NCERT Class 12 Maths · Chapter 3 Matrices · Exercise 3.2 Question 14

✅ Step-by-Step Solution

1. For Part (1) AB, the inner dimensions agree: 2. Its order is 2×2. Use cᵢⱼ=Σₖaᵢₖbₖⱼ.
2. Row 1: \(c_{11}=(5)(2) + (-1)(3)=7,\quad c_{12}=(5)(1) + (-1)(4)=1\).
3. Row 2: \(c_{21}=(6)(2) + (7)(3)=33,\quad c_{22}=(6)(1) + (7)(4)=34\).
4. Part (1) AB = \(\left[\begin{matrix}7 & 1\\33 & 34\end{matrix}\right]\).
5. For Part (1) BA, the inner dimensions agree: 2. Its order is 2×2. Use cᵢⱼ=Σₖaᵢₖbₖⱼ.
6. Row 1: \(c_{11}=(2)(5) + (1)(6)=16,\quad c_{12}=(2)(-1) + (1)(7)=5\).
7. Row 2: \(c_{21}=(3)(5) + (4)(6)=39,\quad c_{22}=(3)(-1) + (4)(7)=25\).
8. Part (1) BA = \(\left[\begin{matrix}16 & 5\\39 & 25\end{matrix}\right]\).
9. For Part (2) AB, the inner dimensions agree: 3. Its order is 3×3. Use cᵢⱼ=Σₖaᵢₖbₖⱼ.
10. Row 1: \(c_{11}=(1)(-1) + (2)(0) + (3)(2)=5,\quad c_{12}=(1)(1) + (2)(-1) + (3)(3)=8,\quad c_{13}=(1)(0) + (2)(1) + (3)(4)=14\).
11. Row 2: \(c_{21}=(0)(-1) + (1)(0) + (0)(2)=0,\quad c_{22}=(0)(1) + (1)(-1) + (0)(3)=-1,\quad c_{23}=(0)(0) + (1)(1) + (0)(4)=1\).
12. Row 3: \(c_{31}=(1)(-1) + (1)(0) + (0)(2)=-1,\quad c_{32}=(1)(1) + (1)(-1) + (0)(3)=0,\quad c_{33}=(1)(0) + (1)(1) + (0)(4)=1\).
13. Part (2) AB = \(\left[\begin{matrix}5 & 8 & 14\\0 & -1 & 1\\-1 & 0 & 1\end{matrix}\right]\).
14. For Part (2) BA, the inner dimensions agree: 3. Its order is 3×3. Use cᵢⱼ=Σₖaᵢₖbₖⱼ.
15. Row 1: \(c_{11}=(-1)(1) + (1)(0) + (0)(1)=-1,\quad c_{12}=(-1)(2) + (1)(1) + (0)(1)=-1,\quad c_{13}=(-1)(3) + (1)(0) + (0)(0)=-3\).
16. Row 2: \(c_{21}=(0)(1) + (-1)(0) + (1)(1)=1,\quad c_{22}=(0)(2) + (-1)(1) + (1)(1)=0,\quad c_{23}=(0)(3) + (-1)(0) + (1)(0)=0\).
17. Row 3: \(c_{31}=(2)(1) + (3)(0) + (4)(1)=6,\quad c_{32}=(2)(2) + (3)(1) + (4)(1)=11,\quad c_{33}=(2)(3) + (3)(0) + (4)(0)=6\).
18. Part (2) BA = \(\left[\begin{matrix}-1 & -1 & -3\\1 & 0 & 0\\6 & 11 & 6\end{matrix}\right]\).
19. Each displayed pair differs in at least one entry, so AB≠BA in both parts.