# Inverse Matrix

#### Yankel

##### Active member
Hello,

I have another question regarding inverse matrices. I need to find

$a^{2}$

given:

$\exists x: \begin{pmatrix} 1 &a \\ 2a &1 \end{pmatrix}^{2}\cdot \begin{pmatrix} 1\\ x \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}$

Any hints or guidance will be appreciated !

Thanks !

#### Deveno

##### Well-known member
MHB Math Scholar
Some random thoughts that occurs to me:

We can see that the matrix:

$\begin{bmatrix}1&2a\\a&1 \end{bmatrix}^2$

is singular. Let's call this matrix $M^2$.

Convince yourself that if $M$ is invertible, so is $M^2$. Since $M^2$ is singular, then, so must $M$ be. Under what conditions will $M$ be singular?

#### Yankel

##### Active member

what makes you think it is singular ? I ran an example setting a=2 on Maple and there was an inverse...

if they tell me there exist x such that...is it right to say that there exist x that solves this linear equation system, meaning that the matrix is invertible ? saying that, is it correct to say that rank of this matrix must equal 2, and that what I need to do here is to find the values of a that gives me that ? If I am correct, I know WHAT to do, but not HOW to do, because of the 17th power up there.

#### Deveno

##### Well-known member
MHB Math Scholar
There are lots of "equivalent" definitions of a singular matrix:

1) A matrix that is not invertible.
2) A matrix $A$ for which there exists some other matrix $B \neq 0$ with $AB = 0$.
3) A matrix $A$ for which there exists some non-zero vector $v$ such that $Av = 0$.
4) A matrix with zero determinant.

It should be clear that definition #3 is easiest to apply, here: since the vector $(1,x)^T$ is non-zero, no matter "what" $x$ may be.

Yes, for SOME values of $a$, the matrix $M$ WILL be invertible. You are not interested in those values, you are interested in the values for which $M$ will NOT be invertible (Hint: look at definition #4).

#### Yankel

##### Active member
basically I am not suppose to use determinants for this one...

I understand what you mean now, for why seeking for values of a for which the matrix is NOT invertible. How should I find them if I have a 17th power - this I am yet to realize

#### Deveno

##### Well-known member
MHB Math Scholar
Not using determinants seems an onerous restriction, as it truly is the most elegant way in this case.

I do not see why you keep mentioning the "17th power" it truly does not seem relevant.

I forgot the mention a 5th equivalent definition of a singular matrix:

One whose columns form a linearly dependent set.

If we apply THIS definition, we must have:

$(2a,1) = (k,ka)$ for some field element (real number? rational? complex? you do not say...) $k$.

Since by direct substitution we see $a = 0$ does not work, we may assume $a \neq 0$.

This leads to the two equations:

$2a = k$
$1 = ka$

That is:

$k = \dfrac{1}{a} = 2a$.

This last equation can be solved for $a^2$.