Respuesta :
Answer:
Option a is "False".
Option b is "True".
Step-by-step explanation:
In point a:
Prove AB=BA
[tex]A=\left[\begin{array}{cc}1&2\\4&5\end{array}\right] \\\\B= \left[\begin{array}{cc}2&1\\8&5\end{array}\right][/tex]
Solve L.H.S part:
[tex]AB= \left[\begin{array}{cc}1&2\\4&5\end{array}\right] \times \left[\begin{array}{cc}2&1\\8&5\end{array}\right][/tex]
[tex]= \left[\begin{array}{cc}2+16 &1+10\\8+40&4+25\end{array}\right]\\\\\\= \left[\begin{array}{cc}18 &11\\48&29\end{array}\right][/tex]
Solve R.H.S part:
[tex]BA= \left[\begin{array}{cc}2&1\\8&5\end{array}\right] \times \left[\begin{array}{cc}1&2\\4&5\end{array}\right][/tex]
[tex]= \left[\begin{array}{cc}2+4 &4+5\\8+20&16+25\end{array}\right]\\ \\ \\= \left[\begin{array}{cc}6 &9\\28&41\end{array}\right][/tex]
So, [tex]AB \neq BA[/tex]
In point b:
[tex]A^{*} =A\\\\(Ak)^{*} =A^k[/tex]
if [tex]k= 1[/tex]it holds the trivially [tex]A^{*}=A ...........(1)[/tex]
by assuming that if it is true for [tex]k=n-1[/tex]
that is
[tex](A^{n-1})^{*}= A^{n-1}............(2) \\\\ if \ k=n\\\\\to (A^{n})^{*} = (A^{n-1} \cdot A^1)^{*} = A^{*}(A^{n-1})^{*} \\\\\to A(A)^{n-1}[/tex] use above equations
[tex]\to A^{n-1+1} \\\\ \to A^{n}[/tex]
that's why it is true.
