Mathematics: True or false: a. A product of two self-adjoint matrices is self-adjoint. b. If A is self-adjoint, then Ak is self-adjoint. Justify your conclusions

Step-by-step explanation:Here the given expression is, = = = (because )=Thus, =...

True or false: a. A product of two self-adjoint matrices

Ответ:

Geo777

12.02.2022

Mathematics

Option a is "False".

Option b is "True".

Step-by-step explanation:

In point a:

Prove AB=BA

$A=\left[\begin{array}{cc}1&2\\4&5\end{array}\right] \\\\B= \left[\begin{array}{cc}2&1\\8&5\end{array}\right]$

Solve L.H.S part:

$AB= \left[\begin{array}{cc}1&2\\4&5\end{array}\right] \times \left[\begin{array}{cc}2&1\\8&5\end{array}\right]$

$= \left[\begin{array}{cc}2+16 &1+10\\8+40&4+25\end{array}\right]\\\\\\= \left[\begin{array}{cc}18 &11\\48&29\end{array}\right]$

Solve R.H.S part:

$BA= \left[\begin{array}{cc}2&1\\8&5\end{array}\right] \times \left[\begin{array}{cc}1&2\\4&5\end{array}\right]$

$= \left[\begin{array}{cc}2+4 &4+5\\8+20&16+25\end{array}\right]\\ \\ \\= \left[\begin{array}{cc}6 &9\\28&41\end{array}\right]$

So, $AB \neq BA$

In point b:

$A^{*} =A\\\\(Ak)^{*} =A^k$

if k= 1 it holds the trivially $A^{*}=A ...........(1)$

by assuming that if it is true for k=n-1

that is

$(A^{n-1})^{*}= A^{n-1}............(2) \\\\ if \ k=n\\\\\to (A^{n})^{*} = (A^{n-1} \cdot A^1)^{*} = A^{*}(A^{n-1})^{*} \\\\\to A(A)^{n-1}$ use above equations

$\to A^{n-1+1} \\\\ \to A^{n}$

that's why it is true.

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Ответ:

0prayforthelost0

02.01.2020

Mathematics

$\frac{x^2\sqrt{x}}{3}$

Step-by-step explanation:

Here the given expression is, $\sqrt{\frac{2x^5}{18} }$

= $\sqrt{\frac{x^5}{9} }$

= $\sqrt{\frac{x^4.x^1}{9} }$ (because $x^{m+n}= x^m.x^n$ )

= $\frac{x^2\sqrt{x} }{3}$

Thus, $\sqrt{\frac{2x^5}{18} }$ = $\frac{x^2\sqrt{x} }{3}$

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True or false: a. A product of two self-adjoint matrices is self-adjoint.
b. If A is self-adjoint, then Ak is self-adjoint.

Justify your conclusions

Ответ:

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