#### Explain Solution RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.5 Question 4 Maths.

$X =\begin{bmatrix} 3 &\frac{3}{2} &\frac{5}{2} \\\\ \frac{3}{2} &4 & 4\\ \\\frac{5}{2} & 4 & 8 \end{bmatrix} and$$Y = \left [ 0 \frac{1}{2} \frac{9}{2} \frac{-1}{2} 0 -1 \frac{-9}{2} 1 0 \right ]$

Hint: Find $X =\frac{1}{2} (A + A^{T}) and \: Y=\frac{1}{2} (A - A^{T})$

Given:$A = \begin{bmatrix} 3 &2 &7 \\ 1 &4 &3 \\ -2 &5 &8 \end{bmatrix}$

Solution:

$Step - 1\rightarrow A =\begin{bmatrix} 3 &2 &7 \\ 1& 4 &3 \\ -2 & 5 &8 \end{bmatrix} and AT =\begin{bmatrix} 3& 1 &-2 \\ 2 &4 &5 \\ 7& 3 &8 \end{bmatrix}$$\rightarrow A =\begin{bmatrix} 3 &2 &7 \\ 1& 4 &3 \\ -2 & 5 &8 \end{bmatrix} and AT =\begin{bmatrix} 3& 1 &-2 \\ 2 &4 &5 \\ 7& 3 &8 \end{bmatrix}$

$Step - 2\rightarrow (A + A^{T}) = \left [ 6\, 3\, 5\, 3\, 8\, 8 \,5\,8 \,16 \right ]$

$Step - 3\rightarrow (A - A^{T}) = \begin{bmatrix} 0 &1 &9 \\ -1& 0 & -2\\ -9& 2 &0 \end{bmatrix}$

$Step - 4 \rightarrow X= (A + A^{T}) = \left [ 3\,\frac{3}{2}\,\frac{5}{2}\,\frac{3}{2}\,4\,4\,\frac{5}{2}\,4\,8 \right ]$

$Y =\frac{1}{2} (A - A^{T})= \begin{bmatrix} 0 &\frac{1}{2} &\frac{9}{2} \\ \\ \frac{-1}{2} & 0 & -1\\ \\ \frac{-9}{2} &1 & 0 \end{bmatrix}$

$Step - 5\rightarrow X^{T} =\begin{bmatrix} 3 &\frac{3}{2} &\frac{5}{2} \\ \frac{3}{2} & 4 & 4\\ \frac{5}{2} &4 & 8 \end{bmatrix}= X and \, Y^{T} =\left [ 0 \frac{1}{2} \frac{9}{2} \frac{-1}{2} 0 -1 \frac{-9}{2} 1 0 \right ]= -Y$

$X$ is symmetric and $Y$ is skew-symmetric. Also, $X + Y = A.$

$\left [ 3\, \frac{3}{2}\, \frac{5}{2}\, \frac{3}{2}\, 4\, 4\, \frac{5}{2} \,4\, 8 \right ] + \left [ 0\, \frac{1}{2} \,\frac{9}{2}\, \frac{-1}{2} -1\, \frac{-9}{2}\, 1\, 0 \right ] = \begin{bmatrix} 3 &2 &7 \\ 1& 4 &3 \\ -2 &5 &8 \end{bmatrix}$

Thus, $A$ is expressed as a sum of symmetric and skew symmetric matrix.