Show simultaneous equations can be expressed in the form Ax = c.
$begingroup$
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 SmithEdgar Smith
61
$endgroup$
add a comment |
$begingroup$
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 SmithEdgar Smith
61
$endgroup$
$begingroup$
What is this first part more than the definition of a matrix and multiplication with a vector?
$endgroup$
– Mark Bennet
Jan 3 at 21:31
add a comment |
$begingroup$
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 SmithEdgar Smith
61
$endgroup$
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 SmithEdgar Smith
61
asked Jan 3 at 21:25
Edgar SmithEdgar Smith
61
asked Jan 3 at 21:25
Edgar SmithEdgar Smith
61
asked Jan 3 at 21:25
Edgar SmithEdgar Smith
61
asked Jan 3 at 21:25
Edgar SmithEdgar Smith
61
61
$begingroup$
What is this first part more than the definition of a matrix and multiplication with a vector?
$endgroup$
– Mark Bennet
Jan 3 at 21:31
add a comment |
$begingroup$
What is this first part more than the definition of a matrix and multiplication with a vector?
$endgroup$
– Mark Bennet
Jan 3 at 21:31
$begingroup$
What is this first part more than the definition of a matrix and multiplication with a vector?
$endgroup$
– Mark Bennet
Jan 3 at 21:31
$begingroup$
What is this first part more than the definition of a matrix and multiplication with a vector?
$endgroup$
– Mark Bennet
Jan 3 at 21:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 4 at 7:27
RileyRiley
1825
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
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 Jan 4 at 7:27
RileyRiley
1825
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 7:27
RileyRiley
1825
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 7:27
RileyRiley
1825
$endgroup$
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 Jan 4 at 7:27
RileyRiley
1825
answered Jan 4 at 7:27
RileyRiley
1825
answered Jan 4 at 7:27
RileyRiley
1825
answered Jan 4 at 7:27
RileyRiley
1825
1825
add a comment |
add a comment |
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$begingroup$
What is this first part more than the definition of a matrix and multiplication with a vector?
$endgroup$
– Mark Bennet
Jan 3 at 21:31