Branlieast questionAnswer the question which is in attachment Irreleva

1.	Branlieast questionAnswer the question which is in attachment Irrelevant answer at a time deleted. No spamming(expecting answers from Maths Aryabatta and copy that, Mr.Magician, Mathsdude)
Answer» $\large\underline{\sf{Solution-}}$ GIVEN that, $\red{\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$ Now, We know $\rm :\longmapsto\:A = \dfrac{1}{2} (2A)$ $\rm :\longmapsto\:A = \dfrac{1}{2} (A + A)$ $\rm :\longmapsto\:A = \dfrac{1}{2} (A + A + A' - A')$ $\rm :\longmapsto\:A = \dfrac{1}{2} (A + A'+ A - A')$ $\rm :\longmapsto\:A = \dfrac{1}{2} (A + A')+ \dfrac{1}{2} (A - A')$ Let we ASSUME that, $\red{\rm :\longmapsto\:A = P + Q}$ where, $\red{\rm :\longmapsto\:P = \dfrac{1}{2} (A + A')}$ and $\green{\rm :\longmapsto\:Q = \dfrac{1}{2} (A - A')}$ Now, Let we FIRST evaluate the value of P. We have $\red{\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$ So, $\red{\rm :\longmapsto\:A' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$ Now, CONSIDER, $\red{\rm :\longmapsto\:A + A' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} + \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$ $\red{\rm :\longmapsto\:A + A' = \begin{gathered}\sf \left[\begin{array}{ccc}12&-4&4\\ - 4&6&-2\\4&-2&6\end{array}\right]\end{gathered}}$ Now, we have $\red{\rm :\longmapsto\:P = \dfrac{1}{2} (A + A')}$ $\red{\bf\implies \:P = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$ Now, $\red{\rm :\longmapsto\:P' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} = P}$ $\red{\bf :\longmapsto\:P \: is \: symmetric}$ Now, we evaluate the value of Q We have, $\green{\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$ So, $\green{\rm :\longmapsto\:A' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$ Now, $\green{\rm :\longmapsto\:A - A' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} - \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$ $\green{\rm :\longmapsto\:A - A' = \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} }$ Now, we have $\green{\rm :\longmapsto\:Q = \dfrac{1}{2} (A - A')}$ $\green{\rm :\longmapsto\:Q = \dfrac{1}{2}\begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} }$ $\green{\bf\implies \:Q = \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} }$ Now, Consider, $\green{\rm :\longmapsto\:Q' = \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} = - Q }$ $\green{\bf :\longmapsto\:Q \: is \: Skew - Symmetric}$ Hence, $\purple{\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} + \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} }$

Branlieast questionAnswer the question which is in attachment Irrelevant answer at a time deleted. No spamming(expecting answers from Maths Aryabatta and copy that, Mr.Magician, Mathsdude)

Answer»

$\large\underline{\sf{Solution-}}$

$\red{\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$

Now, We know

$\rm :\longmapsto\:A = \dfrac{1}{2} (2A)$

$\rm :\longmapsto\:A = \dfrac{1}{2} (A + A)$

$\rm :\longmapsto\:A = \dfrac{1}{2} (A + A + A' - A')$

$\rm :\longmapsto\:A = \dfrac{1}{2} (A + A'+ A - A')$

$\rm :\longmapsto\:A = \dfrac{1}{2} (A + A')+ \dfrac{1}{2} (A - A')$

Let we ASSUME that,

$\red{\rm :\longmapsto\:A = P + Q}$

where,

$\red{\rm :\longmapsto\:P = \dfrac{1}{2} (A + A')}$

and

$\green{\rm :\longmapsto\:Q = \dfrac{1}{2} (A - A')}$

Now,

Let we FIRST evaluate the value of P.

We have

$\red{\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$

So,

$\red{\rm :\longmapsto\:A' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$

Now, CONSIDER,

$\red{\rm :\longmapsto\:A + A' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} + \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$

$\red{\rm :\longmapsto\:A + A' = \begin{gathered}\sf \left[\begin{array}{ccc}12&-4&4\\ - 4&6&-2\\4&-2&6\end{array}\right]\end{gathered}}$

Now, we have

$\red{\rm :\longmapsto\:P = \dfrac{1}{2} (A + A')}$

$\red{\bf\implies \:P = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$

Now,

$\red{\rm :\longmapsto\:P' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} = P}$

$\red{\bf :\longmapsto\:P \: is \: symmetric}$

Now, we evaluate the value of Q

We have,

$\green{\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$

So,

$\green{\rm :\longmapsto\:A' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$

Now,

$\green{\rm :\longmapsto\:A - A' = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} - \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered}}$

$\green{\rm :\longmapsto\:A - A' = \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} }$

Now, we have

$\green{\rm :\longmapsto\:Q = \dfrac{1}{2} (A - A')}$

$\green{\rm :\longmapsto\:Q = \dfrac{1}{2}\begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} }$

$\green{\bf\implies \:Q = \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} }$

Now, Consider,

$\green{\rm :\longmapsto\:Q' = \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} = - Q }$

$\green{\bf :\longmapsto\:Q \: is \: Skew - Symmetric}$

Hence,

$\purple{\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc}6&-2&2\\ - 2&3&-1\\2&-1&3\end{array}\right]\end{gathered} + \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\ 0&0&0\\0&0&0\end{array}\right]\end{gathered} }$