Question

Express the following matrix as the sum of a symmetric and a skew symmetric matrix:. [3 3 -2] [-2 -2 1] [-4 -5 2] These are not different matrices, it's one matrix, I have just used different brackets for different rows.​

Answer 1

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

Given that,

$\begin{gathered}\rm :\longmapsto\:\sf A=\left[\begin{array}{ccc} 3&3& - 2\\ - 2& - 2&1\\ - 4& - 5&2\end{array}\right]\end{gathered} - - - (1)$

We know,

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

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

$\rm \: = \: \: \dfrac{1}{2}(A + A + {A}^{T} - {A}^{T} )$

$\rm \: = \: \: \dfrac{1}{2}(A +{A}^{T}) + \dfrac{1}{2}(A - {A}^{T})$

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

$\bf\implies \:A = P + Q - - (2)$

where,

$\red{\rm :\longmapsto\:P = \dfrac{1}{2}(A +{A}^{T})} - - - (3)$

and

$\red{\rm :\longmapsto\:Q = \dfrac{1}{2}(A - {A}^{T})} - - - (4)$

Now, Given that

$\rm :\longmapsto\:A = \begin{gathered}\:\sf \left[\begin{array}{ccc} 3&3& - 2\\ - 2& - 2&1\\ - 4& - 5&2\end{array}\right]\end{gathered}$

So,

$\rm :\longmapsto\: {A}^{T} = \begin{gathered}\:\sf \left[\begin{array}{ccc} 3& - 2& - 4\\3& - 2& - 5\\ - 2& 1&2\end{array}\right]\end{gathered}$

Now,

Consider,

$\rm :\longmapsto\:A + {A}^{T}$

$\: \rm \: = \:\begin{gathered}\:\sf \left[\begin{array}{ccc} 3&3& - 2\\ - 2& - 2&1\\ - 4& - 5&2\end{array}\right]\end{gathered} + \begin{gathered}\:\sf \left[\begin{array}{ccc} 3& - 2& - 4\\3& - 2& - 5\\ - 2& 1&2\end{array}\right]\end{gathered}$

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

$\rm\implies \:\dfrac{1}{2}(A +{A}^{T}) = \begin{gathered}\:\sf \left[\begin{array}{ccc} 3& \frac{1}{2} & - 3\\ \frac{1}{2} & - 2& - 2\\ - 3& - 2&2\end{array}\right]\end{gathered}$

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

Now, Consider

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

$\bf\implies \:P \: is \: symmetric$

Consider,

$\rm :\longmapsto\:A - {A}^{T}$

$\: \rm \: = \:\begin{gathered}\:\sf \left[\begin{array}{ccc} 3&3& - 2\\ - 2& - 2&1\\ - 4& - 5&2\end{array}\right]\end{gathered} - \begin{gathered}\:\sf \left[\begin{array}{ccc} 3& - 2& - 4\\3& - 2& - 5\\ - 2& 1&2\end{array}\right]\end{gathered}$

$\: \rm \: = \:\begin{gathered}\:\sf \left[\begin{array}{ccc} 0&5&2\\ - 5&0& 6\\ - 2& - 6&0\end{array}\right]\end{gathered}$

$\rm :\implies\:\dfrac{1}{2}(A - {A}^{T}) = \:\begin{gathered}\:\sf \left[\begin{array}{ccc} 0& \frac{5}{2} &1\\ - \frac{5}{2} &0& 3\\ - 1& - 3&0\end{array}\right]\end{gathered}$

$\rm :\implies\:Q = \:\begin{gathered}\:\sf \left[\begin{array}{ccc} 0& \frac{5}{2} &1\\ - \frac{5}{2} &0& 3\\ - 1& - 3&0\end{array}\right]\end{gathered}$

Now,

Consider,

$\rm :\implies\: {Q}^{T} = \:\begin{gathered}\:\sf \left[\begin{array}{ccc} 0& - \frac{5}{2} & - 1\\\frac{5}{2} &0& - 3\\1&3&0\end{array}\right]\end{gathered} = - Q$

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

Hence,

Matrix A

$\begin{gathered}\rm :\longmapsto\:\sf\left[\begin{array}{ccc} 3&3& - 2\\ - 2& - 2&1\\ - 4& - 5&2\end{array}\right]\end{gathered}$

$\: \rm \: = \: \begin{gathered}\:\sf \left[\begin{array}{ccc} 3& \frac{1}{2} & - 3\\ \frac{1}{2} & - 2& - 2\\ - 3& - 2&2\end{array}\right]\end{gathered} + \:\begin{gathered}\:\sf \left[\begin{array}{ccc} 0& \frac{5}{2} &1\\ - \frac{5}{2} &0& 3\\ - 1& - 3&0\end{array}\right]\end{gathered}$