NCERT Class 12 Maths · Chapter 3 Matrices · Exercise 3.3 Question 10

✅ Step-by-Step Solution

1. Every real square A decomposes as S+K, where S=(A+Aᵀ)/2 and K=(A−Aᵀ)/2.
2. Part (1): S = \(\left[\begin{matrix}3 & 3\\3 & -1\end{matrix}\right]\), K = \(\left[\begin{matrix}0 & 2\\-2 & 0\end{matrix}\right]\). Their sum is A; transposition gives Sᵀ=S and Kᵀ=−K.
3. Part (2): S = \(\left[\begin{matrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{matrix}\right]\), K = \(\left[\begin{matrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right]\). Their sum is A; transposition gives Sᵀ=S and Kᵀ=−K.
4. Part (3): S = \(\left[\begin{matrix}3 & \frac{1}{2} & - \frac{5}{2}\\\frac{1}{2} & -2 & -2\\- \frac{5}{2} & -2 & 2\end{matrix}\right]\), K = \(\left[\begin{matrix}0 & \frac{5}{2} & \frac{3}{2}\\- \frac{5}{2} & 0 & 3\\- \frac{3}{2} & -3 & 0\end{matrix}\right]\). Their sum is A; transposition gives Sᵀ=S and Kᵀ=−K.
5. Part (4): S = \(\left[\begin{matrix}1 & 2\\2 & 2\end{matrix}\right]\), K = \(\left[\begin{matrix}0 & 3\\-3 & 0\end{matrix}\right]\). Their sum is A; transposition gives Sᵀ=S and Kᵀ=−K.