When does a matrix have a non-trivial solution?
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Can someone please explain why this theorem is true?
Theorem: If A is the matrix of coefficients of a system of linear equations, then the system has a solution if and only if the rank of the augmented matrix is equal to the rank of the matrix A?
linear-algebra matrices
asked Jan 15 at 22:05
Wilson GuoWilson Guo
312
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add a comment |
$begingroup$
Can someone please explain why this theorem is true?
Theorem: If A is the matrix of coefficients of a system of linear equations, then the system has a solution if and only if the rank of the augmented matrix is equal to the rank of the matrix A?
linear-algebra matrices
asked Jan 15 at 22:05
Wilson GuoWilson Guo
312
$endgroup$
add a comment |
$begingroup$
Can someone please explain why this theorem is true?
Theorem: If A is the matrix of coefficients of a system of linear equations, then the system has a solution if and only if the rank of the augmented matrix is equal to the rank of the matrix A?
linear-algebra matrices
asked Jan 15 at 22:05
Wilson GuoWilson Guo
312
$endgroup$
Can someone please explain why this theorem is true?
Theorem: If A is the matrix of coefficients of a system of linear equations, then the system has a solution if and only if the rank of the augmented matrix is equal to the rank of the matrix A?
linear-algebra matrices
linear-algebra matrices
asked Jan 15 at 22:05
Wilson GuoWilson Guo
312
asked Jan 15 at 22:05
Wilson GuoWilson Guo
312
asked Jan 15 at 22:05
Wilson GuoWilson Guo
312
asked Jan 15 at 22:05
Wilson GuoWilson Guo
312
asked Jan 15 at 22:05
Wilson GuoWilson Guo
312
312
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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Consider the matrix $A$ and the augmented matrix $(,Amid b,)$. Reduce to row-echelon form. The rank (for both of these) is the number of leading (pivot) columns. Now
- the leading columns of $A$ are also leading columns of $(,Amid b,)$;
- so the two ranks are different if and only if the column $b$ becomes a leading column after reduction;
- ...if and only if there is a row of the form $(,0,cdots,0mid c,)$ with $cne0$;
- ...if and only if there is no solution.
answered Jan 15 at 22:10
DavidDavid
68.2k664126
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add a comment |
$begingroup$
The theorem is obvious if the original matrix is in row-reduced echelon form. Reducing it to that form doesn't change the rank, and it doesn't change whether the system has solutions. So it's true for any matrix.
answered Jan 15 at 22:13
ArthurArthur
114k7115197
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
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$begingroup$
Consider the matrix $A$ and the augmented matrix $(,Amid b,)$. Reduce to row-echelon form. The rank (for both of these) is the number of leading (pivot) columns. Now
- the leading columns of $A$ are also leading columns of $(,Amid b,)$;
- so the two ranks are different if and only if the column $b$ becomes a leading column after reduction;
- ...if and only if there is a row of the form $(,0,cdots,0mid c,)$ with $cne0$;
- ...if and only if there is no solution.
answered Jan 15 at 22:10
DavidDavid
68.2k664126
$endgroup$
add a comment |
$begingroup$
Consider the matrix $A$ and the augmented matrix $(,Amid b,)$. Reduce to row-echelon form. The rank (for both of these) is the number of leading (pivot) columns. Now
- the leading columns of $A$ are also leading columns of $(,Amid b,)$;
- so the two ranks are different if and only if the column $b$ becomes a leading column after reduction;
- ...if and only if there is a row of the form $(,0,cdots,0mid c,)$ with $cne0$;
- ...if and only if there is no solution.
answered Jan 15 at 22:10
DavidDavid
68.2k664126
$endgroup$
add a comment |
$begingroup$
Consider the matrix $A$ and the augmented matrix $(,Amid b,)$. Reduce to row-echelon form. The rank (for both of these) is the number of leading (pivot) columns. Now
- the leading columns of $A$ are also leading columns of $(,Amid b,)$;
- so the two ranks are different if and only if the column $b$ becomes a leading column after reduction;
- ...if and only if there is a row of the form $(,0,cdots,0mid c,)$ with $cne0$;
- ...if and only if there is no solution.
answered Jan 15 at 22:10
DavidDavid
68.2k664126
$endgroup$
Consider the matrix $A$ and the augmented matrix $(,Amid b,)$. Reduce to row-echelon form. The rank (for both of these) is the number of leading (pivot) columns. Now
- the leading columns of $A$ are also leading columns of $(,Amid b,)$;
- so the two ranks are different if and only if the column $b$ becomes a leading column after reduction;
- ...if and only if there is a row of the form $(,0,cdots,0mid c,)$ with $cne0$;
- ...if and only if there is no solution.
answered Jan 15 at 22:10
DavidDavid
68.2k664126
answered Jan 15 at 22:10
DavidDavid
68.2k664126
answered Jan 15 at 22:10
DavidDavid
68.2k664126
answered Jan 15 at 22:10
DavidDavid
68.2k664126
68.2k664126
add a comment |
add a comment |
$begingroup$
The theorem is obvious if the original matrix is in row-reduced echelon form. Reducing it to that form doesn't change the rank, and it doesn't change whether the system has solutions. So it's true for any matrix.
answered Jan 15 at 22:13
ArthurArthur
114k7115197
$endgroup$
add a comment |
$begingroup$
The theorem is obvious if the original matrix is in row-reduced echelon form. Reducing it to that form doesn't change the rank, and it doesn't change whether the system has solutions. So it's true for any matrix.
answered Jan 15 at 22:13
ArthurArthur
114k7115197
$endgroup$
add a comment |
$begingroup$
The theorem is obvious if the original matrix is in row-reduced echelon form. Reducing it to that form doesn't change the rank, and it doesn't change whether the system has solutions. So it's true for any matrix.
answered Jan 15 at 22:13
ArthurArthur
114k7115197
$endgroup$
The theorem is obvious if the original matrix is in row-reduced echelon form. Reducing it to that form doesn't change the rank, and it doesn't change whether the system has solutions. So it's true for any matrix.
answered Jan 15 at 22:13
ArthurArthur
114k7115197
answered Jan 15 at 22:13
ArthurArthur
114k7115197
answered Jan 15 at 22:13
ArthurArthur
114k7115197
answered Jan 15 at 22:13
ArthurArthur
114k7115197
114k7115197
add a comment |
add a comment |
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