Identiy matrix with lambda I need some help with my homework in a subject called "Matrices in statistics".
The task is to show that 
$$ P_{ij}(\lambda)^{-1} = P_{ij}\left(\frac{1}{\lambda}\right), $$ where $P_{ij} $ is an identity matrix, where the value of i-th row and j-th column is $\lambda.$
Here's my initial solution (which wasn't enough for the teacher):
--
We need to show, that $$ P_{ij}\left(\frac{1}{\lambda}\right) P_{ij}({\lambda}) = 
P_{ij}({\lambda}) P_{ij}\left(\frac{1}{\lambda}\right) = I_n   $$
So:
1) $$ P_{ij}\left(\frac{1}{\lambda}\right) P_{ij}({\lambda}) = P_{ij}\left(\frac{1}{\lambda} \cdot\lambda\right)   =  P_{ij}(1) = I_n$$
And 
2)  $$  P_{ij}({\lambda})P_{ij}\left(\frac{1}{\lambda}\right) = P_{ij}\left(\lambda\cdot\frac{1}{\lambda}\right)   =  P_{ij}(1) = I_n $$ 
Q.E.D 

I submitted this solution, but my lecturer gave it back and wrote, that I need to supplement this solution. More precisely, I need to show, that
$$ \left(P_{ij}(\lambda)P_{ij}\left(\frac{1}{\lambda}\right)\right)_{kl} = \begin{cases} 1, k=l  \\ 0, k \neq l \end{cases} $$
And that's what gets me confused - I'm not sure, how to do it. 
I would be really thankful, if you could help me with this one! 
 A: Our goal is to show that $P_{ij}(\lambda) P_{ij} (\dfrac{1}{\lambda}) =  P_{ij} (\dfrac{1}{\lambda})P_{ij}(\lambda) = I_n$.  To do this, we use the definition of matrix multiplication.  That is, for $n \times n$ matrices $A$ and $B$, the product $AB$ of these matrices is the matrix $C$ with $ij$ element 
$$
C_{ij} = \sum_{k = 1}^n A_{ik} B_{kj}
$$
If we let $A = P_{ij}(\lambda)$ and $B = P_{ij} (\dfrac{1}{\lambda})$ in this definition, we see that if $i \neq j$, then 
$$
A_{ik} B_{kj} = 0
$$
for all $k$, since these matrices are zero everywhere except for on the diagonal.  This tells us that the sum 
$$
\sum_{k = 1}^n A_{ik} B_{kj}
$$
is equal to zero if $i \neq j$.  That is, the product $AB = P_{ij}(\lambda) P_{ij} (\dfrac{1}{\lambda})$ is a matrix $C$ with $C_{ij} = 0$ for all $i \neq j$. 
Now, if we have $i = j$, then we see that 
$$
\sum_{k = 1}^n A_{ik} B_{kj} = 1
$$
since $A_{ik} B_{kj}$ is zero everywhere except when $k = i = j$.  When $k = i = j$, then we have that $A_{ik}B_{kj} = \lambda \cdot \dfrac{1}{\lambda}$. 
Combining both of these, we see that the product $AB = P_{ij}(\lambda) P_{ij} (\dfrac{1}{\lambda})$ is a matrix $C_{ij}$ with zeros everywhere and $1$s on the diagonal - that is, the product is the identity matrix.  This tells us that $P_{ij}(\lambda) P_{ij} (\dfrac{1}{\lambda}) = I_n$, or equivalently, that $P_{ij}(\lambda)^{-1} = P_{ij} (\dfrac{1}{\lambda})$.  
