Generic expression for the (i,j) element of matrix ABC
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It is given that A is a matrix with dimensions m x n, and C is a matrix with dimensions r x s.
In order for the product ABC to be defined, I assumed that matrix B has dimensions n x r.
I am now asked to write an expression for the (i,j) element of ABC in terms of only the elements of matrices A, B, and C. I don't know how to begin. It's not easy for me to wrap my head around the fact that the matrices have unknown dimensions respresented by variables. If they were defined, I could deduce an expression based off of matrix multiplication rules. Any tips on how to proceed?
linear-algebra matrices
asked Aug 29 at 6:34
Dominic Hicks
275
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up vote
2
down vote
favorite
It is given that A is a matrix with dimensions m x n, and C is a matrix with dimensions r x s.
In order for the product ABC to be defined, I assumed that matrix B has dimensions n x r.
I am now asked to write an expression for the (i,j) element of ABC in terms of only the elements of matrices A, B, and C. I don't know how to begin. It's not easy for me to wrap my head around the fact that the matrices have unknown dimensions respresented by variables. If they were defined, I could deduce an expression based off of matrix multiplication rules. Any tips on how to proceed?
linear-algebra matrices
asked Aug 29 at 6:34
Dominic Hicks
275
Write $ABC = (AB) times C$. Now, express the $(i,j)th$ entry of $ABC$ in terms of the entries of $AB$ and the entries of $C$. Finally,write out the $AB$ entries in terms of the entries of $A$ and $B$, using another variable, to get the general expression. This is nothing but iterated use of the formula for two matrices.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 29 at 6:38
The $(i,j)$-entry of $AB$ is $sum_k=1^n a_ikb_kj$. Now, you need something like that for $ABC$.
– Lord Shark the Unknown
Aug 29 at 6:47
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
It is given that A is a matrix with dimensions m x n, and C is a matrix with dimensions r x s.
In order for the product ABC to be defined, I assumed that matrix B has dimensions n x r.
I am now asked to write an expression for the (i,j) element of ABC in terms of only the elements of matrices A, B, and C. I don't know how to begin. It's not easy for me to wrap my head around the fact that the matrices have unknown dimensions respresented by variables. If they were defined, I could deduce an expression based off of matrix multiplication rules. Any tips on how to proceed?
linear-algebra matrices
asked Aug 29 at 6:34
Dominic Hicks
275
It is given that A is a matrix with dimensions m x n, and C is a matrix with dimensions r x s.
In order for the product ABC to be defined, I assumed that matrix B has dimensions n x r.
I am now asked to write an expression for the (i,j) element of ABC in terms of only the elements of matrices A, B, and C. I don't know how to begin. It's not easy for me to wrap my head around the fact that the matrices have unknown dimensions respresented by variables. If they were defined, I could deduce an expression based off of matrix multiplication rules. Any tips on how to proceed?
linear-algebra matrices
asked Aug 29 at 6:34
Dominic Hicks
275
asked Aug 29 at 6:34
Dominic Hicks
275
asked Aug 29 at 6:34
Dominic Hicks
275
asked Aug 29 at 6:34
Dominic Hicks
275
275
Write $ABC = (AB) times C$. Now, express the $(i,j)th$ entry of $ABC$ in terms of the entries of $AB$ and the entries of $C$. Finally,write out the $AB$ entries in terms of the entries of $A$ and $B$, using another variable, to get the general expression. This is nothing but iterated use of the formula for two matrices.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 29 at 6:38
The $(i,j)$-entry of $AB$ is $sum_k=1^n a_ikb_kj$. Now, you need something like that for $ABC$.
– Lord Shark the Unknown
Aug 29 at 6:47
add a comment |Â
Write $ABC = (AB) times C$. Now, express the $(i,j)th$ entry of $ABC$ in terms of the entries of $AB$ and the entries of $C$. Finally,write out the $AB$ entries in terms of the entries of $A$ and $B$, using another variable, to get the general expression. This is nothing but iterated use of the formula for two matrices.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 29 at 6:38
The $(i,j)$-entry of $AB$ is $sum_k=1^n a_ikb_kj$. Now, you need something like that for $ABC$.
– Lord Shark the Unknown
Aug 29 at 6:47
Write $ABC = (AB) times C$. Now, express the $(i,j)th$ entry of $ABC$ in terms of the entries of $AB$ and the entries of $C$. Finally,write out the $AB$ entries in terms of the entries of $A$ and $B$, using another variable, to get the general expression. This is nothing but iterated use of the formula for two matrices.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 29 at 6:38
Write $ABC = (AB) times C$. Now, express the $(i,j)th$ entry of $ABC$ in terms of the entries of $AB$ and the entries of $C$. Finally,write out the $AB$ entries in terms of the entries of $A$ and $B$, using another variable, to get the general expression. This is nothing but iterated use of the formula for two matrices.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 29 at 6:38
The $(i,j)$-entry of $AB$ is $sum_k=1^n a_ikb_kj$. Now, you need something like that for $ABC$.
– Lord Shark the Unknown
Aug 29 at 6:47
The $(i,j)$-entry of $AB$ is $sum_k=1^n a_ikb_kj$. Now, you need something like that for $ABC$.
– Lord Shark the Unknown
Aug 29 at 6:47
add a comment |Â
1 Answer
1
active
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votes
up vote
1
down vote
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Let $A$ be a $m$-by-$n$ matrix, $B$ a $n$-by-$p$ matrix, and $C$ a $p$-by-$r$ matrix. I'm sure that you're aware that the entry in the $j$th row and $k$th column of $AB$ is given by
$$(AB)_j,k = sum_r=1^n A_j,r B_r,k.$$
Notice this is the product of the $(j,r)$th entry of $A$ and $(r,k)$th entry of $B$. To represent the $(j,k)$th entry of $(AB)C$, we do so by taking the product of the $(j,r)$th entry of $AB$ and $(r,k)$th entry of $C$. The main difference is that our sum is no longer to $n$, but rather to $p$ (can you tell why)?
We then have the following (in a spoiler):
$$((AB)C)_j,k = sum_r=1^p (AB)_j,rC_r,k = sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k. $$
Also, as a quick extra tidbit, technically it's improper to call it quits after the above work. How do you know for instance that $(AB)C=ABC$? What if $A(BC) neq (AB)C$? Which one then would denote $ABC$?
Lucky for us, they're the same! Of course that takes a proof. Can you see how to rearrange the above sum to also show that $(AB)C=A(BC)$? (That is, matrix multiplication is associative!) Here is that in a spoiler:
beginalign* sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k &= sum_r=1^p left( sum_s=1^n A_j,s B_s,r C_r,k right) \ &= sum_s=1^n A_j,s left( sum_r=1^p B_s,r C_r,k right) \ &= sum_s=1^nA_j,s (BC)_s,k\ & = (A(BC))_j,k. endalign*
answered Aug 29 at 7:25
Andrew Tawfeek
1,7901722
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $A$ be a $m$-by-$n$ matrix, $B$ a $n$-by-$p$ matrix, and $C$ a $p$-by-$r$ matrix. I'm sure that you're aware that the entry in the $j$th row and $k$th column of $AB$ is given by
$$(AB)_j,k = sum_r=1^n A_j,r B_r,k.$$
Notice this is the product of the $(j,r)$th entry of $A$ and $(r,k)$th entry of $B$. To represent the $(j,k)$th entry of $(AB)C$, we do so by taking the product of the $(j,r)$th entry of $AB$ and $(r,k)$th entry of $C$. The main difference is that our sum is no longer to $n$, but rather to $p$ (can you tell why)?
We then have the following (in a spoiler):
$$((AB)C)_j,k = sum_r=1^p (AB)_j,rC_r,k = sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k. $$
Also, as a quick extra tidbit, technically it's improper to call it quits after the above work. How do you know for instance that $(AB)C=ABC$? What if $A(BC) neq (AB)C$? Which one then would denote $ABC$?
Lucky for us, they're the same! Of course that takes a proof. Can you see how to rearrange the above sum to also show that $(AB)C=A(BC)$? (That is, matrix multiplication is associative!) Here is that in a spoiler:
beginalign* sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k &= sum_r=1^p left( sum_s=1^n A_j,s B_s,r C_r,k right) \ &= sum_s=1^n A_j,s left( sum_r=1^p B_s,r C_r,k right) \ &= sum_s=1^nA_j,s (BC)_s,k\ & = (A(BC))_j,k. endalign*
answered Aug 29 at 7:25
Andrew Tawfeek
1,7901722
add a comment |Â
up vote
1
down vote
accepted
Let $A$ be a $m$-by-$n$ matrix, $B$ a $n$-by-$p$ matrix, and $C$ a $p$-by-$r$ matrix. I'm sure that you're aware that the entry in the $j$th row and $k$th column of $AB$ is given by
$$(AB)_j,k = sum_r=1^n A_j,r B_r,k.$$
Notice this is the product of the $(j,r)$th entry of $A$ and $(r,k)$th entry of $B$. To represent the $(j,k)$th entry of $(AB)C$, we do so by taking the product of the $(j,r)$th entry of $AB$ and $(r,k)$th entry of $C$. The main difference is that our sum is no longer to $n$, but rather to $p$ (can you tell why)?
We then have the following (in a spoiler):
$$((AB)C)_j,k = sum_r=1^p (AB)_j,rC_r,k = sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k. $$
Also, as a quick extra tidbit, technically it's improper to call it quits after the above work. How do you know for instance that $(AB)C=ABC$? What if $A(BC) neq (AB)C$? Which one then would denote $ABC$?
Lucky for us, they're the same! Of course that takes a proof. Can you see how to rearrange the above sum to also show that $(AB)C=A(BC)$? (That is, matrix multiplication is associative!) Here is that in a spoiler:
beginalign* sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k &= sum_r=1^p left( sum_s=1^n A_j,s B_s,r C_r,k right) \ &= sum_s=1^n A_j,s left( sum_r=1^p B_s,r C_r,k right) \ &= sum_s=1^nA_j,s (BC)_s,k\ & = (A(BC))_j,k. endalign*
answered Aug 29 at 7:25
Andrew Tawfeek
1,7901722
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $A$ be a $m$-by-$n$ matrix, $B$ a $n$-by-$p$ matrix, and $C$ a $p$-by-$r$ matrix. I'm sure that you're aware that the entry in the $j$th row and $k$th column of $AB$ is given by
$$(AB)_j,k = sum_r=1^n A_j,r B_r,k.$$
Notice this is the product of the $(j,r)$th entry of $A$ and $(r,k)$th entry of $B$. To represent the $(j,k)$th entry of $(AB)C$, we do so by taking the product of the $(j,r)$th entry of $AB$ and $(r,k)$th entry of $C$. The main difference is that our sum is no longer to $n$, but rather to $p$ (can you tell why)?
We then have the following (in a spoiler):
$$((AB)C)_j,k = sum_r=1^p (AB)_j,rC_r,k = sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k. $$
Also, as a quick extra tidbit, technically it's improper to call it quits after the above work. How do you know for instance that $(AB)C=ABC$? What if $A(BC) neq (AB)C$? Which one then would denote $ABC$?
Lucky for us, they're the same! Of course that takes a proof. Can you see how to rearrange the above sum to also show that $(AB)C=A(BC)$? (That is, matrix multiplication is associative!) Here is that in a spoiler:
beginalign* sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k &= sum_r=1^p left( sum_s=1^n A_j,s B_s,r C_r,k right) \ &= sum_s=1^n A_j,s left( sum_r=1^p B_s,r C_r,k right) \ &= sum_s=1^nA_j,s (BC)_s,k\ & = (A(BC))_j,k. endalign*
answered Aug 29 at 7:25
Andrew Tawfeek
1,7901722
Let $A$ be a $m$-by-$n$ matrix, $B$ a $n$-by-$p$ matrix, and $C$ a $p$-by-$r$ matrix. I'm sure that you're aware that the entry in the $j$th row and $k$th column of $AB$ is given by
$$(AB)_j,k = sum_r=1^n A_j,r B_r,k.$$
Notice this is the product of the $(j,r)$th entry of $A$ and $(r,k)$th entry of $B$. To represent the $(j,k)$th entry of $(AB)C$, we do so by taking the product of the $(j,r)$th entry of $AB$ and $(r,k)$th entry of $C$. The main difference is that our sum is no longer to $n$, but rather to $p$ (can you tell why)?
We then have the following (in a spoiler):
$$((AB)C)_j,k = sum_r=1^p (AB)_j,rC_r,k = sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k. $$
Also, as a quick extra tidbit, technically it's improper to call it quits after the above work. How do you know for instance that $(AB)C=ABC$? What if $A(BC) neq (AB)C$? Which one then would denote $ABC$?
Lucky for us, they're the same! Of course that takes a proof. Can you see how to rearrange the above sum to also show that $(AB)C=A(BC)$? (That is, matrix multiplication is associative!) Here is that in a spoiler:
beginalign* sum_r=1^p left( sum_s=1^n A_j,s B_s,r right) C_r,k &= sum_r=1^p left( sum_s=1^n A_j,s B_s,r C_r,k right) \ &= sum_s=1^n A_j,s left( sum_r=1^p B_s,r C_r,k right) \ &= sum_s=1^nA_j,s (BC)_s,k\ & = (A(BC))_j,k. endalign*
answered Aug 29 at 7:25
Andrew Tawfeek
1,7901722
edited Aug 29 at 7:36
answered Aug 29 at 7:25
Andrew Tawfeek
1,7901722
answered Aug 29 at 7:25
Andrew Tawfeek
1,7901722
answered Aug 29 at 7:25
Andrew Tawfeek
1,7901722
1,7901722
add a comment |Â
add a comment |Â
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Write $ABC = (AB) times C$. Now, express the $(i,j)th$ entry of $ABC$ in terms of the entries of $AB$ and the entries of $C$. Finally,write out the $AB$ entries in terms of the entries of $A$ and $B$, using another variable, to get the general expression. This is nothing but iterated use of the formula for two matrices.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 29 at 6:38
The $(i,j)$-entry of $AB$ is $sum_k=1^n a_ikb_kj$. Now, you need something like that for $ABC$.
– Lord Shark the Unknown
Aug 29 at 6:47