Question

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Answer 1

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

Given matrix is

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

We know,

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

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

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

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

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

$\rm :\longmapsto\:A = P + Q$

where,

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

and

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

Now,

Consider,

$\rm :\longmapsto\:P$

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

$\rm \: = \: \dfrac{1}{2}\bigg( \begin{gathered}\sf \left[\begin{array}{ccc}6& - 2&2\\ - 2&3& - 1\\ 2& - 1&3\end{array}\right]\end{gathered} + \begin{gathered}\sf \left[\begin{array}{ccc}6& - 2&2\\ - 2&3& - 1\\ 2& - 1&3\end{array}\right]\end{gathered}\bigg)$

$\rm \: = \: \dfrac{1}{2}\begin{gathered}\sf \left[\begin{array}{ccc}12& - 4&4\\ - 4&6& - 2\\ 4& - 2&6\end{array}\right]\end{gathered}$

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

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

Consider,

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

Hence, P is symmetric.

Consider,

$\rm :\longmapsto\:Q$

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

$\rm \: = \: \dfrac{1}{2}\bigg( \begin{gathered}\sf \left[\begin{array}{ccc}6& - 2&2\\ - 2&3& - 1\\ 2& - 1&3\end{array}\right]\end{gathered} - \begin{gathered}\sf \left[\begin{array}{ccc}6& - 2&2\\ - 2&3& - 1\\ 2& - 1&3\end{array}\right]\end{gathered}\bigg)$

$\rm \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\0& 0&0\\ 0&0&0\end{array}\right]\end{gathered}$

Consider,

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

$\bf\implies \:Q \: is \: skew \: symmetric$

Hence,

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

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

Answer 2

Step-by-step explanation:

a severe hereditary form of anemia in which a mutated form of hemoglobin distorts the red blood cells into a crescent shape at low oxygen levels. It is most common among those of African descent.