Class 12 Matrices · Long Answer3 Marksmatrix_multiplication

Class 12 Maths · Matrices · Long Answer #103

For A=\(\begin{bmatrix}0 & 6 & 7\\-6 & 0 & 8\\7 & -8 & 0\end{bmatrix}\), B=\(\begin{bmatrix}0 & 1 & 1\\1 & 0 & 2\\1 & 2 & 0\end{bmatrix}\) and C=\(\begin{bmatrix}2\\-2\\3\end{bmatrix}\), calculate AC, BC and (A+B)C and verify distributivity.

✅ Step-by-Step Solution

Final Answer: AC=[9,12,30]ᵀ; BC=[1,8,−2]ᵀ; (A+B)C=[10,20,28]ᵀ.

Compute AC from the three row products: [0·2+6·(−2)+7·3,−6·2+0·(−2)+8·3,7·2−8·(−2)+0·3]ᵀ=[9,12,30]ᵀ.
Similarly BC=[0·2+1·(−2)+1·3,1·2+0·(−2)+2·3,1·2+2·(−2)+0·3]ᵀ=[1,8,−2]ᵀ.
A+B=[[0,7,8],[−5,0,10],[8,−6,0]], so (A+B)C=[10,20,28]ᵀ by direct multiplication.
Since AC+BC=[9+1,12+8,30−2]ᵀ=[10,20,28]ᵀ=(A+B)C, distributivity is verified.

📋 CBSE Stepwise Marking Scheme

[0.5 marks] Compute AC from the three row products: [0·2+6·(−2)+7·3,−6·2+0·(−2)+8·3,7·2−8·(−2)+0·3]ᵀ=[9,12,30]ᵀ.
[1 mark] Similarly BC=[0·2+1·(−2)+1·3,1·2+0·(−2)+2·3,1·2+2·(−2)+0·3]ᵀ=[1,8,−2]ᵀ.
[1 mark] A+B=[[0,7,8],[−5,0,10],[8,−6,0]], so (A+B)C=[10,20,28]ᵀ by direct multiplication.
[0.5 marks] Since AC+BC=[9+1,12+8,30−2]ᵀ=[10,20,28]ᵀ=(A+B)C, distributivity is verified.

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Class 12 Maths · Matrices · Long Answer #103

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

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