Explain Solution RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.5 Question 7 Maths.

Answer: Symmetric matrix $\begin{bmatrix} 3 &\frac{-3}{2} \\ \frac{-3}{2} & -1 \end{bmatrix}$ and Skew-symmetric matrix $=\begin{bmatrix} 0 &\frac{-5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}$

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

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

Solution:

$Step - 1\rightarrow A^{T} =\begin{bmatrix} 3 &1 \\ -4 &-1 \end{bmatrix}$

$Step - 2\rightarrow (A + A^{T}) = \begin{bmatrix} 6 &-3 \\ -3& -2 \end{bmatrix}$

$Step - 3\rightarrow (A - A^{T}) = \begin{bmatrix} 0 & -5\\ 5& 0 \end{bmatrix}$

$Step - 4\rightarrow P=\frac{1}{2} (A + A^{T}) = \begin{bmatrix} 3 & \frac{-3}{2}\\ \frac{-3}{2} & -1 \end{bmatrix}$

$Q = \frac{1}{2} (A - A^{T}) = \begin{bmatrix} 0 &\frac{-5}{2} \\ \frac{5}{2}& 0 \end{bmatrix}$

$Step - 5\rightarrow P^{T} =\begin{bmatrix} 3 & \frac{-3}{2}\\ \frac{-3}{2}& -1 \end{bmatrix} = P \: and\: Q^{T} = \begin{bmatrix} 0 &\frac{5}{2} \\ \frac{-5}{2}& 0 \end{bmatrix}= -Q$

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

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

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