Column vector as a linear combination?
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Given this matrix we are asked to answer which column vectors can be written as a linear combination of the others, which is easy, you just create another matrix representing the linear combination and see If it has a solution. I've tried to do so and each one had a solution so I concluded every column vector could; nonetheless, the solution given states that the second column can't be written as a linear combination of the others and I wonder If it's due to some mistake I did that I got a wrong conclusion.
begin{pmatrix} -1 & 1 & 3 & -1\ 4 & 8 & 0 & 16\ 2 & 5 & 2 & 9 end{pmatrix}
linear-algebra matrices vectors
asked Jan 2 at 6:13
houda el fezzakhouda el fezzak
274
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add a comment |
$begingroup$
Given this matrix we are asked to answer which column vectors can be written as a linear combination of the others, which is easy, you just create another matrix representing the linear combination and see If it has a solution. I've tried to do so and each one had a solution so I concluded every column vector could; nonetheless, the solution given states that the second column can't be written as a linear combination of the others and I wonder If it's due to some mistake I did that I got a wrong conclusion.
begin{pmatrix} -1 & 1 & 3 & -1\ 4 & 8 & 0 & 16\ 2 & 5 & 2 & 9 end{pmatrix}
linear-algebra matrices vectors
asked Jan 2 at 6:13
houda el fezzakhouda el fezzak
274
$endgroup$
add a comment |
$begingroup$
Given this matrix we are asked to answer which column vectors can be written as a linear combination of the others, which is easy, you just create another matrix representing the linear combination and see If it has a solution. I've tried to do so and each one had a solution so I concluded every column vector could; nonetheless, the solution given states that the second column can't be written as a linear combination of the others and I wonder If it's due to some mistake I did that I got a wrong conclusion.
begin{pmatrix} -1 & 1 & 3 & -1\ 4 & 8 & 0 & 16\ 2 & 5 & 2 & 9 end{pmatrix}
linear-algebra matrices vectors
asked Jan 2 at 6:13
houda el fezzakhouda el fezzak
274
$endgroup$
Given this matrix we are asked to answer which column vectors can be written as a linear combination of the others, which is easy, you just create another matrix representing the linear combination and see If it has a solution. I've tried to do so and each one had a solution so I concluded every column vector could; nonetheless, the solution given states that the second column can't be written as a linear combination of the others and I wonder If it's due to some mistake I did that I got a wrong conclusion.
begin{pmatrix} -1 & 1 & 3 & -1\ 4 & 8 & 0 & 16\ 2 & 5 & 2 & 9 end{pmatrix}
linear-algebra matrices vectors
linear-algebra matrices vectors
asked Jan 2 at 6:13
houda el fezzakhouda el fezzak
274
asked Jan 2 at 6:13
houda el fezzakhouda el fezzak
274
asked Jan 2 at 6:13
houda el fezzakhouda el fezzak
274
asked Jan 2 at 6:13
houda el fezzakhouda el fezzak
274
asked Jan 2 at 6:13
houda el fezzakhouda el fezzak
274
274
add a comment |
add a comment |
1 Answer
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Let's see... $2c_1+c_2=c_4$. The first, second, and fourth columns are linearly dependent, with a relation that can be solved for any one of them in terms of the others. The only one that might not be a linear combination of the others is the third.
If the entries in the three rows are $a,b,c$, the first, second, and fourth columns each satisfy $4a+7b-12c=0$. The third column doesn't; that column gives $-12$ instead.
So then, the third column is the one that can't be written as a linear combination of the others. Looks like mistakes all around.
answered Jan 2 at 6:29
jmerryjmerry
16.8k11633
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1 Answer
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1 Answer
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$begingroup$
Let's see... $2c_1+c_2=c_4$. The first, second, and fourth columns are linearly dependent, with a relation that can be solved for any one of them in terms of the others. The only one that might not be a linear combination of the others is the third.
If the entries in the three rows are $a,b,c$, the first, second, and fourth columns each satisfy $4a+7b-12c=0$. The third column doesn't; that column gives $-12$ instead.
So then, the third column is the one that can't be written as a linear combination of the others. Looks like mistakes all around.
answered Jan 2 at 6:29
jmerryjmerry
16.8k11633
$endgroup$
add a comment |
$begingroup$
Let's see... $2c_1+c_2=c_4$. The first, second, and fourth columns are linearly dependent, with a relation that can be solved for any one of them in terms of the others. The only one that might not be a linear combination of the others is the third.
If the entries in the three rows are $a,b,c$, the first, second, and fourth columns each satisfy $4a+7b-12c=0$. The third column doesn't; that column gives $-12$ instead.
So then, the third column is the one that can't be written as a linear combination of the others. Looks like mistakes all around.
answered Jan 2 at 6:29
jmerryjmerry
16.8k11633
$endgroup$
add a comment |
$begingroup$
Let's see... $2c_1+c_2=c_4$. The first, second, and fourth columns are linearly dependent, with a relation that can be solved for any one of them in terms of the others. The only one that might not be a linear combination of the others is the third.
If the entries in the three rows are $a,b,c$, the first, second, and fourth columns each satisfy $4a+7b-12c=0$. The third column doesn't; that column gives $-12$ instead.
So then, the third column is the one that can't be written as a linear combination of the others. Looks like mistakes all around.
answered Jan 2 at 6:29
jmerryjmerry
16.8k11633
$endgroup$
Let's see... $2c_1+c_2=c_4$. The first, second, and fourth columns are linearly dependent, with a relation that can be solved for any one of them in terms of the others. The only one that might not be a linear combination of the others is the third.
If the entries in the three rows are $a,b,c$, the first, second, and fourth columns each satisfy $4a+7b-12c=0$. The third column doesn't; that column gives $-12$ instead.
So then, the third column is the one that can't be written as a linear combination of the others. Looks like mistakes all around.
answered Jan 2 at 6:29
jmerryjmerry
16.8k11633
answered Jan 2 at 6:29
jmerryjmerry
16.8k11633
answered Jan 2 at 6:29
jmerryjmerry
16.8k11633
answered Jan 2 at 6:29
jmerryjmerry
16.8k11633
16.8k11633
add a comment |
add a comment |
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