Class 12 Matrices · Short Answer2 Marksmatrix_multiplication

Class 12 Maths · Matrices · Short Answer #60

For A=\(\begin{bmatrix}3 & 9 & 0\\1 & 8 & -2\\7 & 5 & 4\end{bmatrix}\) and B=\(\begin{bmatrix}4 & 0 & 2\\7 & 1 & 4\\2 & 2 & 6\end{bmatrix}\), find BᵀAᵀ.

✅ Step-by-Step Solution

Final Answer: \(\begin{bmatrix}75 & 56 & 71\\9 & 4 & 13\\42 & 22 & 58\end{bmatrix}\)

Use BᵀAᵀ=(AB)ᵀ and compute AB=[[75,9,42],[56,4,22],[71,13,58]].
Transpose AB to obtain [[75,56,71],[9,4,13],[42,22,58]].
Therefore BᵀAᵀ=\(\begin{bmatrix}75 & 56 & 71\\9 & 4 & 13\\42 & 22 & 58\end{bmatrix}\).

📋 CBSE Stepwise Marking Scheme

[0.5 marks] Use BᵀAᵀ=(AB)ᵀ and compute AB=[[75,9,42],[56,4,22],[71,13,58]].
[1 mark] Transpose AB to obtain [[75,56,71],[9,4,13],[42,22,58]].
[0.5 marks] Therefore BᵀAᵀ=\(\begin{bmatrix}75 & 56 & 71\\9 & 4 & 13\\42 & 22 & 58\end{bmatrix}\).

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Class 12 Maths · Matrices · Short Answer #60

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 3: Matrices

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