Proving Matrix B=A^-1 | Determinant -10 | Exam Question Solution

[noelo2014]

Homework Statement

A =
[
1 0 1 0
2 3 0 6
0 5 −5 0
0 0 0 2
]

Prove that if there is a 4 × 4-matrix B such that A · B = I₄, then B = A⁻¹.

Homework Equations

The Attempt at a Solution

First of all I got the determinant of the matrix A which is -10
I'm just wondering if there's a shortcut to this problem, I began multiplying the matrix A by B (B is not given, so I used b₁₁, b₂₂, etc. I could equate each term in A*B with the corresponding entry in the identity matrix, which will give loads of equations which I guess could be used to solve for the b's and gives me something which I could multiply by A to get the Identity matrix. I guess this would prove it but this will take me ages, and this is an exam question so I'm guessing there is a simpler way.

If I got this question in an exam I wouldn't dream of doing this, it would take way too long. I'd probably just say that since its determinant is non-zero it must have an inverse and any matrix multiplied by it's inverse is the Identity matrix

Still I'd appreciate if anybody has any other quick way(s)of proving this. Thanks!

 

[jbunniii]

Homework Statement

A =
[
1 0 1 0
2 3 0 6
0 5 −5 0
0 0 0 2
]

Prove that if there is a 4 × 4-matrix B such that A · B = I₄, then B = A⁻¹.

This is actually a very general question which can be answered independently of your specific matrix $A$. If there is a 4x4 matrix $B$ satisfying $AB = I$, then $A$ has a right inverse, so $A$ must be surjective (why?). Therefore $A$ is also injective (why?). What can you conclude?

 

[noelo2014]

I can't really conclude anything. There is only one matrix that can give the identity matrix when multiplied by A and that is A^-1

Therefore B MUST be A^-1

But this is stuff we were thought to assume to be true, anyway I just thought there might have been some neat quick way of proving it.

 

[jbunniii]

I can't really conclude anything. There is only one matrix that can give the identity matrix when multiplied by A and that is A^-1

Therefore B MUST be A^-1

Well, this is a theorem, not an axiom. (And it isn't true if $A$ is not square.) If you have that theorem, then you can just cite it. To prove the theorem, you would proceed along the lines I described in my previous post:

If there exists $B$ such that $AB = I$, then $A$ is surjective, because given any vector $x$ in the target space, we have $x = Ix = ABx = A(Bx)$, i.e., $A$ maps $Bx$ to $x$.

Since $A$ is square, the rank-nullity theorem implies that the null space of $A$ must have zero dimension, so $A$ is also injective. Thus $A$ is bijective, so...

 

[HallsofIvy]

To prove "if AB= I (and A and B are square matrices) then $B= A^{-1}$" you should start with the definition of $A^{-1}$: it is the matrix, if it exists, such that $A^{-1}A= I$ and $AA^{-1}= I$. You already have one of those and must show the other.

 

[Ray Vickson]

Homework Statement

A =
[
1 0 1 0
2 3 0 6
0 5 −5 0
0 0 0 2
]

Prove that if there is a 4 × 4-matrix B such that A · B = I₄, then B = A⁻¹.

Homework Equations

The Attempt at a Solution

First of all I got the determinant of the matrix A which is -10
I'm just wondering if there's a shortcut to this problem, I began multiplying the matrix A by B (B is not given, so I used b₁₁, b₂₂, etc. I could equate each term in A*B with the corresponding entry in the identity matrix, which will give loads of equations which I guess could be used to solve for the b's and gives me something which I could multiply by A to get the Identity matrix. I guess this would prove it but this will take me ages, and this is an exam question so I'm guessing there is a simpler way.

If I got this question in an exam I wouldn't dream of doing this, it would take way too long. I'd probably just say that since its determinant is non-zero it must have an inverse and any matrix multiplied by it's inverse is the Identity matrix

Still I'd appreciate if anybody has any other quick way(s)of proving this. Thanks!

You are given that for an $n \times n$ matrix $A$ there is an $n \times n$ matrix $B$ giving $AB = I$, where $I$ is the $n \times n$ identity matrix. Use the existence of $B$ to conclude the existence of an $n \times n$ matrix $C$ giving $CA = I$; it is not difficult, but PF rules forbid me from telling how to do it. From $AB = I$ and $CA = I$ you need to conclude that $C = B$. Can you do that?

 

[noelo2014]

You are given that for an $n \times n$ matrix $A$ there is an $n \times n$ matrix $B$ giving $AB = I$, where $I$ is the $n \times n$ identity matrix. Use the existence of $B$ to conclude the existence of an $n \times n$ matrix $C$ giving $CA = I$; it is not difficult, but PF rules forbid me from telling how to do it. From $AB = I$ and $CA = I$ you need to conclude that $C = B$. Can you do that?

Looking at my linear Algebra notes, there's something in here that says let "C" be the inverse of A
so

Let C be the inverse of A
so CA=AC=I

I'm given AB=I

B=IB (Identity property)
=> B=(CA)B (Substituton)
=> B=C(AB)
=> B=C(I) (Substitution of AB in place of I because it's given)
=> B=CI
=> B=C

THerefore B=A inverse

 

[Ray Vickson]

Y

Looking at my linear Algebra notes, there's something in here that says let "C" be the inverse of A
so

Let C be the inverse of A
so CA=AC=I

I'm given AB=I

B=IB (Identity property)
=> B=(CA)B (Substituton)
=> B=C(AB)
=> B=C(I) (Substitution of AB in place of I because it's given)
=> B=CI
=> B=C

THerefore B=A inverse

You seem to have skipped over a crucial point: how do you prove that C even exists? That is, given there is a B with AB = I, how do you prove there must also be a C giving CA = I? As you have shown, once you know that C exists you are almost done, and it is easy to finish the problem. The problem is to know why there is such a C! You cannot just say C = A^{-1}, because you are essentially trying to prove that the concept of A^{-1} is meaningful.