Class 12 Determinants · Very Long Answer5 Marksadjugate
Class 12 Maths · Determinants · Very Long Answer #65
For A=\(\begin{bmatrix}\cos\alpha & -\sin\alpha & 0\\\sin\alpha & \cos\alpha & 0\\0 & 0 & 1\end{bmatrix}\), find adj(A) and verify A adj(A)=adj(A)A=det(A)I₃.
✅ Step-by-Step Solution
Final Answer: adj(A)=[[cosα,sinα,0],[−sinα,cosα,0],[0,0,1]]; both products equal I₃.
- Expanding along the last row gives det(A)=cos²α+sin²α=1.
- The signed-minor cofactor matrix is [[cosα,−sinα,0],[sinα,cosα,0],[0,0,1]].
- Transpose it to get adj(A)=[[cosα,sinα,0],[−sinα,cosα,0],[0,0,1]].
- Multiply A by adj(A): the upper 2×2 block has diagonal cos²α+sin²α=1 and off-diagonal zero. Reversing the factors gives the same result; the last diagonal entry stays 1.
- Thus A adj(A)=adj(A)A=I₃=det(A)I₃, proving the identity.
📋 CBSE Stepwise Marking Scheme
- [1 mark] Expanding along the last row gives det(A)=cos²α+sin²α=1.
- [1 mark] The signed-minor cofactor matrix is [[cosα,−sinα,0],[sinα,cosα,0],[0,0,1]].
- [1 mark] Transpose it to get adj(A)=[[cosα,sinα,0],[−sinα,cosα,0],[0,0,1]].
- [1 mark] Multiply A by adj(A): the upper 2×2 block has diagonal cos²α+sin²α=1 and off-diagonal zero. Reversing the factors gives the same result; the last diagonal entry stays 1.
- [1 mark] Thus A adj(A)=adj(A)A=I₃=det(A)I₃, proving the identity.
