Show simultaneous equations can be expressed in the form Ax = c.
Show that any system of simultaneous equations as below can be expressed
a11x1 + a12x2 +...+ a1kxk = c1
a21x1 + a22x2 +...+ a2kxk = c2
ak1x1 + ak2x2 +...+ akkxk = ck
in the form Ax = c for A, a k × k matrix and x, c column vectors.
Suppose that A is invertible. Show that Ax = c has a unique solution.
Suppose that A and B are invertible k × k matrices. Are the matrices AB
and BA invertible? If yes, what are their inverses?
I know that these concepts are true (except the last one I'm not sure about), but how would I show them?
linear-algebra
asked Jan 3 at 21:25
Edgar Smith
61
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Edgar Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Show that any system of simultaneous equations as below can be expressed
a11x1 + a12x2 +...+ a1kxk = c1
a21x1 + a22x2 +...+ a2kxk = c2
ak1x1 + ak2x2 +...+ akkxk = ck
in the form Ax = c for A, a k × k matrix and x, c column vectors.
Suppose that A is invertible. Show that Ax = c has a unique solution.
Suppose that A and B are invertible k × k matrices. Are the matrices AB
and BA invertible? If yes, what are their inverses?
I know that these concepts are true (except the last one I'm not sure about), but how would I show them?
linear-algebra
asked Jan 3 at 21:25
Edgar Smith
61
New contributor
Edgar Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is this first part more than the definition of a matrix and multiplication with a vector?
– Mark Bennet
Jan 3 at 21:31
add a comment |
Show that any system of simultaneous equations as below can be expressed
a11x1 + a12x2 +...+ a1kxk = c1
a21x1 + a22x2 +...+ a2kxk = c2
ak1x1 + ak2x2 +...+ akkxk = ck
in the form Ax = c for A, a k × k matrix and x, c column vectors.
Suppose that A is invertible. Show that Ax = c has a unique solution.
Suppose that A and B are invertible k × k matrices. Are the matrices AB
and BA invertible? If yes, what are their inverses?
I know that these concepts are true (except the last one I'm not sure about), but how would I show them?
linear-algebra
asked Jan 3 at 21:25
Edgar Smith
61
New contributor
Edgar Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Show that any system of simultaneous equations as below can be expressed
a11x1 + a12x2 +...+ a1kxk = c1
a21x1 + a22x2 +...+ a2kxk = c2
ak1x1 + ak2x2 +...+ akkxk = ck
in the form Ax = c for A, a k × k matrix and x, c column vectors.
Suppose that A is invertible. Show that Ax = c has a unique solution.
Suppose that A and B are invertible k × k matrices. Are the matrices AB
and BA invertible? If yes, what are their inverses?
I know that these concepts are true (except the last one I'm not sure about), but how would I show them?
linear-algebra
linear-algebra
asked Jan 3 at 21:25
Edgar Smith
61
New contributor
Edgar Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 3 at 21:25
Edgar Smith
61
New contributor
Edgar Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 3 at 21:25
Edgar Smith
61
New contributor
Edgar Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 3 at 21:25
Edgar Smith
61
asked Jan 3 at 21:25
Edgar Smith
61
61
New contributor
Edgar Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Edgar Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Edgar Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is this first part more than the definition of a matrix and multiplication with a vector?
– Mark Bennet
Jan 3 at 21:31
add a comment |
What is this first part more than the definition of a matrix and multiplication with a vector?
– Mark Bennet
Jan 3 at 21:31
What is this first part more than the definition of a matrix and multiplication with a vector?
– Mark Bennet
Jan 3 at 21:31
What is this first part more than the definition of a matrix and multiplication with a vector?
– Mark Bennet
Jan 3 at 21:31
add a comment |
1 Answer
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active
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For the first, if $A$ is invertible and $Ax_1 = Ax_2 = c$, then $A^{-1}Ax_1 = A^{-1}Ax_2$ so $x_1 = x_2$ which shows uniqueness. (Existence is given by $x_1 = A^{-1}c$.) An equivalent way of seeing this is if you regard $A$ as a linear operator (on a finite dimensional vector space, in this case $mathbb{R}^k$), then being invertible is equivalent to being injective and surjective. Surjectivity then implies the existence of a solution, and injectivity implies its uniqueness.
For the second, the answer is yes, and there are multiple ways to see this. One way is to use the equivalence of invertibility and a nonzero determinant together with the determinant of the product of two matrices, and another way is to explicitly declare $(AB)^{-1} = B^{-1}A^{-1}$ and show that this is indeed inverse to $AB$.
answered 2 days ago
Riley
1625
add a comment |
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1 Answer
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1 Answer
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active
oldest
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For the first, if $A$ is invertible and $Ax_1 = Ax_2 = c$, then $A^{-1}Ax_1 = A^{-1}Ax_2$ so $x_1 = x_2$ which shows uniqueness. (Existence is given by $x_1 = A^{-1}c$.) An equivalent way of seeing this is if you regard $A$ as a linear operator (on a finite dimensional vector space, in this case $mathbb{R}^k$), then being invertible is equivalent to being injective and surjective. Surjectivity then implies the existence of a solution, and injectivity implies its uniqueness.
For the second, the answer is yes, and there are multiple ways to see this. One way is to use the equivalence of invertibility and a nonzero determinant together with the determinant of the product of two matrices, and another way is to explicitly declare $(AB)^{-1} = B^{-1}A^{-1}$ and show that this is indeed inverse to $AB$.
answered 2 days ago
Riley
1625
add a comment |
For the first, if $A$ is invertible and $Ax_1 = Ax_2 = c$, then $A^{-1}Ax_1 = A^{-1}Ax_2$ so $x_1 = x_2$ which shows uniqueness. (Existence is given by $x_1 = A^{-1}c$.) An equivalent way of seeing this is if you regard $A$ as a linear operator (on a finite dimensional vector space, in this case $mathbb{R}^k$), then being invertible is equivalent to being injective and surjective. Surjectivity then implies the existence of a solution, and injectivity implies its uniqueness.
For the second, the answer is yes, and there are multiple ways to see this. One way is to use the equivalence of invertibility and a nonzero determinant together with the determinant of the product of two matrices, and another way is to explicitly declare $(AB)^{-1} = B^{-1}A^{-1}$ and show that this is indeed inverse to $AB$.
answered 2 days ago
Riley
1625
add a comment |
For the first, if $A$ is invertible and $Ax_1 = Ax_2 = c$, then $A^{-1}Ax_1 = A^{-1}Ax_2$ so $x_1 = x_2$ which shows uniqueness. (Existence is given by $x_1 = A^{-1}c$.) An equivalent way of seeing this is if you regard $A$ as a linear operator (on a finite dimensional vector space, in this case $mathbb{R}^k$), then being invertible is equivalent to being injective and surjective. Surjectivity then implies the existence of a solution, and injectivity implies its uniqueness.
For the second, the answer is yes, and there are multiple ways to see this. One way is to use the equivalence of invertibility and a nonzero determinant together with the determinant of the product of two matrices, and another way is to explicitly declare $(AB)^{-1} = B^{-1}A^{-1}$ and show that this is indeed inverse to $AB$.
answered 2 days ago
Riley
1625
For the first, if $A$ is invertible and $Ax_1 = Ax_2 = c$, then $A^{-1}Ax_1 = A^{-1}Ax_2$ so $x_1 = x_2$ which shows uniqueness. (Existence is given by $x_1 = A^{-1}c$.) An equivalent way of seeing this is if you regard $A$ as a linear operator (on a finite dimensional vector space, in this case $mathbb{R}^k$), then being invertible is equivalent to being injective and surjective. Surjectivity then implies the existence of a solution, and injectivity implies its uniqueness.
For the second, the answer is yes, and there are multiple ways to see this. One way is to use the equivalence of invertibility and a nonzero determinant together with the determinant of the product of two matrices, and another way is to explicitly declare $(AB)^{-1} = B^{-1}A^{-1}$ and show that this is indeed inverse to $AB$.
answered 2 days ago
Riley
1625
answered 2 days ago
Riley
1625
answered 2 days ago
Riley
1625
answered 2 days ago
Riley
1625
1625
add a comment |
add a comment |
Edgar Smith is a new contributor. Be nice, and check out our Code of Conduct.
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What is this first part more than the definition of a matrix and multiplication with a vector?
– Mark Bennet
Jan 3 at 21:31