#### Please Solve RD Sharma Class 12 Chapter 4 Algebra of Matrices Exercise Very Short Asnwer Question 45 Maths Textbook Solution.

Answer: $B=\begin{bmatrix} 1 & 1\\ 1 & 3 \end{bmatrix}$

Hint: Here we use the basic concept of symmetric and skew symmetric matrix

Given:

$A=\left [ 1\: 2\: 0\: 3 \right ]$ is written as

B+C where B is symmertric and C is skew symmetric matrix Find B

Solution:

$A=B+C$    .....(i)    $\left [ B=B^{T}symmetric C^{T}=-C skew symmetric \right ]$

$\rightarrow Let\: us\: take\: the\: transpose\: of\: eqn(i)A^{T}=\left ( B+C \right )^{T}A^{T}=B^{T}+C^{T}$

$\! \! \! \! \! \! \! \because\left((A+B)^{T}=A^{T}+B^{T}\right) A^{T}=B-C \quad \ldots .(i i) \text { From }(i) \text { and }(i i) B=\frac{1}{2}\left(A+A^{T}\right) C \\$

$=\frac{1}{2}\left(A-A^{T}\right) B=\frac{1}{2}\left ( \left [ 1\: 2\: 0\: 3 \right ]+\left [ 1\: 0\: 2\: 3 \right ]\right )B=\frac{1}{2}\left [ 1+1\: 2+0\: 0+2\: 3+3 \right ]$

$=\frac{1}{2}\left [ 2\: 2\: 2\: 6 \right ]=\left [ 1\: 1\: 1\: 3 \right ]$