Show that if A ε M_nxn 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.

 

 

Show that if A ε M_nxn 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 [math]A[/math] is [math]n \times n[/math] then [math]\det(tA) = t^n \det(A)[/math]. So since [math]\det(A) \ne 0,\,\,\,t\,\, \ne 0[/math] then [math]t^n \ne 0 \Rightarrow \det(tA) \ne 0[/math]. But you must prove the premise [math]\det(tA) = t^n \det(A)[/math]. Can you do that?

 

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

 

1. [math]AA^{-1}= A^{-1}A =I[/math]

 

2. matrix algebra is associative

 

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

 

See how you get on

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.

Know the feeling. Here's a house point.

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

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

 

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

 

1. [math]AA^{-1}= A^{-1}A =I[/math]

 

2. matrix algebra is associative

 

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

 

See how you get on

 

 

 

It is somewhat simpler to simply note that

 

[math] tA (\frac{1}{t} A^{-1}) = t \frac {1}{t} AA^{-1} = 1 \times I = I[/math]

 

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