#### Explain Solution RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.5 Question 5 Maths.

Answer: Symmetric matrix $= \begin{bmatrix} 4 &\frac{5}{2} &0 \\ \\ \frac{5}{2} & 5 &\frac{5}{2} \\ \\ 0 & \frac{5}{2} & 1 \end{bmatrix}$ and Skew-symmetric matrix $= \begin{bmatrix} 0 &\frac{-1}{2} &-1 \\ \\ \frac{1}{2} & 0 &\frac{9}{2} \\ \\ 1 & \frac{-9}{2} & 0 \end{bmatrix}$

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

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

Solution:

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

$Step - 2\rightarrow (A + A^{T}) = \left [ 8\, 5\, 0\, 5\, 10\, 5\, 0\, 5\, 2 \right ]$

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

$Step - 4\rightarrow P=\frac{1}{2} (A + A^{T}) = \begin{bmatrix} 4 &\frac{5}{2} &0 \\ \\ \frac{5}{2} & 5 &\frac{5}{2} \\ \\ 0 & \frac{5}{2} & 1 \end{bmatrix}$

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

$Step - 5\rightarrow P^{T} = \begin{bmatrix} 4 &\frac{5}{2} &0 \\ \\ \frac{5}{2} & 5 &\frac{5}{2} \\ \\ 0 & \frac{5}{2} & 1 \end{bmatrix}= P \: and \: Q^{T} = \begin{bmatrix} 0 &\frac{-1}{2} &-1 \\ \\ \frac{1}{2} & 0 &\frac{9}{2} \\ \\ 1 & \frac{-9}{2} & 0 \end{bmatrix}= -Q$

Now,$P+ Q = A$ where $P$ is symmetric and $Q$is skew-symmetric.

$\begin{bmatrix} 4 &\frac{5}{2} &0 \\ \\ \frac{5}{2} & 5 &\frac{5}{2} \\ \\ 0 & \frac{5}{2} & 1 \end{bmatrix}$$+ \begin{bmatrix} 0 &\frac{-1}{2} &-1 \\ \\ \frac{1}{2} & 0 &\frac{9}{2} \\ \\ 1 & \frac{-9}{2} & 0 \end{bmatrix}$$= \begin{bmatrix} 4 & 2 &-1 \\ 3 &5 &7 \\ 1 &-2 &1 \end{bmatrix}$

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