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Gauss Jordan elimination problem.
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I am reading about Gauss Jordan elimination and this makes sense:
We want 0s about every leading 1 which seems to make determining the variables a lot easier.
But how does this one work? I can't seem to figure out the intermediate steps:
It looks like to start the 4th row is swapped with the first.
linear-algebra
asked Jul 16 at 17:36
Jwan622
1,61211224
add a comment |Â
up vote
0
down vote
favorite
I am reading about Gauss Jordan elimination and this makes sense:
We want 0s about every leading 1 which seems to make determining the variables a lot easier.
But how does this one work? I can't seem to figure out the intermediate steps:
It looks like to start the 4th row is swapped with the first.
linear-algebra
asked Jul 16 at 17:36
Jwan622
1,61211224
There are many different sequences of steps one can take to get a matrix into RREF... one such algorithm which will always work (though perhaps not the most efficiently in certain scenarios) can be found here. The final result of the row reduction will always be the same regardless of specific steps followed.
– JMoravitz
Jul 16 at 18:39
One such sequence of steps might begin $R_4-R_1mapsto R_4, R_2leftrightarrow R_3,R_1-R_2mapsto R_1,dots$. Another might begin $R_1leftrightarrow R_4,dots$
– JMoravitz
Jul 16 at 18:40
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am reading about Gauss Jordan elimination and this makes sense:
We want 0s about every leading 1 which seems to make determining the variables a lot easier.
But how does this one work? I can't seem to figure out the intermediate steps:
It looks like to start the 4th row is swapped with the first.
linear-algebra
asked Jul 16 at 17:36
Jwan622
1,61211224
I am reading about Gauss Jordan elimination and this makes sense:
We want 0s about every leading 1 which seems to make determining the variables a lot easier.
But how does this one work? I can't seem to figure out the intermediate steps:
It looks like to start the 4th row is swapped with the first.
linear-algebra
asked Jul 16 at 17:36
Jwan622
1,61211224
asked Jul 16 at 17:36
Jwan622
1,61211224
asked Jul 16 at 17:36
Jwan622
1,61211224
asked Jul 16 at 17:36
Jwan622
1,61211224
1,61211224
There are many different sequences of steps one can take to get a matrix into RREF... one such algorithm which will always work (though perhaps not the most efficiently in certain scenarios) can be found here. The final result of the row reduction will always be the same regardless of specific steps followed.
– JMoravitz
Jul 16 at 18:39
One such sequence of steps might begin $R_4-R_1mapsto R_4, R_2leftrightarrow R_3,R_1-R_2mapsto R_1,dots$. Another might begin $R_1leftrightarrow R_4,dots$
– JMoravitz
Jul 16 at 18:40
add a comment |Â
There are many different sequences of steps one can take to get a matrix into RREF... one such algorithm which will always work (though perhaps not the most efficiently in certain scenarios) can be found here. The final result of the row reduction will always be the same regardless of specific steps followed.
– JMoravitz
Jul 16 at 18:39
One such sequence of steps might begin $R_4-R_1mapsto R_4, R_2leftrightarrow R_3,R_1-R_2mapsto R_1,dots$. Another might begin $R_1leftrightarrow R_4,dots$
– JMoravitz
Jul 16 at 18:40
There are many different sequences of steps one can take to get a matrix into RREF... one such algorithm which will always work (though perhaps not the most efficiently in certain scenarios) can be found here. The final result of the row reduction will always be the same regardless of specific steps followed.
– JMoravitz
Jul 16 at 18:39
There are many different sequences of steps one can take to get a matrix into RREF... one such algorithm which will always work (though perhaps not the most efficiently in certain scenarios) can be found here. The final result of the row reduction will always be the same regardless of specific steps followed.
– JMoravitz
Jul 16 at 18:39
One such sequence of steps might begin $R_4-R_1mapsto R_4, R_2leftrightarrow R_3,R_1-R_2mapsto R_1,dots$. Another might begin $R_1leftrightarrow R_4,dots$
– JMoravitz
Jul 16 at 18:40
One such sequence of steps might begin $R_4-R_1mapsto R_4, R_2leftrightarrow R_3,R_1-R_2mapsto R_1,dots$. Another might begin $R_1leftrightarrow R_4,dots$
– JMoravitz
Jul 16 at 18:40
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
In my post here I describe an algorithm which will guarantee that you reach RREF. (Side note: this isn't the algorithm that computers use, but it is similar)
As mentioned in comments above, there are many different sequences of steps you can take to row reduce a matrix completely, but the final result of any row reduction will be exactly the same. That is to say that you may if you wish begin the row reduction process for your specific example with swapping the fourth and first rows and that will still be a fine beginning (though you might not have known ahead of time that this was a particularly useful thing to do).
Following the algorithm I link strictly to a t for your specific matrix begins:
$beginbmatrixcolorred1&1&0&0&0&20\0&0&1&-1&0&-20\0&1&1&0&0&20\1&0&0&0&-1&-10\0&0&0&-1&1&-10endbmatrix$
We first check that the first-row first-column entry (marked in $colorredred$) is nonzero. It is! On to the next step... we check that it is equal to $1$ It is! Now, we zero out everything else in its column by subtracting appropriate multiples of the first row from other rows as necessary.
$R_4-R_1mapsto R_4$
$beginbmatrix colorred1&1&0&0&0&20\colorblue0&0&1&-1&0&-20\colorblue0&1&1&0&0&20\colorblue0&-1&0&0&-1&-30\colorblue0&0&0&-1&1&-10endbmatrix$
Now that that column is taken care of, we can move our attention to the next pivot location, marked below in $colorredred$.
$beginbmatrix 1&1&0&0&0&20\0&colorred0&1&-1&0&-20\0&1&1&0&0&20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We ask the question of it it is zero. It is unfortunately. We then ask the question of if there are any nonzero entries from the same column below it. There is, for example in the third row. We swap the rows then.
$R_2leftrightarrow R_3$
$beginbmatrix1&1&0&0&0&20\0&colorred1&1&0&0&20\0&0&1&-1&0&-20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We then ask the question of if the pivot we are focusing on currently is equal to $1$. It is! Now, we subtract appropriate multiples of that row from other rows in an attempt to zero out the rest of the pivot's column.
$R_1-R_2mapsto R_1$
$R_4+R_2mapsto R_4$
$beginbmatrix1&colorblue0&-1&0&0&0\0&colorred1&1&0&0&20\0&colorblue0&1&-1&0&-20\0&colorblue0&1&0&-1&-10\0&colorblue0&0&-1&1&-10endbmatrix$
Continue in this fashion, positioning pivot points from left-to-right and clearing out all other entries in the columns to be zero, swapping rows if necessary to get a nonzero entry in the pivot location, dividing or multiplying rows to get pivot positions equal to $1$ rather than some other nonzero number, or even skipping columns entirely if no nonzero entry can act as a pivot for that column.
answered Jul 16 at 19:30
JMoravitz
44.3k33481
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
In my post here I describe an algorithm which will guarantee that you reach RREF. (Side note: this isn't the algorithm that computers use, but it is similar)
As mentioned in comments above, there are many different sequences of steps you can take to row reduce a matrix completely, but the final result of any row reduction will be exactly the same. That is to say that you may if you wish begin the row reduction process for your specific example with swapping the fourth and first rows and that will still be a fine beginning (though you might not have known ahead of time that this was a particularly useful thing to do).
Following the algorithm I link strictly to a t for your specific matrix begins:
$beginbmatrixcolorred1&1&0&0&0&20\0&0&1&-1&0&-20\0&1&1&0&0&20\1&0&0&0&-1&-10\0&0&0&-1&1&-10endbmatrix$
We first check that the first-row first-column entry (marked in $colorredred$) is nonzero. It is! On to the next step... we check that it is equal to $1$ It is! Now, we zero out everything else in its column by subtracting appropriate multiples of the first row from other rows as necessary.
$R_4-R_1mapsto R_4$
$beginbmatrix colorred1&1&0&0&0&20\colorblue0&0&1&-1&0&-20\colorblue0&1&1&0&0&20\colorblue0&-1&0&0&-1&-30\colorblue0&0&0&-1&1&-10endbmatrix$
Now that that column is taken care of, we can move our attention to the next pivot location, marked below in $colorredred$.
$beginbmatrix 1&1&0&0&0&20\0&colorred0&1&-1&0&-20\0&1&1&0&0&20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We ask the question of it it is zero. It is unfortunately. We then ask the question of if there are any nonzero entries from the same column below it. There is, for example in the third row. We swap the rows then.
$R_2leftrightarrow R_3$
$beginbmatrix1&1&0&0&0&20\0&colorred1&1&0&0&20\0&0&1&-1&0&-20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We then ask the question of if the pivot we are focusing on currently is equal to $1$. It is! Now, we subtract appropriate multiples of that row from other rows in an attempt to zero out the rest of the pivot's column.
$R_1-R_2mapsto R_1$
$R_4+R_2mapsto R_4$
$beginbmatrix1&colorblue0&-1&0&0&0\0&colorred1&1&0&0&20\0&colorblue0&1&-1&0&-20\0&colorblue0&1&0&-1&-10\0&colorblue0&0&-1&1&-10endbmatrix$
Continue in this fashion, positioning pivot points from left-to-right and clearing out all other entries in the columns to be zero, swapping rows if necessary to get a nonzero entry in the pivot location, dividing or multiplying rows to get pivot positions equal to $1$ rather than some other nonzero number, or even skipping columns entirely if no nonzero entry can act as a pivot for that column.
answered Jul 16 at 19:30
JMoravitz
44.3k33481
add a comment |Â
up vote
1
down vote
accepted
In my post here I describe an algorithm which will guarantee that you reach RREF. (Side note: this isn't the algorithm that computers use, but it is similar)
As mentioned in comments above, there are many different sequences of steps you can take to row reduce a matrix completely, but the final result of any row reduction will be exactly the same. That is to say that you may if you wish begin the row reduction process for your specific example with swapping the fourth and first rows and that will still be a fine beginning (though you might not have known ahead of time that this was a particularly useful thing to do).
Following the algorithm I link strictly to a t for your specific matrix begins:
$beginbmatrixcolorred1&1&0&0&0&20\0&0&1&-1&0&-20\0&1&1&0&0&20\1&0&0&0&-1&-10\0&0&0&-1&1&-10endbmatrix$
We first check that the first-row first-column entry (marked in $colorredred$) is nonzero. It is! On to the next step... we check that it is equal to $1$ It is! Now, we zero out everything else in its column by subtracting appropriate multiples of the first row from other rows as necessary.
$R_4-R_1mapsto R_4$
$beginbmatrix colorred1&1&0&0&0&20\colorblue0&0&1&-1&0&-20\colorblue0&1&1&0&0&20\colorblue0&-1&0&0&-1&-30\colorblue0&0&0&-1&1&-10endbmatrix$
Now that that column is taken care of, we can move our attention to the next pivot location, marked below in $colorredred$.
$beginbmatrix 1&1&0&0&0&20\0&colorred0&1&-1&0&-20\0&1&1&0&0&20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We ask the question of it it is zero. It is unfortunately. We then ask the question of if there are any nonzero entries from the same column below it. There is, for example in the third row. We swap the rows then.
$R_2leftrightarrow R_3$
$beginbmatrix1&1&0&0&0&20\0&colorred1&1&0&0&20\0&0&1&-1&0&-20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We then ask the question of if the pivot we are focusing on currently is equal to $1$. It is! Now, we subtract appropriate multiples of that row from other rows in an attempt to zero out the rest of the pivot's column.
$R_1-R_2mapsto R_1$
$R_4+R_2mapsto R_4$
$beginbmatrix1&colorblue0&-1&0&0&0\0&colorred1&1&0&0&20\0&colorblue0&1&-1&0&-20\0&colorblue0&1&0&-1&-10\0&colorblue0&0&-1&1&-10endbmatrix$
Continue in this fashion, positioning pivot points from left-to-right and clearing out all other entries in the columns to be zero, swapping rows if necessary to get a nonzero entry in the pivot location, dividing or multiplying rows to get pivot positions equal to $1$ rather than some other nonzero number, or even skipping columns entirely if no nonzero entry can act as a pivot for that column.
answered Jul 16 at 19:30
JMoravitz
44.3k33481
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In my post here I describe an algorithm which will guarantee that you reach RREF. (Side note: this isn't the algorithm that computers use, but it is similar)
As mentioned in comments above, there are many different sequences of steps you can take to row reduce a matrix completely, but the final result of any row reduction will be exactly the same. That is to say that you may if you wish begin the row reduction process for your specific example with swapping the fourth and first rows and that will still be a fine beginning (though you might not have known ahead of time that this was a particularly useful thing to do).
Following the algorithm I link strictly to a t for your specific matrix begins:
$beginbmatrixcolorred1&1&0&0&0&20\0&0&1&-1&0&-20\0&1&1&0&0&20\1&0&0&0&-1&-10\0&0&0&-1&1&-10endbmatrix$
We first check that the first-row first-column entry (marked in $colorredred$) is nonzero. It is! On to the next step... we check that it is equal to $1$ It is! Now, we zero out everything else in its column by subtracting appropriate multiples of the first row from other rows as necessary.
$R_4-R_1mapsto R_4$
$beginbmatrix colorred1&1&0&0&0&20\colorblue0&0&1&-1&0&-20\colorblue0&1&1&0&0&20\colorblue0&-1&0&0&-1&-30\colorblue0&0&0&-1&1&-10endbmatrix$
Now that that column is taken care of, we can move our attention to the next pivot location, marked below in $colorredred$.
$beginbmatrix 1&1&0&0&0&20\0&colorred0&1&-1&0&-20\0&1&1&0&0&20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We ask the question of it it is zero. It is unfortunately. We then ask the question of if there are any nonzero entries from the same column below it. There is, for example in the third row. We swap the rows then.
$R_2leftrightarrow R_3$
$beginbmatrix1&1&0&0&0&20\0&colorred1&1&0&0&20\0&0&1&-1&0&-20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We then ask the question of if the pivot we are focusing on currently is equal to $1$. It is! Now, we subtract appropriate multiples of that row from other rows in an attempt to zero out the rest of the pivot's column.
$R_1-R_2mapsto R_1$
$R_4+R_2mapsto R_4$
$beginbmatrix1&colorblue0&-1&0&0&0\0&colorred1&1&0&0&20\0&colorblue0&1&-1&0&-20\0&colorblue0&1&0&-1&-10\0&colorblue0&0&-1&1&-10endbmatrix$
Continue in this fashion, positioning pivot points from left-to-right and clearing out all other entries in the columns to be zero, swapping rows if necessary to get a nonzero entry in the pivot location, dividing or multiplying rows to get pivot positions equal to $1$ rather than some other nonzero number, or even skipping columns entirely if no nonzero entry can act as a pivot for that column.
answered Jul 16 at 19:30
JMoravitz
44.3k33481
In my post here I describe an algorithm which will guarantee that you reach RREF. (Side note: this isn't the algorithm that computers use, but it is similar)
As mentioned in comments above, there are many different sequences of steps you can take to row reduce a matrix completely, but the final result of any row reduction will be exactly the same. That is to say that you may if you wish begin the row reduction process for your specific example with swapping the fourth and first rows and that will still be a fine beginning (though you might not have known ahead of time that this was a particularly useful thing to do).
Following the algorithm I link strictly to a t for your specific matrix begins:
$beginbmatrixcolorred1&1&0&0&0&20\0&0&1&-1&0&-20\0&1&1&0&0&20\1&0&0&0&-1&-10\0&0&0&-1&1&-10endbmatrix$
We first check that the first-row first-column entry (marked in $colorredred$) is nonzero. It is! On to the next step... we check that it is equal to $1$ It is! Now, we zero out everything else in its column by subtracting appropriate multiples of the first row from other rows as necessary.
$R_4-R_1mapsto R_4$
$beginbmatrix colorred1&1&0&0&0&20\colorblue0&0&1&-1&0&-20\colorblue0&1&1&0&0&20\colorblue0&-1&0&0&-1&-30\colorblue0&0&0&-1&1&-10endbmatrix$
Now that that column is taken care of, we can move our attention to the next pivot location, marked below in $colorredred$.
$beginbmatrix 1&1&0&0&0&20\0&colorred0&1&-1&0&-20\0&1&1&0&0&20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We ask the question of it it is zero. It is unfortunately. We then ask the question of if there are any nonzero entries from the same column below it. There is, for example in the third row. We swap the rows then.
$R_2leftrightarrow R_3$
$beginbmatrix1&1&0&0&0&20\0&colorred1&1&0&0&20\0&0&1&-1&0&-20\0&-1&0&0&-1&-30\0&0&0&-1&1&-10endbmatrix$
We then ask the question of if the pivot we are focusing on currently is equal to $1$. It is! Now, we subtract appropriate multiples of that row from other rows in an attempt to zero out the rest of the pivot's column.
$R_1-R_2mapsto R_1$
$R_4+R_2mapsto R_4$
$beginbmatrix1&colorblue0&-1&0&0&0\0&colorred1&1&0&0&20\0&colorblue0&1&-1&0&-20\0&colorblue0&1&0&-1&-10\0&colorblue0&0&-1&1&-10endbmatrix$
Continue in this fashion, positioning pivot points from left-to-right and clearing out all other entries in the columns to be zero, swapping rows if necessary to get a nonzero entry in the pivot location, dividing or multiplying rows to get pivot positions equal to $1$ rather than some other nonzero number, or even skipping columns entirely if no nonzero entry can act as a pivot for that column.
answered Jul 16 at 19:30
JMoravitz
44.3k33481
answered Jul 16 at 19:30
JMoravitz
44.3k33481
answered Jul 16 at 19:30
JMoravitz
44.3k33481
answered Jul 16 at 19:30
JMoravitz
44.3k33481
44.3k33481
add a comment |Â
add a comment |Â
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There are many different sequences of steps one can take to get a matrix into RREF... one such algorithm which will always work (though perhaps not the most efficiently in certain scenarios) can be found here. The final result of the row reduction will always be the same regardless of specific steps followed.
– JMoravitz
Jul 16 at 18:39
One such sequence of steps might begin $R_4-R_1mapsto R_4, R_2leftrightarrow R_3,R_1-R_2mapsto R_1,dots$. Another might begin $R_1leftrightarrow R_4,dots$
– JMoravitz
Jul 16 at 18:40