NCERT Class 12 Maths · Chapter 3 Matrices · Exercise 3.3 Question 5

✅ Step-by-Step Solution

1. For Part (1) AB, the inner dimensions agree: 1. Its order is 3×3. Use cᵢⱼ=Σₖaᵢₖbₖⱼ.
2. Row 1: \(c_{11}=(1)(-1)=-1,\quad c_{12}=(1)(2)=2,\quad c_{13}=(1)(1)=1\).
3. Row 2: \(c_{21}=(-4)(-1)=4,\quad c_{22}=(-4)(2)=-8,\quad c_{23}=(-4)(1)=-4\).
4. Row 3: \(c_{31}=(3)(-1)=-3,\quad c_{32}=(3)(2)=6,\quad c_{33}=(3)(1)=3\).
5. Part (1) AB = \(\left[\begin{matrix}-1 & 2 & 1\\4 & -8 & -4\\-3 & 6 & 3\end{matrix}\right]\).
6. Transpose this product to obtain \(\left[\begin{matrix}-1 & 4 & -3\\2 & -8 & 6\\1 & -4 & 3\end{matrix}\right]\).
7. The reversed transpose product BᵀAᵀ=\(\left[\begin{matrix}-1 & 4 & -3\\2 & -8 & 6\\1 & -4 & 3\end{matrix}\right]\) is identical.
8. For Part (2) AB, the inner dimensions agree: 1. Its order is 3×3. Use cᵢⱼ=Σₖaᵢₖbₖⱼ.
9. Row 1: \(c_{11}=(0)(1)=0,\quad c_{12}=(0)(5)=0,\quad c_{13}=(0)(7)=0\).
10. Row 2: \(c_{21}=(1)(1)=1,\quad c_{22}=(1)(5)=5,\quad c_{23}=(1)(7)=7\).
11. Row 3: \(c_{31}=(2)(1)=2,\quad c_{32}=(2)(5)=10,\quad c_{33}=(2)(7)=14\).
12. Part (2) AB = \(\left[\begin{matrix}0 & 0 & 0\\1 & 5 & 7\\2 & 10 & 14\end{matrix}\right]\).
13. Transpose this product to obtain \(\left[\begin{matrix}0 & 1 & 2\\0 & 5 & 10\\0 & 7 & 14\end{matrix}\right]\).
14. The reversed transpose product BᵀAᵀ=\(\left[\begin{matrix}0 & 1 & 2\\0 & 5 & 10\\0 & 7 & 14\end{matrix}\right]\) is identical.