Mixing
Inverse matrix of matrix (all rows equal) plus identity matrix
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Let $A$ be a matrix where all rows are equal, for example,
$$A=left[beginarrayccc
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3
endarrayright]$$
Then what is the inverse of the matrix $B=I+A$, where $I$ is the identity matrix? For example,
$$B=left[beginarrayccc
a_1+1 & a_2 & a_3 \
a_1 & a_2+1 & a_3 \
a_1 & a_2 & a_3+1
endarrayright]$$
I have a conjecture, which computation has so far confirmed:
$$B^-1=I-fracAmboxtr(A)+1$$
Why is this true?
linear-algebra matrices inverse
edited Jul 21 at 10:32
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 5:18
user159452
320110
add a comment |Â
up vote
8
down vote
favorite
Let $A$ be a matrix where all rows are equal, for example,
$$A=left[beginarrayccc
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3
endarrayright]$$
Then what is the inverse of the matrix $B=I+A$, where $I$ is the identity matrix? For example,
$$B=left[beginarrayccc
a_1+1 & a_2 & a_3 \
a_1 & a_2+1 & a_3 \
a_1 & a_2 & a_3+1
endarrayright]$$
I have a conjecture, which computation has so far confirmed:
$$B^-1=I-fracAmboxtr(A)+1$$
Why is this true?
linear-algebra matrices inverse
edited Jul 21 at 10:32
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 5:18
user159452
320110
1
Your matrix is not symmetric, unless $a_1=a_2=a_3$.
– Suzet
Jul 21 at 5:20
Thanks. Edited in question.
– user159452
Jul 21 at 5:28
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Let $A$ be a matrix where all rows are equal, for example,
$$A=left[beginarrayccc
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3
endarrayright]$$
Then what is the inverse of the matrix $B=I+A$, where $I$ is the identity matrix? For example,
$$B=left[beginarrayccc
a_1+1 & a_2 & a_3 \
a_1 & a_2+1 & a_3 \
a_1 & a_2 & a_3+1
endarrayright]$$
I have a conjecture, which computation has so far confirmed:
$$B^-1=I-fracAmboxtr(A)+1$$
Why is this true?
linear-algebra matrices inverse
edited Jul 21 at 10:32
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 5:18
user159452
320110
Let $A$ be a matrix where all rows are equal, for example,
$$A=left[beginarrayccc
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3
endarrayright]$$
Then what is the inverse of the matrix $B=I+A$, where $I$ is the identity matrix? For example,
$$B=left[beginarrayccc
a_1+1 & a_2 & a_3 \
a_1 & a_2+1 & a_3 \
a_1 & a_2 & a_3+1
endarrayright]$$
I have a conjecture, which computation has so far confirmed:
$$B^-1=I-fracAmboxtr(A)+1$$
Why is this true?
linear-algebra matrices inverse
edited Jul 21 at 10:32
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 5:18
user159452
320110
edited Jul 21 at 10:32
Rodrigo de Azevedo
12.6k41751
edited Jul 21 at 10:32
Rodrigo de Azevedo
12.6k41751
edited Jul 21 at 10:32
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Jul 21 at 5:18
user159452
320110
asked Jul 21 at 5:18
user159452
320110
asked Jul 21 at 5:18
user159452
320110
320110
1
Your matrix is not symmetric, unless $a_1=a_2=a_3$.
– Suzet
Jul 21 at 5:20
Thanks. Edited in question.
– user159452
Jul 21 at 5:28
add a comment |Â
1
Your matrix is not symmetric, unless $a_1=a_2=a_3$.
– Suzet
Jul 21 at 5:20
Thanks. Edited in question.
– user159452
Jul 21 at 5:28
1
1
Your matrix is not symmetric, unless $a_1=a_2=a_3$.
– Suzet
Jul 21 at 5:20
Your matrix is not symmetric, unless $a_1=a_2=a_3$.
– Suzet
Jul 21 at 5:20
Thanks. Edited in question.
– user159452
Jul 21 at 5:28
Thanks. Edited in question.
– user159452
Jul 21 at 5:28
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
14
down vote
$A=ea^T$ where $a=beginbmatrix a_1 \ a_2 \ a_3 endbmatrix$ and $e = beginbmatrix 1 \ 1 \ 1endbmatrix.$
Now, we can use the Sherman-Morrison formula, which states that a matrix $C$ is invertible, then $C+uv^T$ is invertible if $1+v^TC^-1u ne 0$ and $$(C+uv^T)^-1=C^-1-fracC^-1uv^TC^-11+v^TC^-1u.$$
$B=I+ea^T$, hence $B$ is invertible if $1+a^Te=1+sum_i=1^3a_i=1+trace(A) ne 0.$ and
$$B^-1=I-fracea^T1+trace(A)=I-fracA1+trace(A)$$
edited Jul 25 at 3:47
Parcly Taxel
33.6k136588
answered Jul 21 at 5:33
Siong Thye Goh
77.6k134795
Thank you! This is beautiful. Just what I needed, and I would have never figured it out myself.
– user159452
Jul 21 at 11:04
add a comment |Â
up vote
7
down vote
In the following, let us assume that $mboxtr(A)+1not = 0$.
We have $B=A+I$. Let us compute $C:=(A+I)left(I-fracAmboxtr(A)+1right)$ and see if we get the identity matrix.
$$C=-frac1mboxtr(A)+1A^2+fracmboxtr(A)mboxtr(A)+1A+I$$
so that to get the desired result, we need to prove that $A^2=mboxtr(A)A$.
The $i,j$ coefficient of $A^2$ is given by $$sum_k=1^n A_i,kA_k,j=sum_k=1^n a_ka_j=mboxtr(A)a_j=mboxtr(A)A_i,j $$
So that indeed, we obtain the desired identity.
NB : To justify that the matrix $B=A+I$ is invertible if and only if $mboxtr(A)+1not = 0$, note that the above computation actually establishes the identity $$(mboxtr(A)+1)I=(A+I)((mboxtr(A)+1)I-A)$$
from which the equivalence follows.
answered Jul 21 at 5:36
Suzet
2,211527
Thank you, Suzet.
– user159452
Jul 21 at 11:05
add a comment |Â
up vote
2
down vote
The inverse of $B$:
$$B^-1=fractextadj(B)det(B).$$
The determinant of $B$:
$$det(B)=beginvmatrixa_1+1 & a_2 & a_3 \
a_1 & a_2+1 & a_3 \
a_1 & a_2 & a_3+1endvmatrix=
beginvmatrixa_1+1 & a_2 & a_3 \
-1 & 1 & 0 \
-1 & 0 & 1endvmatrix=a_1+a_2+a_3+1.$$
The adjugate of $B$:
$$textadj(B)=textC^T=\
beginpmatrix
(a_2+1)(a_3+1)-a_2a_3 & -a_2(a_3+1)+a_2a_3 & a_2a_3-a_3(a_2+1) \
-a_1(a_3+1)+a_1a_3 & (a_1+1)(a_3+1)-a_1a_3 & -a_3(a_1+1)+a_1a_3 \
a_1a_2-a_1(a_2+1) & -a_2(a_1+1)+a_1a_2 & (a_1+1)(a_2+1)-a_1a_2
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1-a_1 & -a_2 & -a_3 \
-a_1 & a_1+a_2+a_3+1-a_2 & -a_3 \
-a_1 & -a_2 & a_1+a_2+a_3+1-a_3
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1 & 0 & 0 \
0 & a_1+a_2+a_3+1 & 0 \
0 & 0 & a_1+a_2+a_3+1
endpmatrix-
beginpmatrix
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3
endpmatrix.
$$
Note: $C^T$ is the transpose of the cofactor matrix of $B$.
Hence, the result follows.
answered Jul 25 at 8:55
farruhota
13.7k2632
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
$A=ea^T$ where $a=beginbmatrix a_1 \ a_2 \ a_3 endbmatrix$ and $e = beginbmatrix 1 \ 1 \ 1endbmatrix.$
Now, we can use the Sherman-Morrison formula, which states that a matrix $C$ is invertible, then $C+uv^T$ is invertible if $1+v^TC^-1u ne 0$ and $$(C+uv^T)^-1=C^-1-fracC^-1uv^TC^-11+v^TC^-1u.$$
$B=I+ea^T$, hence $B$ is invertible if $1+a^Te=1+sum_i=1^3a_i=1+trace(A) ne 0.$ and
$$B^-1=I-fracea^T1+trace(A)=I-fracA1+trace(A)$$
edited Jul 25 at 3:47
Parcly Taxel
33.6k136588
answered Jul 21 at 5:33
Siong Thye Goh
77.6k134795
Thank you! This is beautiful. Just what I needed, and I would have never figured it out myself.
– user159452
Jul 21 at 11:04
add a comment |Â
up vote
14
down vote
$A=ea^T$ where $a=beginbmatrix a_1 \ a_2 \ a_3 endbmatrix$ and $e = beginbmatrix 1 \ 1 \ 1endbmatrix.$
Now, we can use the Sherman-Morrison formula, which states that a matrix $C$ is invertible, then $C+uv^T$ is invertible if $1+v^TC^-1u ne 0$ and $$(C+uv^T)^-1=C^-1-fracC^-1uv^TC^-11+v^TC^-1u.$$
$B=I+ea^T$, hence $B$ is invertible if $1+a^Te=1+sum_i=1^3a_i=1+trace(A) ne 0.$ and
$$B^-1=I-fracea^T1+trace(A)=I-fracA1+trace(A)$$
edited Jul 25 at 3:47
Parcly Taxel
33.6k136588
answered Jul 21 at 5:33
Siong Thye Goh
77.6k134795
Thank you! This is beautiful. Just what I needed, and I would have never figured it out myself.
– user159452
Jul 21 at 11:04
add a comment |Â
up vote
14
down vote
up vote
14
down vote
$A=ea^T$ where $a=beginbmatrix a_1 \ a_2 \ a_3 endbmatrix$ and $e = beginbmatrix 1 \ 1 \ 1endbmatrix.$
Now, we can use the Sherman-Morrison formula, which states that a matrix $C$ is invertible, then $C+uv^T$ is invertible if $1+v^TC^-1u ne 0$ and $$(C+uv^T)^-1=C^-1-fracC^-1uv^TC^-11+v^TC^-1u.$$
$B=I+ea^T$, hence $B$ is invertible if $1+a^Te=1+sum_i=1^3a_i=1+trace(A) ne 0.$ and
$$B^-1=I-fracea^T1+trace(A)=I-fracA1+trace(A)$$
edited Jul 25 at 3:47
Parcly Taxel
33.6k136588
answered Jul 21 at 5:33
Siong Thye Goh
77.6k134795
$A=ea^T$ where $a=beginbmatrix a_1 \ a_2 \ a_3 endbmatrix$ and $e = beginbmatrix 1 \ 1 \ 1endbmatrix.$
Now, we can use the Sherman-Morrison formula, which states that a matrix $C$ is invertible, then $C+uv^T$ is invertible if $1+v^TC^-1u ne 0$ and $$(C+uv^T)^-1=C^-1-fracC^-1uv^TC^-11+v^TC^-1u.$$
$B=I+ea^T$, hence $B$ is invertible if $1+a^Te=1+sum_i=1^3a_i=1+trace(A) ne 0.$ and
$$B^-1=I-fracea^T1+trace(A)=I-fracA1+trace(A)$$
edited Jul 25 at 3:47
Parcly Taxel
33.6k136588
answered Jul 21 at 5:33
Siong Thye Goh
77.6k134795
edited Jul 25 at 3:47
Parcly Taxel
33.6k136588
edited Jul 25 at 3:47
Parcly Taxel
33.6k136588
edited Jul 25 at 3:47
Parcly Taxel
33.6k136588
33.6k136588
answered Jul 21 at 5:33
Siong Thye Goh
77.6k134795
answered Jul 21 at 5:33
Siong Thye Goh
77.6k134795
answered Jul 21 at 5:33
Siong Thye Goh
77.6k134795
77.6k134795
Thank you! This is beautiful. Just what I needed, and I would have never figured it out myself.
– user159452
Jul 21 at 11:04
add a comment |Â
Thank you! This is beautiful. Just what I needed, and I would have never figured it out myself.
– user159452
Jul 21 at 11:04
Thank you! This is beautiful. Just what I needed, and I would have never figured it out myself.
– user159452
Jul 21 at 11:04
Thank you! This is beautiful. Just what I needed, and I would have never figured it out myself.
– user159452
Jul 21 at 11:04
add a comment |Â
up vote
7
down vote
In the following, let us assume that $mboxtr(A)+1not = 0$.
We have $B=A+I$. Let us compute $C:=(A+I)left(I-fracAmboxtr(A)+1right)$ and see if we get the identity matrix.
$$C=-frac1mboxtr(A)+1A^2+fracmboxtr(A)mboxtr(A)+1A+I$$
so that to get the desired result, we need to prove that $A^2=mboxtr(A)A$.
The $i,j$ coefficient of $A^2$ is given by $$sum_k=1^n A_i,kA_k,j=sum_k=1^n a_ka_j=mboxtr(A)a_j=mboxtr(A)A_i,j $$
So that indeed, we obtain the desired identity.
NB : To justify that the matrix $B=A+I$ is invertible if and only if $mboxtr(A)+1not = 0$, note that the above computation actually establishes the identity $$(mboxtr(A)+1)I=(A+I)((mboxtr(A)+1)I-A)$$
from which the equivalence follows.
answered Jul 21 at 5:36
Suzet
2,211527
Thank you, Suzet.
– user159452
Jul 21 at 11:05
add a comment |Â
up vote
7
down vote
In the following, let us assume that $mboxtr(A)+1not = 0$.
We have $B=A+I$. Let us compute $C:=(A+I)left(I-fracAmboxtr(A)+1right)$ and see if we get the identity matrix.
$$C=-frac1mboxtr(A)+1A^2+fracmboxtr(A)mboxtr(A)+1A+I$$
so that to get the desired result, we need to prove that $A^2=mboxtr(A)A$.
The $i,j$ coefficient of $A^2$ is given by $$sum_k=1^n A_i,kA_k,j=sum_k=1^n a_ka_j=mboxtr(A)a_j=mboxtr(A)A_i,j $$
So that indeed, we obtain the desired identity.
NB : To justify that the matrix $B=A+I$ is invertible if and only if $mboxtr(A)+1not = 0$, note that the above computation actually establishes the identity $$(mboxtr(A)+1)I=(A+I)((mboxtr(A)+1)I-A)$$
from which the equivalence follows.
answered Jul 21 at 5:36
Suzet
2,211527
Thank you, Suzet.
– user159452
Jul 21 at 11:05
add a comment |Â
up vote
7
down vote
up vote
7
down vote
In the following, let us assume that $mboxtr(A)+1not = 0$.
We have $B=A+I$. Let us compute $C:=(A+I)left(I-fracAmboxtr(A)+1right)$ and see if we get the identity matrix.
$$C=-frac1mboxtr(A)+1A^2+fracmboxtr(A)mboxtr(A)+1A+I$$
so that to get the desired result, we need to prove that $A^2=mboxtr(A)A$.
The $i,j$ coefficient of $A^2$ is given by $$sum_k=1^n A_i,kA_k,j=sum_k=1^n a_ka_j=mboxtr(A)a_j=mboxtr(A)A_i,j $$
So that indeed, we obtain the desired identity.
NB : To justify that the matrix $B=A+I$ is invertible if and only if $mboxtr(A)+1not = 0$, note that the above computation actually establishes the identity $$(mboxtr(A)+1)I=(A+I)((mboxtr(A)+1)I-A)$$
from which the equivalence follows.
answered Jul 21 at 5:36
Suzet
2,211527
In the following, let us assume that $mboxtr(A)+1not = 0$.
We have $B=A+I$. Let us compute $C:=(A+I)left(I-fracAmboxtr(A)+1right)$ and see if we get the identity matrix.
$$C=-frac1mboxtr(A)+1A^2+fracmboxtr(A)mboxtr(A)+1A+I$$
so that to get the desired result, we need to prove that $A^2=mboxtr(A)A$.
The $i,j$ coefficient of $A^2$ is given by $$sum_k=1^n A_i,kA_k,j=sum_k=1^n a_ka_j=mboxtr(A)a_j=mboxtr(A)A_i,j $$
So that indeed, we obtain the desired identity.
NB : To justify that the matrix $B=A+I$ is invertible if and only if $mboxtr(A)+1not = 0$, note that the above computation actually establishes the identity $$(mboxtr(A)+1)I=(A+I)((mboxtr(A)+1)I-A)$$
from which the equivalence follows.
answered Jul 21 at 5:36
Suzet
2,211527
edited Jul 21 at 5:41
answered Jul 21 at 5:36
Suzet
2,211527
answered Jul 21 at 5:36
Suzet
2,211527
answered Jul 21 at 5:36
Suzet
2,211527
2,211527
Thank you, Suzet.
– user159452
Jul 21 at 11:05
add a comment |Â
Thank you, Suzet.
– user159452
Jul 21 at 11:05
Thank you, Suzet.
– user159452
Jul 21 at 11:05
Thank you, Suzet.
– user159452
Jul 21 at 11:05
add a comment |Â
up vote
2
down vote
The inverse of $B$:
$$B^-1=fractextadj(B)det(B).$$
The determinant of $B$:
$$det(B)=beginvmatrixa_1+1 & a_2 & a_3 \
a_1 & a_2+1 & a_3 \
a_1 & a_2 & a_3+1endvmatrix=
beginvmatrixa_1+1 & a_2 & a_3 \
-1 & 1 & 0 \
-1 & 0 & 1endvmatrix=a_1+a_2+a_3+1.$$
The adjugate of $B$:
$$textadj(B)=textC^T=\
beginpmatrix
(a_2+1)(a_3+1)-a_2a_3 & -a_2(a_3+1)+a_2a_3 & a_2a_3-a_3(a_2+1) \
-a_1(a_3+1)+a_1a_3 & (a_1+1)(a_3+1)-a_1a_3 & -a_3(a_1+1)+a_1a_3 \
a_1a_2-a_1(a_2+1) & -a_2(a_1+1)+a_1a_2 & (a_1+1)(a_2+1)-a_1a_2
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1-a_1 & -a_2 & -a_3 \
-a_1 & a_1+a_2+a_3+1-a_2 & -a_3 \
-a_1 & -a_2 & a_1+a_2+a_3+1-a_3
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1 & 0 & 0 \
0 & a_1+a_2+a_3+1 & 0 \
0 & 0 & a_1+a_2+a_3+1
endpmatrix-
beginpmatrix
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3
endpmatrix.
$$
Note: $C^T$ is the transpose of the cofactor matrix of $B$.
Hence, the result follows.
answered Jul 25 at 8:55
farruhota
13.7k2632
add a comment |Â
up vote
2
down vote
The inverse of $B$:
$$B^-1=fractextadj(B)det(B).$$
The determinant of $B$:
$$det(B)=beginvmatrixa_1+1 & a_2 & a_3 \
a_1 & a_2+1 & a_3 \
a_1 & a_2 & a_3+1endvmatrix=
beginvmatrixa_1+1 & a_2 & a_3 \
-1 & 1 & 0 \
-1 & 0 & 1endvmatrix=a_1+a_2+a_3+1.$$
The adjugate of $B$:
$$textadj(B)=textC^T=\
beginpmatrix
(a_2+1)(a_3+1)-a_2a_3 & -a_2(a_3+1)+a_2a_3 & a_2a_3-a_3(a_2+1) \
-a_1(a_3+1)+a_1a_3 & (a_1+1)(a_3+1)-a_1a_3 & -a_3(a_1+1)+a_1a_3 \
a_1a_2-a_1(a_2+1) & -a_2(a_1+1)+a_1a_2 & (a_1+1)(a_2+1)-a_1a_2
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1-a_1 & -a_2 & -a_3 \
-a_1 & a_1+a_2+a_3+1-a_2 & -a_3 \
-a_1 & -a_2 & a_1+a_2+a_3+1-a_3
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1 & 0 & 0 \
0 & a_1+a_2+a_3+1 & 0 \
0 & 0 & a_1+a_2+a_3+1
endpmatrix-
beginpmatrix
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3
endpmatrix.
$$
Note: $C^T$ is the transpose of the cofactor matrix of $B$.
Hence, the result follows.
answered Jul 25 at 8:55
farruhota
13.7k2632
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The inverse of $B$:
$$B^-1=fractextadj(B)det(B).$$
The determinant of $B$:
$$det(B)=beginvmatrixa_1+1 & a_2 & a_3 \
a_1 & a_2+1 & a_3 \
a_1 & a_2 & a_3+1endvmatrix=
beginvmatrixa_1+1 & a_2 & a_3 \
-1 & 1 & 0 \
-1 & 0 & 1endvmatrix=a_1+a_2+a_3+1.$$
The adjugate of $B$:
$$textadj(B)=textC^T=\
beginpmatrix
(a_2+1)(a_3+1)-a_2a_3 & -a_2(a_3+1)+a_2a_3 & a_2a_3-a_3(a_2+1) \
-a_1(a_3+1)+a_1a_3 & (a_1+1)(a_3+1)-a_1a_3 & -a_3(a_1+1)+a_1a_3 \
a_1a_2-a_1(a_2+1) & -a_2(a_1+1)+a_1a_2 & (a_1+1)(a_2+1)-a_1a_2
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1-a_1 & -a_2 & -a_3 \
-a_1 & a_1+a_2+a_3+1-a_2 & -a_3 \
-a_1 & -a_2 & a_1+a_2+a_3+1-a_3
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1 & 0 & 0 \
0 & a_1+a_2+a_3+1 & 0 \
0 & 0 & a_1+a_2+a_3+1
endpmatrix-
beginpmatrix
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3
endpmatrix.
$$
Note: $C^T$ is the transpose of the cofactor matrix of $B$.
Hence, the result follows.
answered Jul 25 at 8:55
farruhota
13.7k2632
The inverse of $B$:
$$B^-1=fractextadj(B)det(B).$$
The determinant of $B$:
$$det(B)=beginvmatrixa_1+1 & a_2 & a_3 \
a_1 & a_2+1 & a_3 \
a_1 & a_2 & a_3+1endvmatrix=
beginvmatrixa_1+1 & a_2 & a_3 \
-1 & 1 & 0 \
-1 & 0 & 1endvmatrix=a_1+a_2+a_3+1.$$
The adjugate of $B$:
$$textadj(B)=textC^T=\
beginpmatrix
(a_2+1)(a_3+1)-a_2a_3 & -a_2(a_3+1)+a_2a_3 & a_2a_3-a_3(a_2+1) \
-a_1(a_3+1)+a_1a_3 & (a_1+1)(a_3+1)-a_1a_3 & -a_3(a_1+1)+a_1a_3 \
a_1a_2-a_1(a_2+1) & -a_2(a_1+1)+a_1a_2 & (a_1+1)(a_2+1)-a_1a_2
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1-a_1 & -a_2 & -a_3 \
-a_1 & a_1+a_2+a_3+1-a_2 & -a_3 \
-a_1 & -a_2 & a_1+a_2+a_3+1-a_3
endpmatrix=\
beginpmatrix
a_1+a_2+a_3+1 & 0 & 0 \
0 & a_1+a_2+a_3+1 & 0 \
0 & 0 & a_1+a_2+a_3+1
endpmatrix-
beginpmatrix
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3 \
a_1 & a_2 & a_3
endpmatrix.
$$
Note: $C^T$ is the transpose of the cofactor matrix of $B$.
Hence, the result follows.
answered Jul 25 at 8:55
farruhota
13.7k2632
answered Jul 25 at 8:55
farruhota
13.7k2632
answered Jul 25 at 8:55
farruhota
13.7k2632
answered Jul 25 at 8:55
farruhota
13.7k2632
13.7k2632
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1
Your matrix is not symmetric, unless $a_1=a_2=a_3$.
– Suzet
Jul 21 at 5:20
Thanks. Edited in question.
– user159452
Jul 21 at 5:28