3begin{vmatrix}
            color{red}1 & 0 & 0 & 0 &0 \
            0 & color{blue}1 & 0 & 0 &0 \
            0 & 0 & color{brown}1 & 0& 0 \
            0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
            end{vmatrix}=-1ne 0.$$
            Consider the expanded matrix and perform the same row interchanges:
            $$begin{vmatrix}
            0 & color{blue}1 & 0 & 0 &0 \
            color{red}1 & 0 & 0 & 0 &0 \
            0 & 0 & 0 & 0&color{green}1 \
            0 & 0 & color{brown}1 & 0 &0 \0&0&0&color{purple}1&0
            end{vmatrix}begin{vmatrix}
            1 & 0 & 0 & 0 &0 \
            0 & 1 & 0 & 0 &0 \
            0 & 0 & 1 & 0&0 \
            0 & 0 & 0 & 1 &0 \0&0&0&0&0
            end{vmatrix} Rightarrow begin{vmatrix}
            color{red}1 & 0 & 0 & 0 &0 \
            0 & color{blue}1 & 0 & 0 &0 \
            0 & 0 & color{brown}1 & 0& 0 \
            0 & 0 & 0 & color{purple}1&0 \0&0&0&0&color{green}1
            end{vmatrix}begin{vmatrix}
            0 & 1 & 0 & 0 &0 \
            1 & 0 & 0 & 0 &0 \
            0 & 0 & 0 & 1&0 \
            0 & 0 & 0 & 0 &1 \
            0 & 0 & 1 & 0 &0
            end{vmatrix} $$
            Hence, the inverse matrix is:
            $$A^{-1}=begin{vmatrix}
            0 & 1 & 0 & 0 &0 \
            1 & 0 & 0 & 0 &0 \
            0 & 0 & 0 & 1&0 \
            0 & 0 & 0 & 0 &1 \
            0 & 0 & 1 & 0 &0
            end{vmatrix}$$







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            share|cite|improve this answer










            answered Jan 7 at 17:08









            farruhotafarruhota

            19.9k2738




            19.9k2738























                3












                $begingroup$

                Note that the given matrix is a so-called permutation matrix, the inverse of it is the permutation matrix associated to the inverse permutation.




                Later edit: The matrix $A$ works on a "generic vector" $x$ with components $x_1,dots, x_5$ as below by mapping it into a vector with components $x_{sigma(1)}, dots,x_{sigma(5)}$ for a particular permutation $sigma$. Explicitly:



                $$
                A
                begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
                =
                begin{bmatrix}
                0 & 1 & 0 & 0 & 0 \
                1 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 1 \
                0 & 0 & 1 & 0 & 0 \
                0 & 0 & 0 & 1 & 0
                end{bmatrix}
                begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
                =
                begin{bmatrix}x_2\x_1\x_5\x_3\x_4end{bmatrix}
                $$
                The corresponding permutation $sigma$ (with the inverse $tau$) is of course
                $$
                sigma
                =
                binom{12345}{21534}=(12)(354) ,
                qquadtext{ with }
                tau:=sigma^{-1}=
                binom{12345}{21453}=(12)(345) .
                $$
                Now associate in the natural way the permutation matrix for $tau$...



                Note: It is important to see at a glance, that this matrix is a permutation matrix. (Exactly one $1$ entry on each row and/or column.) This is part of a structure. (One has a mapping / a representation of the symmetric group into the general linear group of the corresponding dimension over $Bbb Z$, or $Bbb Q$ if we prefer a field.) "Other answers" like writing $A$ as a $(2+3)times(2+3)$ block matrix, then doing the work in smaller pieces, may also lead to a good end, but for my taste, if i see a structural answer, and a computational one (Gauss scheme, block split, adjoint matrix, etc...) the structural one wins for me.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  can u elaborate ur answer
                  $endgroup$
                  – jasmine
                  Jan 7 at 11:47
















                3












                $begingroup$

                Note that the given matrix is a so-called permutation matrix, the inverse of it is the permutation matrix associated to the inverse permutation.




                Later edit: The matrix $A$ works on a "generic vector" $x$ with components $x_1,dots, x_5$ as below by mapping it into a vector with components $x_{sigma(1)}, dots,x_{sigma(5)}$ for a particular permutation $sigma$. Explicitly:



                $$
                A
                begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
                =
                begin{bmatrix}
                0 & 1 & 0 & 0 & 0 \
                1 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 1 \
                0 & 0 & 1 & 0 & 0 \
                0 & 0 & 0 & 1 & 0
                end{bmatrix}
                begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
                =
                begin{bmatrix}x_2\x_1\x_5\x_3\x_4end{bmatrix}
                $$
                The corresponding permutation $sigma$ (with the inverse $tau$) is of course
                $$
                sigma
                =
                binom{12345}{21534}=(12)(354) ,
                qquadtext{ with }
                tau:=sigma^{-1}=
                binom{12345}{21453}=(12)(345) .
                $$
                Now associate in the natural way the permutation matrix for $tau$...



                Note: It is important to see at a glance, that this matrix is a permutation matrix. (Exactly one $1$ entry on each row and/or column.) This is part of a structure. (One has a mapping / a representation of the symmetric group into the general linear group of the corresponding dimension over $Bbb Z$, or $Bbb Q$ if we prefer a field.) "Other answers" like writing $A$ as a $(2+3)times(2+3)$ block matrix, then doing the work in smaller pieces, may also lead to a good end, but for my taste, if i see a structural answer, and a computational one (Gauss scheme, block split, adjoint matrix, etc...) the structural one wins for me.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  can u elaborate ur answer
                  $endgroup$
                  – jasmine
                  Jan 7 at 11:47














                3












                3








                3





                $begingroup$

                Note that the given matrix is a so-called permutation matrix, the inverse of it is the permutation matrix associated to the inverse permutation.




                Later edit: The matrix $A$ works on a "generic vector" $x$ with components $x_1,dots, x_5$ as below by mapping it into a vector with components $x_{sigma(1)}, dots,x_{sigma(5)}$ for a particular permutation $sigma$. Explicitly:



                $$
                A
                begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
                =
                begin{bmatrix}
                0 & 1 & 0 & 0 & 0 \
                1 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 1 \
                0 & 0 & 1 & 0 & 0 \
                0 & 0 & 0 & 1 & 0
                end{bmatrix}
                begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
                =
                begin{bmatrix}x_2\x_1\x_5\x_3\x_4end{bmatrix}
                $$
                The corresponding permutation $sigma$ (with the inverse $tau$) is of course
                $$
                sigma
                =
                binom{12345}{21534}=(12)(354) ,
                qquadtext{ with }
                tau:=sigma^{-1}=
                binom{12345}{21453}=(12)(345) .
                $$
                Now associate in the natural way the permutation matrix for $tau$...



                Note: It is important to see at a glance, that this matrix is a permutation matrix. (Exactly one $1$ entry on each row and/or column.) This is part of a structure. (One has a mapping / a representation of the symmetric group into the general linear group of the corresponding dimension over $Bbb Z$, or $Bbb Q$ if we prefer a field.) "Other answers" like writing $A$ as a $(2+3)times(2+3)$ block matrix, then doing the work in smaller pieces, may also lead to a good end, but for my taste, if i see a structural answer, and a computational one (Gauss scheme, block split, adjoint matrix, etc...) the structural one wins for me.






                share|cite|improve this answer











                $endgroup$



                Note that the given matrix is a so-called permutation matrix, the inverse of it is the permutation matrix associated to the inverse permutation.




                Later edit: The matrix $A$ works on a "generic vector" $x$ with components $x_1,dots, x_5$ as below by mapping it into a vector with components $x_{sigma(1)}, dots,x_{sigma(5)}$ for a particular permutation $sigma$. Explicitly:



                $$
                A
                begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
                =
                begin{bmatrix}
                0 & 1 & 0 & 0 & 0 \
                1 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 1 \
                0 & 0 & 1 & 0 & 0 \
                0 & 0 & 0 & 1 & 0
                end{bmatrix}
                begin{bmatrix}x_1\x_2\x_3\x_4\x_5end{bmatrix}
                =
                begin{bmatrix}x_2\x_1\x_5\x_3\x_4end{bmatrix}
                $$
                The corresponding permutation $sigma$ (with the inverse $tau$) is of course
                $$
                sigma
                =
                binom{12345}{21534}=(12)(354) ,
                qquadtext{ with }
                tau:=sigma^{-1}=
                binom{12345}{21453}=(12)(345) .
                $$
                Now associate in the natural way the permutation matrix for $tau$...



                Note: It is important to see at a glance, that this matrix is a permutation matrix. (Exactly one $1$ entry on each row and/or column.) This is part of a structure. (One has a mapping / a representation of the symmetric group into the general linear group of the corresponding dimension over $Bbb Z$, or $Bbb Q$ if we prefer a field.) "Other answers" like writing $A$ as a $(2+3)times(2+3)$ block matrix, then doing the work in smaller pieces, may also lead to a good end, but for my taste, if i see a structural answer, and a computational one (Gauss scheme, block split, adjoint matrix, etc...) the structural one wins for me.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 7 at 17:46

























                answered Jan 7 at 11:36









                dan_fuleadan_fulea

                6,4581312




                6,4581312












                • $begingroup$
                  can u elaborate ur answer
                  $endgroup$
                  – jasmine
                  Jan 7 at 11:47


















                • $begingroup$
                  can u elaborate ur answer
                  $endgroup$
                  – jasmine
                  Jan 7 at 11:47
















                $begingroup$
                can u elaborate ur answer
                $endgroup$
                – jasmine
                Jan 7 at 11:47




                $begingroup$
                can u elaborate ur answer
                $endgroup$
                – jasmine
                Jan 7 at 11:47











                2












                $begingroup$

                This matrix is a permutation matrix and also orthogonal. What do you know about inverses of orthogonal matrices?






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  This matrix is a permutation matrix and also orthogonal. What do you know about inverses of orthogonal matrices?






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    This matrix is a permutation matrix and also orthogonal. What do you know about inverses of orthogonal matrices?






                    share|cite|improve this answer









                    $endgroup$



                    This matrix is a permutation matrix and also orthogonal. What do you know about inverses of orthogonal matrices?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 7 at 11:40









                    Andreas H.Andreas H.

                    1,26166




                    1,26166























                        1












                        $begingroup$

                        The determinant of the matrix is not $0$, the determinant is $1$.



                        Hint:



                        Take a look at $A^2, A^3,A^4,dots$.






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          eigenvalue are $0$...i thinks det $A =0$ @Xum
                          $endgroup$
                          – jasmine
                          Jan 7 at 11:35










                        • $begingroup$
                          @jasmine None of the eigenvalues of the matrix is $0$. And the determinant is $1$.
                          $endgroup$
                          – 5xum
                          Jan 7 at 11:43
















                        1












                        $begingroup$

                        The determinant of the matrix is not $0$, the determinant is $1$.



                        Hint:



                        Take a look at $A^2, A^3,A^4,dots$.






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          eigenvalue are $0$...i thinks det $A =0$ @Xum
                          $endgroup$
                          – jasmine
                          Jan 7 at 11:35










                        • $begingroup$
                          @jasmine None of the eigenvalues of the matrix is $0$. And the determinant is $1$.
                          $endgroup$
                          – 5xum
                          Jan 7 at 11:43














                        1












                        1








                        1





                        $begingroup$

                        The determinant of the matrix is not $0$, the determinant is $1$.



                        Hint:



                        Take a look at $A^2, A^3,A^4,dots$.






                        share|cite|improve this answer









                        $endgroup$



                        The determinant of the matrix is not $0$, the determinant is $1$.



                        Hint:



                        Take a look at $A^2, A^3,A^4,dots$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 7 at 11:34









                        5xum5xum

                        90.2k394161




                        90.2k394161












                        • $begingroup$
                          eigenvalue are $0$...i thinks det $A =0$ @Xum
                          $endgroup$
                          – jasmine
                          Jan 7 at 11:35










                        • $begingroup$
                          @jasmine None of the eigenvalues of the matrix is $0$. And the determinant is $1$.
                          $endgroup$
                          – 5xum
                          Jan 7 at 11:43


















                        • $begingroup$
                          eigenvalue are $0$...i thinks det $A =0$ @Xum
                          $endgroup$
                          – jasmine
                          Jan 7 at 11:35










                        • $begingroup$
                          @jasmine None of the eigenvalues of the matrix is $0$. And the determinant is $1$.
                          $endgroup$
                          – 5xum
                          Jan 7 at 11:43
















                        $begingroup$
                        eigenvalue are $0$...i thinks det $A =0$ @Xum
                        $endgroup$
                        – jasmine
                        Jan 7 at 11:35




                        $begingroup$
                        eigenvalue are $0$...i thinks det $A =0$ @Xum
                        $endgroup$
                        – jasmine
                        Jan 7 at 11:35












                        $begingroup$
                        @jasmine None of the eigenvalues of the matrix is $0$. And the determinant is $1$.
                        $endgroup$
                        – 5xum
                        Jan 7 at 11:43




                        $begingroup$
                        @jasmine None of the eigenvalues of the matrix is $0$. And the determinant is $1$.
                        $endgroup$
                        – 5xum
                        Jan 7 at 11:43











                        1












                        $begingroup$

                        Note that this matrix is obtained from the identity matrix by exchanging columns (or rows), and that columns (and rows) are still linearly independent, so immediately we have that the determinant is not $0$. Even more precisely, we immediately have that the determinant is $1$ or $-1$. (Immediately= without doing any calculation)






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Note that this matrix is obtained from the identity matrix by exchanging columns (or rows), and that columns (and rows) are still linearly independent, so immediately we have that the determinant is not $0$. Even more precisely, we immediately have that the determinant is $1$ or $-1$. (Immediately= without doing any calculation)






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Note that this matrix is obtained from the identity matrix by exchanging columns (or rows), and that columns (and rows) are still linearly independent, so immediately we have that the determinant is not $0$. Even more precisely, we immediately have that the determinant is $1$ or $-1$. (Immediately= without doing any calculation)






                            share|cite|improve this answer









                            $endgroup$



                            Note that this matrix is obtained from the identity matrix by exchanging columns (or rows), and that columns (and rows) are still linearly independent, so immediately we have that the determinant is not $0$. Even more precisely, we immediately have that the determinant is $1$ or $-1$. (Immediately= without doing any calculation)







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 7 at 11:38









                            MinatoMinato

                            467212




                            467212






























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