Calculate inverse of matrix with -1 on diagonal and 1 on the rest












1












$begingroup$


Calculate the inverse of the matrix



begin{bmatrix}
-1& 1& ...& ...&1 \
1& -1& 1& ... &1 \
...& ...& ...& ...&1 \
1&1 &1 & ...&1 \
1& 1 &1 &1 &-1
end{bmatrix}



$-1$ on the diagonal and $1$ on the rest.



The key I think is to perform a sequence of elementary transformations on



[ A | $I_{n}$ ] until we get [ $I_{n}$ | $A^{-1}$ ] but that seems to be complicated.










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  • $begingroup$
    Your matrix can be expressed as $A+uv^top$ (see this answer. Then use the Sherman-Morrison formula.
    $endgroup$
    – StubbornAtom
    Nov 30 '18 at 18:50
















1












$begingroup$


Calculate the inverse of the matrix



begin{bmatrix}
-1& 1& ...& ...&1 \
1& -1& 1& ... &1 \
...& ...& ...& ...&1 \
1&1 &1 & ...&1 \
1& 1 &1 &1 &-1
end{bmatrix}



$-1$ on the diagonal and $1$ on the rest.



The key I think is to perform a sequence of elementary transformations on



[ A | $I_{n}$ ] until we get [ $I_{n}$ | $A^{-1}$ ] but that seems to be complicated.










share|cite|improve this question









$endgroup$












  • $begingroup$
    Your matrix can be expressed as $A+uv^top$ (see this answer. Then use the Sherman-Morrison formula.
    $endgroup$
    – StubbornAtom
    Nov 30 '18 at 18:50














1












1








1


1



$begingroup$


Calculate the inverse of the matrix



begin{bmatrix}
-1& 1& ...& ...&1 \
1& -1& 1& ... &1 \
...& ...& ...& ...&1 \
1&1 &1 & ...&1 \
1& 1 &1 &1 &-1
end{bmatrix}



$-1$ on the diagonal and $1$ on the rest.



The key I think is to perform a sequence of elementary transformations on



[ A | $I_{n}$ ] until we get [ $I_{n}$ | $A^{-1}$ ] but that seems to be complicated.










share|cite|improve this question









$endgroup$




Calculate the inverse of the matrix



begin{bmatrix}
-1& 1& ...& ...&1 \
1& -1& 1& ... &1 \
...& ...& ...& ...&1 \
1&1 &1 & ...&1 \
1& 1 &1 &1 &-1
end{bmatrix}



$-1$ on the diagonal and $1$ on the rest.



The key I think is to perform a sequence of elementary transformations on



[ A | $I_{n}$ ] until we get [ $I_{n}$ | $A^{-1}$ ] but that seems to be complicated.







matrices inverse






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asked Nov 30 '18 at 18:34









SADBOYSSADBOYS

4288




4288












  • $begingroup$
    Your matrix can be expressed as $A+uv^top$ (see this answer. Then use the Sherman-Morrison formula.
    $endgroup$
    – StubbornAtom
    Nov 30 '18 at 18:50


















  • $begingroup$
    Your matrix can be expressed as $A+uv^top$ (see this answer. Then use the Sherman-Morrison formula.
    $endgroup$
    – StubbornAtom
    Nov 30 '18 at 18:50
















$begingroup$
Your matrix can be expressed as $A+uv^top$ (see this answer. Then use the Sherman-Morrison formula.
$endgroup$
– StubbornAtom
Nov 30 '18 at 18:50




$begingroup$
Your matrix can be expressed as $A+uv^top$ (see this answer. Then use the Sherman-Morrison formula.
$endgroup$
– StubbornAtom
Nov 30 '18 at 18:50










3 Answers
3






active

oldest

votes


















3












$begingroup$

Let $e$ be the all one vector. We have



$$A=-2I+ee^T$$
By Sheman-Morrison formula:
begin{align}A^{-1}&=(-2I+ee^T)^{-1}\&=-frac12I-frac{-left(frac12Iright)ee^Tleft(-frac12Iright)}{1+e^Tleft( -frac12Iright)e} \
&=-frac12I-frac{frac14ee^T}{1-frac{n}2} \
&=-frac12I-frac{ee^T}{4-2n}
end{align}



Hence, the off diagonal entries are $-frac1{4-2n}$ and the diagonal entries are $-frac12-frac1{4-2n}$.



Remark: If $n=2$, the matrix is not invertible.






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  • $begingroup$
    Because the determinant is $-2^{n-1}(n-2)$ I think, thanks for the help^^
    $endgroup$
    – SADBOYS
    Nov 30 '18 at 22:56



















2












$begingroup$

The inverse also has a constant $a$ on the main diagonal and $b$ everywhere else. The constants do depend on the matrix size $n;$ as noted, for $n=2$ there is no inverse. Just try the thing for $n=3$ and $n=4$ and $n=5.$ By that time you should have it, for $n geq 3$



for $n=3:$
$$
left(
begin{array}{rrr}
-1 & 1&1 \
1 & -1&1 \
1 & 1&-1 \
end{array}
right)
left(
begin{array}{ccc}
a & b&b \
b & a&b \
b & b&a \
end{array}
right)=
left(
begin{array}{rrr}
1 & 0&0 \
0 & 1&0 \
0 & 0&1 \
end{array}
right)
$$



for $n=4,$ different $a,b:$
$$
left(
begin{array}{rrrr}
-1 & 1&1&1 \
1 & -1&1&1 \
1 & 1&-1&1 \
1 & 1&1&-1 \
end{array}
right)
left(
begin{array}{cccc}
a & b&b&b \
b & a&b&b \
b & b&a&b \
b & b&b&a \
end{array}
right)=
left(
begin{array}{rrrr}
1 & 0&0&0 \
0 & 1&0&0 \
0 & 0&1&0 \
0 & 0&0&1 \
end{array}
right)
$$



for $n=5,$ still different $a,b:$
$$
left(
begin{array}{rrrrr}
-1 & 1&1&1&1 \
1 & -1&1&1&1 \
1 & 1&-1&1&1 \
1 & 1&1&-1&1 \
1&1&1&1&-1 \
end{array}
right)
left(
begin{array}{ccccc}
a & b&b&b &b\
b & a&b&b&b \
b & b&a&b &b \
b & b&b&a &b \
b&b&b&b&a \
end{array}
right)=
left(
begin{array}{rrrrr}
1 & 0&0&0 &0 \
0 & 1&0&0&0 \
0 & 0&1&0 &0 \
0 & 0&0&1 &0 \
0&0&0&0&1
end{array}
right)
$$






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    1












    $begingroup$

    It's not too bad...
    $$begin{array}{c}-1\-1\vdots\*
    end{array}left[begin{array}{cccc|cccc}
    -1&1&cdots&1 &1\
    1&-1&cdots&1 &&1\
    vdots&vdots&ddots&vdots &&&ddots\
    1&1&cdots&-1 &&&&1
    end{array}right] implies$$



    $$begin{array}{c}*\*\vdots\small 1/2end{array}
    left[begin{array}{cccc|cccc}
    -2&&&2 &1&&&-1\
    &-2&&2 &&1&&-1\
    &&ddots& &&&ddots\
    1&1&&-1 &&&&1
    end{array}right] implies$$



    $$begin{array}{c}*small -1/2\*small -1/2\vdots\* small ^1!/_{!n-2}end{array}
    left[begin{array}{cccc|cccc}
    -2&&&2 &1&&&-1\
    &-2&&2 &&1&&-1\
    &&ddots& &&&ddots\
    &&&n-2 &small 1/2&small 1/2&&small-^{(n-3)!}/_{!2}
    end{array}right] implies$$



    $$begin{array}{c}1\1\vdots\*end{array}
    left[begin{array}{cccc|cccc}
    1&&&-1 &small -1/2&&&small 1/2\
    &1&&-1 &&small -1/2&&small 1/2\
    &&ddots& &&&ddots\
    &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
    end{array}right] implies$$



    $$begin{array}{c} \ \ \ end{array}
    left[begin{array}{cccc|cccc}
    1&&& &small -^1!/_{!2}+ ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
    &1&& &small ^1!/_{!2(n-2)}&small -^1!/_{!2}+ ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
    &&ddots& &vdots&vdots&ddots&vdots\
    &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
    end{array}right]$$






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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Let $e$ be the all one vector. We have



      $$A=-2I+ee^T$$
      By Sheman-Morrison formula:
      begin{align}A^{-1}&=(-2I+ee^T)^{-1}\&=-frac12I-frac{-left(frac12Iright)ee^Tleft(-frac12Iright)}{1+e^Tleft( -frac12Iright)e} \
      &=-frac12I-frac{frac14ee^T}{1-frac{n}2} \
      &=-frac12I-frac{ee^T}{4-2n}
      end{align}



      Hence, the off diagonal entries are $-frac1{4-2n}$ and the diagonal entries are $-frac12-frac1{4-2n}$.



      Remark: If $n=2$, the matrix is not invertible.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Because the determinant is $-2^{n-1}(n-2)$ I think, thanks for the help^^
        $endgroup$
        – SADBOYS
        Nov 30 '18 at 22:56
















      3












      $begingroup$

      Let $e$ be the all one vector. We have



      $$A=-2I+ee^T$$
      By Sheman-Morrison formula:
      begin{align}A^{-1}&=(-2I+ee^T)^{-1}\&=-frac12I-frac{-left(frac12Iright)ee^Tleft(-frac12Iright)}{1+e^Tleft( -frac12Iright)e} \
      &=-frac12I-frac{frac14ee^T}{1-frac{n}2} \
      &=-frac12I-frac{ee^T}{4-2n}
      end{align}



      Hence, the off diagonal entries are $-frac1{4-2n}$ and the diagonal entries are $-frac12-frac1{4-2n}$.



      Remark: If $n=2$, the matrix is not invertible.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Because the determinant is $-2^{n-1}(n-2)$ I think, thanks for the help^^
        $endgroup$
        – SADBOYS
        Nov 30 '18 at 22:56














      3












      3








      3





      $begingroup$

      Let $e$ be the all one vector. We have



      $$A=-2I+ee^T$$
      By Sheman-Morrison formula:
      begin{align}A^{-1}&=(-2I+ee^T)^{-1}\&=-frac12I-frac{-left(frac12Iright)ee^Tleft(-frac12Iright)}{1+e^Tleft( -frac12Iright)e} \
      &=-frac12I-frac{frac14ee^T}{1-frac{n}2} \
      &=-frac12I-frac{ee^T}{4-2n}
      end{align}



      Hence, the off diagonal entries are $-frac1{4-2n}$ and the diagonal entries are $-frac12-frac1{4-2n}$.



      Remark: If $n=2$, the matrix is not invertible.






      share|cite|improve this answer









      $endgroup$



      Let $e$ be the all one vector. We have



      $$A=-2I+ee^T$$
      By Sheman-Morrison formula:
      begin{align}A^{-1}&=(-2I+ee^T)^{-1}\&=-frac12I-frac{-left(frac12Iright)ee^Tleft(-frac12Iright)}{1+e^Tleft( -frac12Iright)e} \
      &=-frac12I-frac{frac14ee^T}{1-frac{n}2} \
      &=-frac12I-frac{ee^T}{4-2n}
      end{align}



      Hence, the off diagonal entries are $-frac1{4-2n}$ and the diagonal entries are $-frac12-frac1{4-2n}$.



      Remark: If $n=2$, the matrix is not invertible.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 30 '18 at 18:46









      Siong Thye GohSiong Thye Goh

      100k1465117




      100k1465117












      • $begingroup$
        Because the determinant is $-2^{n-1}(n-2)$ I think, thanks for the help^^
        $endgroup$
        – SADBOYS
        Nov 30 '18 at 22:56


















      • $begingroup$
        Because the determinant is $-2^{n-1}(n-2)$ I think, thanks for the help^^
        $endgroup$
        – SADBOYS
        Nov 30 '18 at 22:56
















      $begingroup$
      Because the determinant is $-2^{n-1}(n-2)$ I think, thanks for the help^^
      $endgroup$
      – SADBOYS
      Nov 30 '18 at 22:56




      $begingroup$
      Because the determinant is $-2^{n-1}(n-2)$ I think, thanks for the help^^
      $endgroup$
      – SADBOYS
      Nov 30 '18 at 22:56











      2












      $begingroup$

      The inverse also has a constant $a$ on the main diagonal and $b$ everywhere else. The constants do depend on the matrix size $n;$ as noted, for $n=2$ there is no inverse. Just try the thing for $n=3$ and $n=4$ and $n=5.$ By that time you should have it, for $n geq 3$



      for $n=3:$
      $$
      left(
      begin{array}{rrr}
      -1 & 1&1 \
      1 & -1&1 \
      1 & 1&-1 \
      end{array}
      right)
      left(
      begin{array}{ccc}
      a & b&b \
      b & a&b \
      b & b&a \
      end{array}
      right)=
      left(
      begin{array}{rrr}
      1 & 0&0 \
      0 & 1&0 \
      0 & 0&1 \
      end{array}
      right)
      $$



      for $n=4,$ different $a,b:$
      $$
      left(
      begin{array}{rrrr}
      -1 & 1&1&1 \
      1 & -1&1&1 \
      1 & 1&-1&1 \
      1 & 1&1&-1 \
      end{array}
      right)
      left(
      begin{array}{cccc}
      a & b&b&b \
      b & a&b&b \
      b & b&a&b \
      b & b&b&a \
      end{array}
      right)=
      left(
      begin{array}{rrrr}
      1 & 0&0&0 \
      0 & 1&0&0 \
      0 & 0&1&0 \
      0 & 0&0&1 \
      end{array}
      right)
      $$



      for $n=5,$ still different $a,b:$
      $$
      left(
      begin{array}{rrrrr}
      -1 & 1&1&1&1 \
      1 & -1&1&1&1 \
      1 & 1&-1&1&1 \
      1 & 1&1&-1&1 \
      1&1&1&1&-1 \
      end{array}
      right)
      left(
      begin{array}{ccccc}
      a & b&b&b &b\
      b & a&b&b&b \
      b & b&a&b &b \
      b & b&b&a &b \
      b&b&b&b&a \
      end{array}
      right)=
      left(
      begin{array}{rrrrr}
      1 & 0&0&0 &0 \
      0 & 1&0&0&0 \
      0 & 0&1&0 &0 \
      0 & 0&0&1 &0 \
      0&0&0&0&1
      end{array}
      right)
      $$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        The inverse also has a constant $a$ on the main diagonal and $b$ everywhere else. The constants do depend on the matrix size $n;$ as noted, for $n=2$ there is no inverse. Just try the thing for $n=3$ and $n=4$ and $n=5.$ By that time you should have it, for $n geq 3$



        for $n=3:$
        $$
        left(
        begin{array}{rrr}
        -1 & 1&1 \
        1 & -1&1 \
        1 & 1&-1 \
        end{array}
        right)
        left(
        begin{array}{ccc}
        a & b&b \
        b & a&b \
        b & b&a \
        end{array}
        right)=
        left(
        begin{array}{rrr}
        1 & 0&0 \
        0 & 1&0 \
        0 & 0&1 \
        end{array}
        right)
        $$



        for $n=4,$ different $a,b:$
        $$
        left(
        begin{array}{rrrr}
        -1 & 1&1&1 \
        1 & -1&1&1 \
        1 & 1&-1&1 \
        1 & 1&1&-1 \
        end{array}
        right)
        left(
        begin{array}{cccc}
        a & b&b&b \
        b & a&b&b \
        b & b&a&b \
        b & b&b&a \
        end{array}
        right)=
        left(
        begin{array}{rrrr}
        1 & 0&0&0 \
        0 & 1&0&0 \
        0 & 0&1&0 \
        0 & 0&0&1 \
        end{array}
        right)
        $$



        for $n=5,$ still different $a,b:$
        $$
        left(
        begin{array}{rrrrr}
        -1 & 1&1&1&1 \
        1 & -1&1&1&1 \
        1 & 1&-1&1&1 \
        1 & 1&1&-1&1 \
        1&1&1&1&-1 \
        end{array}
        right)
        left(
        begin{array}{ccccc}
        a & b&b&b &b\
        b & a&b&b&b \
        b & b&a&b &b \
        b & b&b&a &b \
        b&b&b&b&a \
        end{array}
        right)=
        left(
        begin{array}{rrrrr}
        1 & 0&0&0 &0 \
        0 & 1&0&0&0 \
        0 & 0&1&0 &0 \
        0 & 0&0&1 &0 \
        0&0&0&0&1
        end{array}
        right)
        $$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          The inverse also has a constant $a$ on the main diagonal and $b$ everywhere else. The constants do depend on the matrix size $n;$ as noted, for $n=2$ there is no inverse. Just try the thing for $n=3$ and $n=4$ and $n=5.$ By that time you should have it, for $n geq 3$



          for $n=3:$
          $$
          left(
          begin{array}{rrr}
          -1 & 1&1 \
          1 & -1&1 \
          1 & 1&-1 \
          end{array}
          right)
          left(
          begin{array}{ccc}
          a & b&b \
          b & a&b \
          b & b&a \
          end{array}
          right)=
          left(
          begin{array}{rrr}
          1 & 0&0 \
          0 & 1&0 \
          0 & 0&1 \
          end{array}
          right)
          $$



          for $n=4,$ different $a,b:$
          $$
          left(
          begin{array}{rrrr}
          -1 & 1&1&1 \
          1 & -1&1&1 \
          1 & 1&-1&1 \
          1 & 1&1&-1 \
          end{array}
          right)
          left(
          begin{array}{cccc}
          a & b&b&b \
          b & a&b&b \
          b & b&a&b \
          b & b&b&a \
          end{array}
          right)=
          left(
          begin{array}{rrrr}
          1 & 0&0&0 \
          0 & 1&0&0 \
          0 & 0&1&0 \
          0 & 0&0&1 \
          end{array}
          right)
          $$



          for $n=5,$ still different $a,b:$
          $$
          left(
          begin{array}{rrrrr}
          -1 & 1&1&1&1 \
          1 & -1&1&1&1 \
          1 & 1&-1&1&1 \
          1 & 1&1&-1&1 \
          1&1&1&1&-1 \
          end{array}
          right)
          left(
          begin{array}{ccccc}
          a & b&b&b &b\
          b & a&b&b&b \
          b & b&a&b &b \
          b & b&b&a &b \
          b&b&b&b&a \
          end{array}
          right)=
          left(
          begin{array}{rrrrr}
          1 & 0&0&0 &0 \
          0 & 1&0&0&0 \
          0 & 0&1&0 &0 \
          0 & 0&0&1 &0 \
          0&0&0&0&1
          end{array}
          right)
          $$






          share|cite|improve this answer











          $endgroup$



          The inverse also has a constant $a$ on the main diagonal and $b$ everywhere else. The constants do depend on the matrix size $n;$ as noted, for $n=2$ there is no inverse. Just try the thing for $n=3$ and $n=4$ and $n=5.$ By that time you should have it, for $n geq 3$



          for $n=3:$
          $$
          left(
          begin{array}{rrr}
          -1 & 1&1 \
          1 & -1&1 \
          1 & 1&-1 \
          end{array}
          right)
          left(
          begin{array}{ccc}
          a & b&b \
          b & a&b \
          b & b&a \
          end{array}
          right)=
          left(
          begin{array}{rrr}
          1 & 0&0 \
          0 & 1&0 \
          0 & 0&1 \
          end{array}
          right)
          $$



          for $n=4,$ different $a,b:$
          $$
          left(
          begin{array}{rrrr}
          -1 & 1&1&1 \
          1 & -1&1&1 \
          1 & 1&-1&1 \
          1 & 1&1&-1 \
          end{array}
          right)
          left(
          begin{array}{cccc}
          a & b&b&b \
          b & a&b&b \
          b & b&a&b \
          b & b&b&a \
          end{array}
          right)=
          left(
          begin{array}{rrrr}
          1 & 0&0&0 \
          0 & 1&0&0 \
          0 & 0&1&0 \
          0 & 0&0&1 \
          end{array}
          right)
          $$



          for $n=5,$ still different $a,b:$
          $$
          left(
          begin{array}{rrrrr}
          -1 & 1&1&1&1 \
          1 & -1&1&1&1 \
          1 & 1&-1&1&1 \
          1 & 1&1&-1&1 \
          1&1&1&1&-1 \
          end{array}
          right)
          left(
          begin{array}{ccccc}
          a & b&b&b &b\
          b & a&b&b&b \
          b & b&a&b &b \
          b & b&b&a &b \
          b&b&b&b&a \
          end{array}
          right)=
          left(
          begin{array}{rrrrr}
          1 & 0&0&0 &0 \
          0 & 1&0&0&0 \
          0 & 0&1&0 &0 \
          0 & 0&0&1 &0 \
          0&0&0&0&1
          end{array}
          right)
          $$







          share|cite|improve this answer














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          share|cite|improve this answer








          edited Nov 30 '18 at 19:21

























          answered Nov 30 '18 at 19:04









          Will JagyWill Jagy

          102k5101199




          102k5101199























              1












              $begingroup$

              It's not too bad...
              $$begin{array}{c}-1\-1\vdots\*
              end{array}left[begin{array}{cccc|cccc}
              -1&1&cdots&1 &1\
              1&-1&cdots&1 &&1\
              vdots&vdots&ddots&vdots &&&ddots\
              1&1&cdots&-1 &&&&1
              end{array}right] implies$$



              $$begin{array}{c}*\*\vdots\small 1/2end{array}
              left[begin{array}{cccc|cccc}
              -2&&&2 &1&&&-1\
              &-2&&2 &&1&&-1\
              &&ddots& &&&ddots\
              1&1&&-1 &&&&1
              end{array}right] implies$$



              $$begin{array}{c}*small -1/2\*small -1/2\vdots\* small ^1!/_{!n-2}end{array}
              left[begin{array}{cccc|cccc}
              -2&&&2 &1&&&-1\
              &-2&&2 &&1&&-1\
              &&ddots& &&&ddots\
              &&&n-2 &small 1/2&small 1/2&&small-^{(n-3)!}/_{!2}
              end{array}right] implies$$



              $$begin{array}{c}1\1\vdots\*end{array}
              left[begin{array}{cccc|cccc}
              1&&&-1 &small -1/2&&&small 1/2\
              &1&&-1 &&small -1/2&&small 1/2\
              &&ddots& &&&ddots\
              &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
              end{array}right] implies$$



              $$begin{array}{c} \ \ \ end{array}
              left[begin{array}{cccc|cccc}
              1&&& &small -^1!/_{!2}+ ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
              &1&& &small ^1!/_{!2(n-2)}&small -^1!/_{!2}+ ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
              &&ddots& &vdots&vdots&ddots&vdots\
              &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
              end{array}right]$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                It's not too bad...
                $$begin{array}{c}-1\-1\vdots\*
                end{array}left[begin{array}{cccc|cccc}
                -1&1&cdots&1 &1\
                1&-1&cdots&1 &&1\
                vdots&vdots&ddots&vdots &&&ddots\
                1&1&cdots&-1 &&&&1
                end{array}right] implies$$



                $$begin{array}{c}*\*\vdots\small 1/2end{array}
                left[begin{array}{cccc|cccc}
                -2&&&2 &1&&&-1\
                &-2&&2 &&1&&-1\
                &&ddots& &&&ddots\
                1&1&&-1 &&&&1
                end{array}right] implies$$



                $$begin{array}{c}*small -1/2\*small -1/2\vdots\* small ^1!/_{!n-2}end{array}
                left[begin{array}{cccc|cccc}
                -2&&&2 &1&&&-1\
                &-2&&2 &&1&&-1\
                &&ddots& &&&ddots\
                &&&n-2 &small 1/2&small 1/2&&small-^{(n-3)!}/_{!2}
                end{array}right] implies$$



                $$begin{array}{c}1\1\vdots\*end{array}
                left[begin{array}{cccc|cccc}
                1&&&-1 &small -1/2&&&small 1/2\
                &1&&-1 &&small -1/2&&small 1/2\
                &&ddots& &&&ddots\
                &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
                end{array}right] implies$$



                $$begin{array}{c} \ \ \ end{array}
                left[begin{array}{cccc|cccc}
                1&&& &small -^1!/_{!2}+ ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
                &1&& &small ^1!/_{!2(n-2)}&small -^1!/_{!2}+ ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
                &&ddots& &vdots&vdots&ddots&vdots\
                &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
                end{array}right]$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  It's not too bad...
                  $$begin{array}{c}-1\-1\vdots\*
                  end{array}left[begin{array}{cccc|cccc}
                  -1&1&cdots&1 &1\
                  1&-1&cdots&1 &&1\
                  vdots&vdots&ddots&vdots &&&ddots\
                  1&1&cdots&-1 &&&&1
                  end{array}right] implies$$



                  $$begin{array}{c}*\*\vdots\small 1/2end{array}
                  left[begin{array}{cccc|cccc}
                  -2&&&2 &1&&&-1\
                  &-2&&2 &&1&&-1\
                  &&ddots& &&&ddots\
                  1&1&&-1 &&&&1
                  end{array}right] implies$$



                  $$begin{array}{c}*small -1/2\*small -1/2\vdots\* small ^1!/_{!n-2}end{array}
                  left[begin{array}{cccc|cccc}
                  -2&&&2 &1&&&-1\
                  &-2&&2 &&1&&-1\
                  &&ddots& &&&ddots\
                  &&&n-2 &small 1/2&small 1/2&&small-^{(n-3)!}/_{!2}
                  end{array}right] implies$$



                  $$begin{array}{c}1\1\vdots\*end{array}
                  left[begin{array}{cccc|cccc}
                  1&&&-1 &small -1/2&&&small 1/2\
                  &1&&-1 &&small -1/2&&small 1/2\
                  &&ddots& &&&ddots\
                  &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
                  end{array}right] implies$$



                  $$begin{array}{c} \ \ \ end{array}
                  left[begin{array}{cccc|cccc}
                  1&&& &small -^1!/_{!2}+ ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
                  &1&& &small ^1!/_{!2(n-2)}&small -^1!/_{!2}+ ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
                  &&ddots& &vdots&vdots&ddots&vdots\
                  &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
                  end{array}right]$$






                  share|cite|improve this answer









                  $endgroup$



                  It's not too bad...
                  $$begin{array}{c}-1\-1\vdots\*
                  end{array}left[begin{array}{cccc|cccc}
                  -1&1&cdots&1 &1\
                  1&-1&cdots&1 &&1\
                  vdots&vdots&ddots&vdots &&&ddots\
                  1&1&cdots&-1 &&&&1
                  end{array}right] implies$$



                  $$begin{array}{c}*\*\vdots\small 1/2end{array}
                  left[begin{array}{cccc|cccc}
                  -2&&&2 &1&&&-1\
                  &-2&&2 &&1&&-1\
                  &&ddots& &&&ddots\
                  1&1&&-1 &&&&1
                  end{array}right] implies$$



                  $$begin{array}{c}*small -1/2\*small -1/2\vdots\* small ^1!/_{!n-2}end{array}
                  left[begin{array}{cccc|cccc}
                  -2&&&2 &1&&&-1\
                  &-2&&2 &&1&&-1\
                  &&ddots& &&&ddots\
                  &&&n-2 &small 1/2&small 1/2&&small-^{(n-3)!}/_{!2}
                  end{array}right] implies$$



                  $$begin{array}{c}1\1\vdots\*end{array}
                  left[begin{array}{cccc|cccc}
                  1&&&-1 &small -1/2&&&small 1/2\
                  &1&&-1 &&small -1/2&&small 1/2\
                  &&ddots& &&&ddots\
                  &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
                  end{array}right] implies$$



                  $$begin{array}{c} \ \ \ end{array}
                  left[begin{array}{cccc|cccc}
                  1&&& &small -^1!/_{!2}+ ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
                  &1&& &small ^1!/_{!2(n-2)}&small -^1!/_{!2}+ ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
                  &&ddots& &vdots&vdots&ddots&vdots\
                  &&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
                  end{array}right]$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 20:30









                  I like SerenaI like Serena

                  4,0421721




                  4,0421721






























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