Matrices — Maths STD 12 Science — Question

Therefore, A' = $\left[\begin{array}{ccc} {3} & {-2} & {-4} \\ {3} & {-2} & {-5} \\ {-1} & {1} & {2} \end{array}\right]$
Now, on adding A and A’ we will get,
$A+A^{\prime}=\left[\begin{array}{ccc} {3} & {3} & {-1} \\ {-2} & {-2} & {1} \\ {-4} & {-5} & {2} \end{array}\right]+\left[\begin{array}{ccc} {3} & {-2} & {-4} \\ {3} & {-2} & {-5} \\ {-1} & {1} & {2} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{ccc} {3+3} & {3+(-2)} & {-1+(-4)} \\ {-2+3} & {-2+(-2)} & {1+(-5)} \\ {-4+(-1)} & {-5+1} & {2+2} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{rrr} {6} & {1} & {-5} \\ {1} & {-4} & {-4} \\ {-5} & {-4} & {4} \end{array}\right]$
Now, Let M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$
Therefore, M = $\frac{1}{2}\left[\begin{array}{ccc} {6} & {1} & {-5} \\ {1} & {-4} & {-4} \\ {-5} & {-4} & {4} \end{array}\right]$
$\Rightarrow \mathrm{M}=\left[\begin{array}{ccc} {3} & {\frac{1}{2}} & {\frac{-5}{2}} \\ {\frac{1}{2}} & {-2} & {-2} \\ {\frac{-5}{2}} & {-2} & {2} \end{array}\right]$
Now, $\mathrm{M}^{\prime}=\left[\begin{array}{ccc} {3} & {\frac{1}{2}} & {\frac{-5}{2}} \\ {\frac{1}{2}} & {-2} & {-2} \\ {\frac{-5}{2}} & {-2} & {2} \end{array}\right]$
$\Rightarrow$ M' = M
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as M' = M
Now, on subtracting A’ from A we will get,
$A-A^{\prime}=\left[\begin{array}{ccc} {3} & {3} & {-1} \\ {-2} & {-2} & {1} \\ {-4} & {-5} & {2} \end{array}\right]-\left[\begin{array}{ccc} {3} & {-2} & {-4} \\ {3} & {-2} & {-5} \\ {-1} & {1} & {2} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{ccc} {3-3} & {3-(-2)} & {-1-(-4)} \\ {-2-3} & {-2-(-2)} & {1-(-5)} \\ {-4-(-1)} & {-5-1} & {2-2} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{ccc} {0} & {5} & {3} \\ {-5} & {0} & {6} \\ {-3} & {-6} & {0} \end{array}\right]$
Now, Let N = $\frac{1}{2}\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$
Therefore, N = $\frac{1}{2}\left[\begin{array}{ccc} {0} & {5} & {3} \\ {-5} & {0} & {6} \\ {-3} & {-6} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}=\left[\begin{array}{ccc} {0} & {\frac{5}{2}} & {\frac{3}{2}} \\ {\frac{-5}{2}} & {0} & {3} \\ {\frac{-3}{2}} & {-3} & {0} \end{array}\right]$
Now, $\mathrm{N}^{\prime}=\left[\begin{array}{ccc} {0} & {\frac{-5}{2}} & {\frac{-3}{2}} \\ {\frac{5}{2}} & {0} & {-3} \\ {\frac{3}{2}} & {3} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}^{\prime}=-\mathrm{N}$
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as N' = - N
w, Add M and N, we get,
$M+N=\left[\begin{array}{ccc} {3} & {\frac{1}{2}} & {\frac{-5}{2}} \\ {\frac{1}{2}} & {-2} & {-2} \\ {\frac{-5}{2}} & {-2} & {2} \end{array}\right]+\left[\begin{array}{ccc} {0} & {\frac{5}{2}} & {\frac{3}{2}} \\ {\frac{-5}{2}} & {0} & {3} \\ {\frac{-3}{2}} & {-3} & {0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{ccc} {3+0} & {\frac{1}{2}+\frac{5}{2}} & {\frac{-5}{2}+\frac{3}{2}} \\ {\frac{1}{2}+\frac{-5}{2}} & {-2+0} & {-2+3} \\ {\frac{-5}{2}+\frac{-3}{2}} & {-2+(-3)} & {0+2} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{ccc} {3} & {\frac{6}{2}} & {\frac{-2}{2}} \\ {\frac{-4}{2}} & {-2} & {1} \\ {\frac{-8}{2}} & {-5} & {2} \end{array}\right]$
So we see here, M + N = $\left[\begin{array}{ccc} {3} & {3} & {-1} \\ {-2} & {-2} & {1} \\ {-4} & {-5} & {2} \end{array}\right]=A$

Thus, A is represented as the sum of a symmetric matrix M and a skew-symmetric matrix N.