Augmented matrix in equation form
$begingroup$
I recently learned about using an augmented matrix to find the inverse of a matrix by putting the entire identity matrix on the right side of an augmented matrix and getting the left side to reduced row-echelon form. How would you represent that as a system of equations though?
For example,
$$
left[begin{array}{rr|rr}
1 & 2 & 1 & 0 \
3 & 4 & 0 & 1
end{array}right]
$$
could be written as
$$
x_1 + 2x_2 = ? \
3x_1 + 4x_2 = ?
$$
linear-algebra matrices
asked Jan 17 at 5:15
tautau
1275
$endgroup$
add a comment |
$begingroup$
I recently learned about using an augmented matrix to find the inverse of a matrix by putting the entire identity matrix on the right side of an augmented matrix and getting the left side to reduced row-echelon form. How would you represent that as a system of equations though?
For example,
$$
left[begin{array}{rr|rr}
1 & 2 & 1 & 0 \
3 & 4 & 0 & 1
end{array}right]
$$
could be written as
$$
x_1 + 2x_2 = ? \
3x_1 + 4x_2 = ?
$$
linear-algebra matrices
asked Jan 17 at 5:15
tautau
1275
$endgroup$
2
$begingroup$
There are actually two systems of equations here with different right hand sides and the same coefficients on the left hand side.
$endgroup$
– Brian Borchers
Jan 17 at 5:19
$begingroup$
could please give a brief example and provide it as an answer to this question?
$endgroup$
– tau
Jan 17 at 18:33
$begingroup$
Matrices, augmented or otherwise, don’t always represent systems of equations. Here, augmenting the matrix is just a convenient way to apply the same row operations to the identity.
$endgroup$
– amd
Jan 17 at 22:56
$begingroup$
so it is NOT 2 systems of equations? @amd
$endgroup$
– tau
Jan 18 at 1:55
add a comment |
$begingroup$
I recently learned about using an augmented matrix to find the inverse of a matrix by putting the entire identity matrix on the right side of an augmented matrix and getting the left side to reduced row-echelon form. How would you represent that as a system of equations though?
For example,
$$
left[begin{array}{rr|rr}
1 & 2 & 1 & 0 \
3 & 4 & 0 & 1
end{array}right]
$$
could be written as
$$
x_1 + 2x_2 = ? \
3x_1 + 4x_2 = ?
$$
linear-algebra matrices
asked Jan 17 at 5:15
tautau
1275
$endgroup$
I recently learned about using an augmented matrix to find the inverse of a matrix by putting the entire identity matrix on the right side of an augmented matrix and getting the left side to reduced row-echelon form. How would you represent that as a system of equations though?
For example,
$$
left[begin{array}{rr|rr}
1 & 2 & 1 & 0 \
3 & 4 & 0 & 1
end{array}right]
$$
could be written as
$$
x_1 + 2x_2 = ? \
3x_1 + 4x_2 = ?
$$
linear-algebra matrices
linear-algebra matrices
asked Jan 17 at 5:15
tautau
1275
asked Jan 17 at 5:15
tautau
1275
asked Jan 17 at 5:15
tautau
1275
asked Jan 17 at 5:15
tautau
1275
asked Jan 17 at 5:15
tautau
1275
1275
2
$begingroup$
There are actually two systems of equations here with different right hand sides and the same coefficients on the left hand side.
$endgroup$
– Brian Borchers
Jan 17 at 5:19
$begingroup$
could please give a brief example and provide it as an answer to this question?
$endgroup$
– tau
Jan 17 at 18:33
$begingroup$
Matrices, augmented or otherwise, don’t always represent systems of equations. Here, augmenting the matrix is just a convenient way to apply the same row operations to the identity.
$endgroup$
– amd
Jan 17 at 22:56
$begingroup$
so it is NOT 2 systems of equations? @amd
$endgroup$
– tau
Jan 18 at 1:55
add a comment |
2
$begingroup$
There are actually two systems of equations here with different right hand sides and the same coefficients on the left hand side.
$endgroup$
– Brian Borchers
Jan 17 at 5:19
$begingroup$
could please give a brief example and provide it as an answer to this question?
$endgroup$
– tau
Jan 17 at 18:33
$begingroup$
Matrices, augmented or otherwise, don’t always represent systems of equations. Here, augmenting the matrix is just a convenient way to apply the same row operations to the identity.
$endgroup$
– amd
Jan 17 at 22:56
$begingroup$
so it is NOT 2 systems of equations? @amd
$endgroup$
– tau
Jan 18 at 1:55
2
2
$begingroup$
There are actually two systems of equations here with different right hand sides and the same coefficients on the left hand side.
$endgroup$
– Brian Borchers
Jan 17 at 5:19
$begingroup$
There are actually two systems of equations here with different right hand sides and the same coefficients on the left hand side.
$endgroup$
– Brian Borchers
Jan 17 at 5:19
$begingroup$
could please give a brief example and provide it as an answer to this question?
$endgroup$
– tau
Jan 17 at 18:33
$begingroup$
could please give a brief example and provide it as an answer to this question?
$endgroup$
– tau
Jan 17 at 18:33
$begingroup$
Matrices, augmented or otherwise, don’t always represent systems of equations. Here, augmenting the matrix is just a convenient way to apply the same row operations to the identity.
$endgroup$
– amd
Jan 17 at 22:56
$begingroup$
Matrices, augmented or otherwise, don’t always represent systems of equations. Here, augmenting the matrix is just a convenient way to apply the same row operations to the identity.
$endgroup$
– amd
Jan 17 at 22:56
$begingroup$
so it is NOT 2 systems of equations? @amd
$endgroup$
– tau
Jan 18 at 1:55
$begingroup$
so it is NOT 2 systems of equations? @amd
$endgroup$
– tau
Jan 18 at 1:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Sometimes a matrix is just a matrix. How one might interpret an augmented matrix depends a lot on context.
Looking at this from the point of view of this method of computing a matrix inverse, to me it’s more of a bookkeeping device that a representation of any particular system of linear equations. Row-reducing a matrix is equivalent to left-multiplication by some invertible matrix. Because of the way that matrix multiplication is defined this can be “distributed” among portions of the augmented matrix. In particular, you have $$Eleft[Amid Iright] = left[EAmid Eright] = [Imid E].$$ The left-hand side of the result of row-reduction tells us that $EA=I$, or $E=A^{-1}$, which is what is on the right-hand side of the reduced matrix. Augmenting $A$ with the identity matrix is a way of ensuring that the same elementary row operations are automatically applied to both.
Now, one can certainly cook up systems of linear equations that the augmented matrix might represent. As noted in the comments to your question, since the augmentation consists of two columns, it’s tempting to say that it represents two systems of equations, one for each column. However, the context is matrix inversion, which instead suggests to me the matrix equation: $$begin{bmatrix}1&2\3&4end{bmatrix} begin{bmatrix}m_{11}&m_{12}\m_{21}&m_{22}end{bmatrix} = begin{bmatrix}1&0\0&1end{bmatrix},$$ which can be written as the system $$begin{align} m_{11}+2m_{21} &= 1 \ 3m_{11}+4m_{21} &= 0 \ m_{12}+2m_{22} &= 0 \ 3m_{12}+4m_{22} &= 1.end{align}$$ This system would more usually be rendered as the augmented matrix $$left[begin{array}{cccc|c}1&2&0&0 & 1 \ 3&4&0&0 & 0 \ 0&0&1&2 & 0 \ 0&0&3&4 & 1end{array}right]$$ but the repetitive block structure means that you’ll end up applying the same row operations to the last two rows that you do to the first, so this matrix can be collapsed into the more compact form in your question. Note that this system of equations can be broken up into two independent systems of equations, so you could certainly think of it in those terms, too.
answered Jan 18 at 21:37
amdamd
31.7k21052
$endgroup$
add a comment |
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$begingroup$
Sometimes a matrix is just a matrix. How one might interpret an augmented matrix depends a lot on context.
Looking at this from the point of view of this method of computing a matrix inverse, to me it’s more of a bookkeeping device that a representation of any particular system of linear equations. Row-reducing a matrix is equivalent to left-multiplication by some invertible matrix. Because of the way that matrix multiplication is defined this can be “distributed” among portions of the augmented matrix. In particular, you have $$Eleft[Amid Iright] = left[EAmid Eright] = [Imid E].$$ The left-hand side of the result of row-reduction tells us that $EA=I$, or $E=A^{-1}$, which is what is on the right-hand side of the reduced matrix. Augmenting $A$ with the identity matrix is a way of ensuring that the same elementary row operations are automatically applied to both.
Now, one can certainly cook up systems of linear equations that the augmented matrix might represent. As noted in the comments to your question, since the augmentation consists of two columns, it’s tempting to say that it represents two systems of equations, one for each column. However, the context is matrix inversion, which instead suggests to me the matrix equation: $$begin{bmatrix}1&2\3&4end{bmatrix} begin{bmatrix}m_{11}&m_{12}\m_{21}&m_{22}end{bmatrix} = begin{bmatrix}1&0\0&1end{bmatrix},$$ which can be written as the system $$begin{align} m_{11}+2m_{21} &= 1 \ 3m_{11}+4m_{21} &= 0 \ m_{12}+2m_{22} &= 0 \ 3m_{12}+4m_{22} &= 1.end{align}$$ This system would more usually be rendered as the augmented matrix $$left[begin{array}{cccc|c}1&2&0&0 & 1 \ 3&4&0&0 & 0 \ 0&0&1&2 & 0 \ 0&0&3&4 & 1end{array}right]$$ but the repetitive block structure means that you’ll end up applying the same row operations to the last two rows that you do to the first, so this matrix can be collapsed into the more compact form in your question. Note that this system of equations can be broken up into two independent systems of equations, so you could certainly think of it in those terms, too.
answered Jan 18 at 21:37
amdamd
31.7k21052
$endgroup$
add a comment |
$begingroup$
Sometimes a matrix is just a matrix. How one might interpret an augmented matrix depends a lot on context.
Looking at this from the point of view of this method of computing a matrix inverse, to me it’s more of a bookkeeping device that a representation of any particular system of linear equations. Row-reducing a matrix is equivalent to left-multiplication by some invertible matrix. Because of the way that matrix multiplication is defined this can be “distributed” among portions of the augmented matrix. In particular, you have $$Eleft[Amid Iright] = left[EAmid Eright] = [Imid E].$$ The left-hand side of the result of row-reduction tells us that $EA=I$, or $E=A^{-1}$, which is what is on the right-hand side of the reduced matrix. Augmenting $A$ with the identity matrix is a way of ensuring that the same elementary row operations are automatically applied to both.
Now, one can certainly cook up systems of linear equations that the augmented matrix might represent. As noted in the comments to your question, since the augmentation consists of two columns, it’s tempting to say that it represents two systems of equations, one for each column. However, the context is matrix inversion, which instead suggests to me the matrix equation: $$begin{bmatrix}1&2\3&4end{bmatrix} begin{bmatrix}m_{11}&m_{12}\m_{21}&m_{22}end{bmatrix} = begin{bmatrix}1&0\0&1end{bmatrix},$$ which can be written as the system $$begin{align} m_{11}+2m_{21} &= 1 \ 3m_{11}+4m_{21} &= 0 \ m_{12}+2m_{22} &= 0 \ 3m_{12}+4m_{22} &= 1.end{align}$$ This system would more usually be rendered as the augmented matrix $$left[begin{array}{cccc|c}1&2&0&0 & 1 \ 3&4&0&0 & 0 \ 0&0&1&2 & 0 \ 0&0&3&4 & 1end{array}right]$$ but the repetitive block structure means that you’ll end up applying the same row operations to the last two rows that you do to the first, so this matrix can be collapsed into the more compact form in your question. Note that this system of equations can be broken up into two independent systems of equations, so you could certainly think of it in those terms, too.
answered Jan 18 at 21:37
amdamd
31.7k21052
$endgroup$
add a comment |
$begingroup$
Sometimes a matrix is just a matrix. How one might interpret an augmented matrix depends a lot on context.
Looking at this from the point of view of this method of computing a matrix inverse, to me it’s more of a bookkeeping device that a representation of any particular system of linear equations. Row-reducing a matrix is equivalent to left-multiplication by some invertible matrix. Because of the way that matrix multiplication is defined this can be “distributed” among portions of the augmented matrix. In particular, you have $$Eleft[Amid Iright] = left[EAmid Eright] = [Imid E].$$ The left-hand side of the result of row-reduction tells us that $EA=I$, or $E=A^{-1}$, which is what is on the right-hand side of the reduced matrix. Augmenting $A$ with the identity matrix is a way of ensuring that the same elementary row operations are automatically applied to both.
Now, one can certainly cook up systems of linear equations that the augmented matrix might represent. As noted in the comments to your question, since the augmentation consists of two columns, it’s tempting to say that it represents two systems of equations, one for each column. However, the context is matrix inversion, which instead suggests to me the matrix equation: $$begin{bmatrix}1&2\3&4end{bmatrix} begin{bmatrix}m_{11}&m_{12}\m_{21}&m_{22}end{bmatrix} = begin{bmatrix}1&0\0&1end{bmatrix},$$ which can be written as the system $$begin{align} m_{11}+2m_{21} &= 1 \ 3m_{11}+4m_{21} &= 0 \ m_{12}+2m_{22} &= 0 \ 3m_{12}+4m_{22} &= 1.end{align}$$ This system would more usually be rendered as the augmented matrix $$left[begin{array}{cccc|c}1&2&0&0 & 1 \ 3&4&0&0 & 0 \ 0&0&1&2 & 0 \ 0&0&3&4 & 1end{array}right]$$ but the repetitive block structure means that you’ll end up applying the same row operations to the last two rows that you do to the first, so this matrix can be collapsed into the more compact form in your question. Note that this system of equations can be broken up into two independent systems of equations, so you could certainly think of it in those terms, too.
answered Jan 18 at 21:37
amdamd
31.7k21052
$endgroup$
Sometimes a matrix is just a matrix. How one might interpret an augmented matrix depends a lot on context.
Looking at this from the point of view of this method of computing a matrix inverse, to me it’s more of a bookkeeping device that a representation of any particular system of linear equations. Row-reducing a matrix is equivalent to left-multiplication by some invertible matrix. Because of the way that matrix multiplication is defined this can be “distributed” among portions of the augmented matrix. In particular, you have $$Eleft[Amid Iright] = left[EAmid Eright] = [Imid E].$$ The left-hand side of the result of row-reduction tells us that $EA=I$, or $E=A^{-1}$, which is what is on the right-hand side of the reduced matrix. Augmenting $A$ with the identity matrix is a way of ensuring that the same elementary row operations are automatically applied to both.
Now, one can certainly cook up systems of linear equations that the augmented matrix might represent. As noted in the comments to your question, since the augmentation consists of two columns, it’s tempting to say that it represents two systems of equations, one for each column. However, the context is matrix inversion, which instead suggests to me the matrix equation: $$begin{bmatrix}1&2\3&4end{bmatrix} begin{bmatrix}m_{11}&m_{12}\m_{21}&m_{22}end{bmatrix} = begin{bmatrix}1&0\0&1end{bmatrix},$$ which can be written as the system $$begin{align} m_{11}+2m_{21} &= 1 \ 3m_{11}+4m_{21} &= 0 \ m_{12}+2m_{22} &= 0 \ 3m_{12}+4m_{22} &= 1.end{align}$$ This system would more usually be rendered as the augmented matrix $$left[begin{array}{cccc|c}1&2&0&0 & 1 \ 3&4&0&0 & 0 \ 0&0&1&2 & 0 \ 0&0&3&4 & 1end{array}right]$$ but the repetitive block structure means that you’ll end up applying the same row operations to the last two rows that you do to the first, so this matrix can be collapsed into the more compact form in your question. Note that this system of equations can be broken up into two independent systems of equations, so you could certainly think of it in those terms, too.
answered Jan 18 at 21:37
amdamd
31.7k21052
answered Jan 18 at 21:37
amdamd
31.7k21052
answered Jan 18 at 21:37
amdamd
31.7k21052
answered Jan 18 at 21:37
amdamd
31.7k21052
31.7k21052
add a comment |
add a comment |
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2
$begingroup$
There are actually two systems of equations here with different right hand sides and the same coefficients on the left hand side.
$endgroup$
– Brian Borchers
Jan 17 at 5:19
$begingroup$
could please give a brief example and provide it as an answer to this question?
$endgroup$
– tau
Jan 17 at 18:33
$begingroup$
Matrices, augmented or otherwise, don’t always represent systems of equations. Here, augmenting the matrix is just a convenient way to apply the same row operations to the identity.
$endgroup$
– amd
Jan 17 at 22:56
$begingroup$
so it is NOT 2 systems of equations? @amd
$endgroup$
– tau
Jan 18 at 1:55