In matrices while performing a row operation, am i allow to multiply a row by a zero or negative number?

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In matrices, while performing a row operation, am I allow to multiply a row by a zero or negative number? Please help thanks!
matrices matrix-equations
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In matrices, while performing a row operation, am I allow to multiply a row by a zero or negative number? Please help thanks!
matrices matrix-equations
1
negative number yes, zero - usually not.
– Matti P.
Sep 10 at 12:06
@MattiP. Thanks, so i can still perform zero multiplication?
– Jon Wick
Sep 10 at 12:21
Well, first of all, your question lacks a lot of context. I don't know what you're trying to calculate. I assume that you're solving a linear system of equations with the elimination method. If you multiply a row by zero, that row then only represents the equation $0=0$, effectively taking out information. So ... in general I don't recommend multiplying by zero.
– Matti P.
Sep 10 at 12:24
If you are allowed to multiply a row by zero, then it follows that you are allowed to multiply every row by zero, which gets you the all-zero matrix. What do you do then, Jon?
– Gerry Myerson
Sep 10 at 12:55
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up vote
0
down vote
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up vote
0
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favorite
In matrices, while performing a row operation, am I allow to multiply a row by a zero or negative number? Please help thanks!
matrices matrix-equations
In matrices, while performing a row operation, am I allow to multiply a row by a zero or negative number? Please help thanks!
matrices matrix-equations
matrices matrix-equations
asked Sep 10 at 12:00
Jon Wick
64
64
1
negative number yes, zero - usually not.
– Matti P.
Sep 10 at 12:06
@MattiP. Thanks, so i can still perform zero multiplication?
– Jon Wick
Sep 10 at 12:21
Well, first of all, your question lacks a lot of context. I don't know what you're trying to calculate. I assume that you're solving a linear system of equations with the elimination method. If you multiply a row by zero, that row then only represents the equation $0=0$, effectively taking out information. So ... in general I don't recommend multiplying by zero.
– Matti P.
Sep 10 at 12:24
If you are allowed to multiply a row by zero, then it follows that you are allowed to multiply every row by zero, which gets you the all-zero matrix. What do you do then, Jon?
– Gerry Myerson
Sep 10 at 12:55
add a comment |Â
1
negative number yes, zero - usually not.
– Matti P.
Sep 10 at 12:06
@MattiP. Thanks, so i can still perform zero multiplication?
– Jon Wick
Sep 10 at 12:21
Well, first of all, your question lacks a lot of context. I don't know what you're trying to calculate. I assume that you're solving a linear system of equations with the elimination method. If you multiply a row by zero, that row then only represents the equation $0=0$, effectively taking out information. So ... in general I don't recommend multiplying by zero.
– Matti P.
Sep 10 at 12:24
If you are allowed to multiply a row by zero, then it follows that you are allowed to multiply every row by zero, which gets you the all-zero matrix. What do you do then, Jon?
– Gerry Myerson
Sep 10 at 12:55
1
1
negative number yes, zero - usually not.
– Matti P.
Sep 10 at 12:06
negative number yes, zero - usually not.
– Matti P.
Sep 10 at 12:06
@MattiP. Thanks, so i can still perform zero multiplication?
– Jon Wick
Sep 10 at 12:21
@MattiP. Thanks, so i can still perform zero multiplication?
– Jon Wick
Sep 10 at 12:21
Well, first of all, your question lacks a lot of context. I don't know what you're trying to calculate. I assume that you're solving a linear system of equations with the elimination method. If you multiply a row by zero, that row then only represents the equation $0=0$, effectively taking out information. So ... in general I don't recommend multiplying by zero.
– Matti P.
Sep 10 at 12:24
Well, first of all, your question lacks a lot of context. I don't know what you're trying to calculate. I assume that you're solving a linear system of equations with the elimination method. If you multiply a row by zero, that row then only represents the equation $0=0$, effectively taking out information. So ... in general I don't recommend multiplying by zero.
– Matti P.
Sep 10 at 12:24
If you are allowed to multiply a row by zero, then it follows that you are allowed to multiply every row by zero, which gets you the all-zero matrix. What do you do then, Jon?
– Gerry Myerson
Sep 10 at 12:55
If you are allowed to multiply a row by zero, then it follows that you are allowed to multiply every row by zero, which gets you the all-zero matrix. What do you do then, Jon?
– Gerry Myerson
Sep 10 at 12:55
add a comment |Â
1 Answer
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4
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There are three types of row operations:
- Multiplying a row by a constant $cne0$
- Adding to a row a different row multiplied by a constant $d$
- Swapping two rows
The third type is of no concern with your question.
Why the limitations on the first two types?
The idea is that row operations should be reversible, so when we perform a row operation, we can perform another one that “undoes†the effect of the previous one. The main reason is that such row operations don't change the solution set of a linear system (translated into matrix form), exactly because the operations are reversible.
Multiplying a row by $0$ is not reversible: this should be clear because doing this operation on all rows will yield the null matrix and we surely lose information. Thus is disallowed.
Adding to a row itself multiplied by $-1$ is the same as multiplying the row by $0$, so in general it is disallowed to add a row to itself multiplied by a scalar.
How do you reverse an operation of the first kind? By multiplying the same row by $c^-1$. What for an operation of the second kind? Say we add to row $i$ row $j$ multiplied by $d$; we reverse this by adding to row $i$ row $j$ multiplied by $-d$.
The constants $c$ and $d$ can be anything (but for the first type $c$ must be nonzero). Even $d=0$ is allowed in operations of the second kind: it is just doing nothing, which is of course reversed by doing nothing again.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
There are three types of row operations:
- Multiplying a row by a constant $cne0$
- Adding to a row a different row multiplied by a constant $d$
- Swapping two rows
The third type is of no concern with your question.
Why the limitations on the first two types?
The idea is that row operations should be reversible, so when we perform a row operation, we can perform another one that “undoes†the effect of the previous one. The main reason is that such row operations don't change the solution set of a linear system (translated into matrix form), exactly because the operations are reversible.
Multiplying a row by $0$ is not reversible: this should be clear because doing this operation on all rows will yield the null matrix and we surely lose information. Thus is disallowed.
Adding to a row itself multiplied by $-1$ is the same as multiplying the row by $0$, so in general it is disallowed to add a row to itself multiplied by a scalar.
How do you reverse an operation of the first kind? By multiplying the same row by $c^-1$. What for an operation of the second kind? Say we add to row $i$ row $j$ multiplied by $d$; we reverse this by adding to row $i$ row $j$ multiplied by $-d$.
The constants $c$ and $d$ can be anything (but for the first type $c$ must be nonzero). Even $d=0$ is allowed in operations of the second kind: it is just doing nothing, which is of course reversed by doing nothing again.
add a comment |Â
up vote
4
down vote
accepted
There are three types of row operations:
- Multiplying a row by a constant $cne0$
- Adding to a row a different row multiplied by a constant $d$
- Swapping two rows
The third type is of no concern with your question.
Why the limitations on the first two types?
The idea is that row operations should be reversible, so when we perform a row operation, we can perform another one that “undoes†the effect of the previous one. The main reason is that such row operations don't change the solution set of a linear system (translated into matrix form), exactly because the operations are reversible.
Multiplying a row by $0$ is not reversible: this should be clear because doing this operation on all rows will yield the null matrix and we surely lose information. Thus is disallowed.
Adding to a row itself multiplied by $-1$ is the same as multiplying the row by $0$, so in general it is disallowed to add a row to itself multiplied by a scalar.
How do you reverse an operation of the first kind? By multiplying the same row by $c^-1$. What for an operation of the second kind? Say we add to row $i$ row $j$ multiplied by $d$; we reverse this by adding to row $i$ row $j$ multiplied by $-d$.
The constants $c$ and $d$ can be anything (but for the first type $c$ must be nonzero). Even $d=0$ is allowed in operations of the second kind: it is just doing nothing, which is of course reversed by doing nothing again.
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
There are three types of row operations:
- Multiplying a row by a constant $cne0$
- Adding to a row a different row multiplied by a constant $d$
- Swapping two rows
The third type is of no concern with your question.
Why the limitations on the first two types?
The idea is that row operations should be reversible, so when we perform a row operation, we can perform another one that “undoes†the effect of the previous one. The main reason is that such row operations don't change the solution set of a linear system (translated into matrix form), exactly because the operations are reversible.
Multiplying a row by $0$ is not reversible: this should be clear because doing this operation on all rows will yield the null matrix and we surely lose information. Thus is disallowed.
Adding to a row itself multiplied by $-1$ is the same as multiplying the row by $0$, so in general it is disallowed to add a row to itself multiplied by a scalar.
How do you reverse an operation of the first kind? By multiplying the same row by $c^-1$. What for an operation of the second kind? Say we add to row $i$ row $j$ multiplied by $d$; we reverse this by adding to row $i$ row $j$ multiplied by $-d$.
The constants $c$ and $d$ can be anything (but for the first type $c$ must be nonzero). Even $d=0$ is allowed in operations of the second kind: it is just doing nothing, which is of course reversed by doing nothing again.
There are three types of row operations:
- Multiplying a row by a constant $cne0$
- Adding to a row a different row multiplied by a constant $d$
- Swapping two rows
The third type is of no concern with your question.
Why the limitations on the first two types?
The idea is that row operations should be reversible, so when we perform a row operation, we can perform another one that “undoes†the effect of the previous one. The main reason is that such row operations don't change the solution set of a linear system (translated into matrix form), exactly because the operations are reversible.
Multiplying a row by $0$ is not reversible: this should be clear because doing this operation on all rows will yield the null matrix and we surely lose information. Thus is disallowed.
Adding to a row itself multiplied by $-1$ is the same as multiplying the row by $0$, so in general it is disallowed to add a row to itself multiplied by a scalar.
How do you reverse an operation of the first kind? By multiplying the same row by $c^-1$. What for an operation of the second kind? Say we add to row $i$ row $j$ multiplied by $d$; we reverse this by adding to row $i$ row $j$ multiplied by $-d$.
The constants $c$ and $d$ can be anything (but for the first type $c$ must be nonzero). Even $d=0$ is allowed in operations of the second kind: it is just doing nothing, which is of course reversed by doing nothing again.
answered Sep 10 at 12:49


egreg
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167k1281190
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1
negative number yes, zero - usually not.
– Matti P.
Sep 10 at 12:06
@MattiP. Thanks, so i can still perform zero multiplication?
– Jon Wick
Sep 10 at 12:21
Well, first of all, your question lacks a lot of context. I don't know what you're trying to calculate. I assume that you're solving a linear system of equations with the elimination method. If you multiply a row by zero, that row then only represents the equation $0=0$, effectively taking out information. So ... in general I don't recommend multiplying by zero.
– Matti P.
Sep 10 at 12:24
If you are allowed to multiply a row by zero, then it follows that you are allowed to multiply every row by zero, which gets you the all-zero matrix. What do you do then, Jon?
– Gerry Myerson
Sep 10 at 12:55