Krdytkyu

This excersice took place in class I had today. The exercise was the following:

Let the regular matrix $A$,$B$:

$$A^2 = left[ begin{matrix} 2 &-2 & 2\ -2 & 2 & 2 \ 4&4&-4 end{matrix} right] hspace{2cm} B = left[ begin{matrix} 0 &-1 & 2\ 0 & 2 & 0 \ 1&3&-1 end{matrix} right]$$

Knowing that $(A^2)^{-1} = A^{-1}B$, find A.

My Attempt

During the exam I tried the following:

$(A^2)^{-1} = A^{-1}B Leftrightarrow A^2(A^2)^{-1} = A^2A^{-1}B Leftrightarrow I_n = AB Leftrightarrow A = B^{-1}$

And procedeed to find the inverse of B, which is:

$A = B^{-1} = left[ begin{matrix} frac{1}{2} &frac{-5}{4} & 1\ 0 & frac{1}{2} & 0 \ frac{1}{2}& frac{1}{4}&0 end{matrix} right]$

But, as I tend to make silly mistakes, I figured out that it would be a better way to do find $A$ just by using matrix products. We had $I_n = AB Leftrightarrow A = A^2B$. Using this method I got that

$A = A^2B = left[ begin{matrix} 2 & 0 & 2\ 2 & 12 & -6 \ -4& -8&12 end{matrix} right]$

But as you can see the $2$ matrix are different. Where is the mistake in my logic?

Another thing I noticed is that $I_n = AB Leftrightarrow A*I_n*B = A*(AB)*B Leftrightarrow I_n = AB = A^2B^2$, but $A^2B^2$ isn't the identity matrix, so maybe the exercise is wrong, but I rather not rush into that thinking.

Long story short, matrix multiplication is not commutative.

Te equation $A=A^2B$ is not correct. From the equation $$(A^2)^{-1} = A^{-1}B$$

you have to multiply it by $A$ from the left to get
$$A(A^2)^{-1} = B$$

and the multiply it by $A^2$ from the right and you get

$$A=Bcdot A^2$$

which is not the same as $A=A^2B$.

However, even using $A=BA^2$ results in two different calculations of what $A$ is, and this seems to me like a mistake in the exercise. Indeed, from $(A^2)^{-1} = A^{-1}B$, it should follow that $I=A^2A^{-1}B=AB$, and from that, it should follow that $A=B^{-1}$. However, it is not true that $(B^{-1})^2=A^2$ which means that $A=B^{-1}$ cannot hold, which is a contradiction.

Use the properties of determinant to show that the given relationship does not hold.
$$det ((A^2)^{-1})=-frac1{64}=det (A^{-1})cdot(-4)$$
from the given relationship and properties of determinant. So $det (A)=64cdot 4$ and this cannot be true.

The problem is broken, here is a precise statement showing how it is broken.

Let $B=left[ begin{matrix} 0 &-1 & 2\ 0 & 2 & 0 \ 1&3&-1 end{matrix} right].$

For every invertible matrix $A$, at least one of the following are false:

1. $A^2 = left[ begin{matrix} 2 &-2 & 2\ -2 & 2 & 2 \ 4&4&-4 end{matrix} right]$




2. $(A^2)^{-1}=A^{-1}B$

Proof.
If 2. holds, then
$$
det bigl((A^2)^{-1}bigl)=det (A^{-1}B).
$$

By direct computation, $det B=-4$. On the other hand,
$$
det left[ begin{matrix} 2 &-2 & 2\ -2 & 2 & 2 \ 4&4&-4 end{matrix} right]=-64.
$$
Consequently if 1. and 2. both hold, then $(-64)^{-1}=(det A)^{-1}cdot (-4),text{ or equivalently, } det A=256.$ But this contradicts 1. Thus, it there does not exist any invertible matrix satisfying both 1. and 2. simultaneously.