CBSE Class 12 Mathematics 2026 Main Paper 65(B) · Q325 Marksmatrix_inverse
CBSE Class 12 Mathematics 2026 Main Paper 65(B) · Question 32 (Matrices)
Find the inverse of A=[[1,−1,1],[2,1,−3],[1,1,1]], then solve A(x,y,z)ᵀ=(4,0,2)ᵀ. OR For A=[[3,2],[7,5]], B=[[6,7],[8,9]], show (AB)⁻¹=B⁻¹A⁻¹.
Alternative 1 5 Marks
Find the inverse of A=[[1,−1,1],[2,1,−3],[1,1,1]], then solve A(x,y,z)ᵀ=(4,0,2)ᵀ.
Answer: A⁻¹=[[2/5,1/5,1/5],[−1/2,0,1/2],[1/10,−1/5,3/10]]; (x,y,z)=(2,−1,1).
- The determinant is 10. Compute the cofactors and transpose them.
- The adjugate rows are (4,2,2),(−5,0,5),(1,−2,3). Divide by 10.
- Multiply the inverse by (4,0,2)ᵀ and check each original equation.
Alternative 2 5 Marks
For A=[[3,2],[7,5]], B=[[6,7],[8,9]], show (AB)⁻¹=B⁻¹A⁻¹.
Answer: Both equal [[−94,39],[82,−34]]/2.
- The inverses are A⁻¹=[[5,−2],[−7,3]] and B⁻¹=[[-9,7],[8,−6]]/2.
- Compute AB=[[34,39],[82,94]], whose determinant is −2.
- Its inverse is [[−94,39],[82,−34]]/2, equal to the product B⁻¹A⁻¹.