NCERT Class 12 Maths · Chapter 3 Matrices · Exercise 3.3 Question 2

✅ Step-by-Step Solution

1. First A = \(\left[\begin{matrix}-1 & 2 & 3\\5 & 7 & 9\\-2 & 1 & 1\end{matrix}\right]\), B = \(\left[\begin{matrix}-4 & 1 & -5\\1 & 2 & 0\\1 & 3 & 1\end{matrix}\right]\).
2. Aᵀ = \(\left[\begin{matrix}-1 & 5 & -2\\2 & 7 & 1\\3 & 9 & 1\end{matrix}\right]\), Bᵀ = \(\left[\begin{matrix}-4 & 1 & 1\\1 & 2 & 3\\-5 & 0 & 1\end{matrix}\right]\).
3. (i) A+B=\(\left[\begin{matrix}-5 & 3 & -2\\6 & 9 & 9\\-1 & 4 & 2\end{matrix}\right]\), so (A+B)ᵀ=\(\left[\begin{matrix}-5 & 6 & -1\\3 & 9 & 4\\-2 & 9 & 2\end{matrix}\right]\)=Aᵀ+Bᵀ.
4. (ii) A−B=\(\left[\begin{matrix}3 & 1 & 8\\4 & 5 & 9\\-3 & -2 & 0\end{matrix}\right]\), so (A−B)ᵀ=\(\left[\begin{matrix}3 & 4 & -3\\1 & 5 & -2\\8 & 9 & 0\end{matrix}\right]\)=Aᵀ−Bᵀ.