Question

Let A =3 2 7
1 4 3
- 2 5 8
Find matrices X and Y such that X + Y = A, where X is a symmetric
and Y is a skew-symmetric matrix.​

Answer 1

Given :-

$\begin{gathered}\sf A=\left[\begin{array}{ccc}3&2&7\\1&4&3\\ - 2&5&8\end{array}\right]\end{gathered}$

Now,

$\rm :\longmapsto\:A$

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

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

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

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

$\sf \: = \: \: X + Y$

where,

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

and

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

Now,

$\rm :\longmapsto\:\begin{gathered}\sf A=\left[\begin{array}{ccc}3&2&7\\1&4&3\\ - 2&5&8\end{array}\right]\end{gathered}$

So,

$\rm :\longmapsto\:\begin{gathered}\sf {A}^{T} =\left[\begin{array}{ccc}3&1& - 2\\2&4&5\\ 7&3&8\end{array}\right]\end{gathered}$

Now,

Consider,

$\rm :\longmapsto\:\begin{gathered}\sf A + {A}^{T} =\left[\begin{array}{ccc}3&2&7\\1&4&3\\ - 2&5&8\end{array}\right]\end{gathered} + \begin{gathered}\sf \left[\begin{array}{ccc}3&1& - 2\\2&4&5\\ 7&3&8\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf A + {A}^{T} =\left[\begin{array}{ccc}6&3&5\\3&8&8\\ 5&8&16\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf \dfrac{1}{2} (A + {A}^{T}) =\left[\begin{array}{ccc}3& \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} &4&4\\ \frac{5}{2} &4&8\end{array}\right]\end{gathered}$

$\bf\implies \:\:\begin{gathered}\sf X =\left[\begin{array}{ccc}3& \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} &4&4\\ \frac{5}{2} &4&8\end{array}\right]\end{gathered}$

Now,

$\bf\implies \:\:\begin{gathered}\sf {X}^{T} =\left[\begin{array}{ccc}3& \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} &4&4\\ \frac{5}{2} &4&8\end{array}\right]\end{gathered} = X$

$\bf\implies \:X \: is \: symmetric.$

Now,

Consider,

$\rm :\longmapsto\:\begin{gathered}\sf A - {A}^{T} =\left[\begin{array}{ccc}3&2&7\\1&4&3\\ - 2&5&8\end{array}\right]\end{gathered} - \begin{gathered}\sf \left[\begin{array}{ccc}3&1& - 2\\2&4&5\\ 7&3&8\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf A - {A}^{T} =\left[\begin{array}{ccc}0&1&9\\ - 1&0& - 2\\ - 9&2&0\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf \dfrac{1}{2}( A - {A}^{T}) =\left[\begin{array}{ccc}0& \frac{1}{2} &\frac{9}{2} \\ - \frac{1}{2} &0& - 1\\ - \frac{9}{2} &1&0\end{array}\right]\end{gathered}$

$\bf\implies \:\:\begin{gathered}\sf Y =\left[\begin{array}{ccc}0& \frac{1}{2} &\frac{9}{2} \\ - \frac{1}{2} &0& - 1\\ - \frac{9}{2} &1&0\end{array}\right]\end{gathered}$

$\bf\implies \:\:\begin{gathered}\sf {Y}^{T} =\left[\begin{array}{ccc}0& - \frac{1}{2} & - \frac{9}{2} \\ \frac{1}{2} &0& 1\\ \frac{9}{2} & - 1&0\end{array}\right]\end{gathered} = - Y$

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

Consider,

$\:\:\begin{gathered}\sf X + Y =\left[\begin{array}{ccc}3& \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} &4&4\\ \frac{5}{2} &4&8\end{array}\right]\end{gathered} + \begin{gathered}\sf \left[\begin{array}{ccc}0& \frac{1}{2} &\frac{9}{2} \\ - \frac{1}{2} &0& - 1\\ - \frac{9}{2} &1&0\end{array}\right]\end{gathered}$

$\rm :\implies\:\begin{gathered}\sf X + Y=\left[\begin{array}{ccc}3&2&7\\1&4&3\\ - 2&5&8\end{array}\right]\end{gathered}$

$\bf\implies \:X + Y = A$

Answer 2

Answer:

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