Class 12 Determinants · Long Answer3 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Long Answer #95

Let A=\(\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}\), B=\(\begin{bmatrix}4 & -6\\-2 & 4\end{bmatrix}\). Compute AB, then use it to solve 2x+y=4 and 3x+2y=1.

✅ Step-by-Step Solution

Final Answer: AB=2I₂; x=7,y=−10.

Multiply to obtain AB=[[8−6,−12+12],[4−4,−6+8]]=2I₂, so A⁻¹=B/2.
The system coefficient matrix is Aᵀ=[[2,1],[3,2]], hence its inverse is (A⁻¹)ᵀ=Bᵀ/2.
Thus [x,y]ᵀ=(1/2)[[4,−2],[−6,4]][4,1]ᵀ=(1/2)[14,−20]ᵀ=[7,−10]ᵀ.
Therefore AB=2I₂ and the solution is x=7,y=−10.

📋 CBSE Stepwise Marking Scheme

[0.5 marks] Multiply to obtain AB=[[8−6,−12+12],[4−4,−6+8]]=2I₂, so A⁻¹=B/2.
[1 mark] The system coefficient matrix is Aᵀ=[[2,1],[3,2]], hence its inverse is (A⁻¹)ᵀ=Bᵀ/2.
[1 mark] Thus [x,y]ᵀ=(1/2)[[4,−2],[−6,4]][4,1]ᵀ=(1/2)[14,−20]ᵀ=[7,−10]ᵀ.
[0.5 marks] Therefore AB=2I₂ and the solution is x=7,y=−10.

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Class 12 Maths · Determinants · Long Answer #95

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 4: Determinants

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