Inverse Matrices

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Show that if A ε Mnxn is nonsingular and t ≠ 0, then tA is nonsingular and (tA)-1 = (1/t)A-1.

I need to show an intense proof of this statement. Although I can grasp the concept in my head, I am unsure as to the mathematical reasons and theorems that prove this true.

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Show that if A ε Mnxn is nonsingular and t ≠ 0, then tA is nonsingular and (tA)-1 = (1/t)A-1.

I am not going to do this for you, as it really isn't that hard. But I will give a few pointers.

Note that since $A$ is $n \times n$ then $\det(tA) = t^n \det(A)$. So since $\det(A) \ne 0,\,\,\,t\,\, \ne 0$ then $t^n \ne 0 \Rightarrow \det(tA) \ne 0$. But you must prove the premise $\det(tA) = t^n \det(A)$. Can you do that?

For the second part, namely $(tA)^{-1} = \frac{1}{t}A^{-1}$ you need only to prove that $(AB)^{-1} = B^{-1}A^{-1}$, remembering that you can treat $t$ as a $1 \times 1$ matrix. Recall that

1. $AA^{-1}= A^{-1}A =I$

2. matrix algebra is associative

3. if $x$ is then treated as an element in a commutative ring, here most likely a field, then $xA = Ax$.

See how you get on

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How disheartening it is to try and help a poster who then doesn't even acknowledge one's efforts, let alone act on them. Worse, it is downright rude. Ah well.

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Know the feeling. Here's a house point.

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It happens, there isn't much that can be done about it beyond being satisfied that you made a worthy effort.

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I am not going to do this for you, as it really isn't that hard. But I will give a few pointers.

Note that since $A$ is $n \times n$ then $\det(tA) = t^n \det(A)$. So since $\det(A) \ne 0,\,\,\,t\,\, \ne 0$ then $t^n \ne 0 \Rightarrow \det(tA) \ne 0$. But you must prove the premise $\det(tA) = t^n \det(A)$. Can you do that?

For the second part, namely $(tA)^{-1} = \frac{1}{t}A^{-1}$ you need only to prove that $(AB)^{-1} = B^{-1}A^{-1}$, remembering that you can treat $t$ as a $1 \times 1$ matrix. Recall that

1. $AA^{-1}= A^{-1}A =I$

2. matrix algebra is associative

3. if $x$ is then treated as an element in a commutative ring, here most likely a field, then $xA = Ax$.

See how you get on

It is somewhat simpler to simply note that

$tA (\frac{1}{t} A^{-1}) = t \frac {1}{t} AA^{-1} = 1 \times I = I$

This works for linear operators in general and not just matrices on a finite-dimensional space.

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