Answer:
Entries of I^k are are also identity elements.
Step-by-step explanation:
a) For the 2×2 identity matrix I, show that I² =I
[tex]I^{2}=\left[\begin{array}{cc}1&0\\0&1\end{array}\right] \times \left[\begin{array}{cc}1&0\\0&1\end{array}\right] \\\\=\left[\begin{array}{cc}1\times 1+0\times 0&1\times 0+0\times 1\\0\times 1+1\times 0&0\times 0+1\times1\end{array}\right] \\\\=\left[\begin{array}{cc}1&0\\0&1\end{array}\right][/tex]
Hence proved I² =I
b) For the n×n identity matrix I, show that I² =I
n×n identity matrix is as shown in figure
Elements of identity matrix are
[tex]\delta I_{ij}=1\quad if\quad i=j\\\delta I_{ij}=0\quad if\quad i\ne j\\[/tex]
As square of 1 is equal to 1 so for n×n identity matrix I, I² =I
(c) what do you think the enteries of Ik are?
As mentioned above
[tex]\delta I_{ij}=1\quad if\quad i=j\\\delta I_{ij}=0\quad if\quad i\ne j\\[/tex]
Any power of 1 is equal to 1 so kth power of 1 is also 1. According to this Ik=I
