CBSE Class 12 Mathematics 2026 Main Paper 65/1/3 · Q345 Marksmatrices
CBSE Class 12 Mathematics 2026 Main Paper 65/1/3 · Question 34 (Matrices)
If A = [[0,2,1],[-2,-1,-2],[1,-1,0]], find A⁻¹ and use it to solve the system: −2y + z = 7, 2x − y − z = 8, x − 2y = 10. OR If [[3,-1,sin3x],[-7,4,cos2x],[-11,7,2]] is a singular matrix, then find all values of x where x ∈ [0, π/2].
Alternative 1 5 Marks
If A = [[0,2,1],[-2,-1,-2],[1,-1,0]], find A⁻¹ and use it to solve the system: −2y + z = 7, 2x − y − z = 8, x − 2y = 10.
Answer: A⁻¹ = [[2,1,3],[2,1,2],[−3,−2,−4]]; x=0, y=−5, z=−3.
- det(A) = 0·((−1)(0)−(−2)(−1)) − 2·((−2)(0)−(−2)(1)) + 1·((−2)(−1)−(−1)(1)) = 0(−2) − 2(2) + 1(3) = −1 ≠ 0.
- Computing cofactors: C11=−2, C12=−2, C13=3, C21=−1, C22=−1, C23=2, C31=−3, C32=−2, C33=4. adj(A) = transpose of cofactor matrix = [[−2,−1,−3],[−2,−1,−2],[3,2,4]]. A⁻¹ = (1/|A|) adj(A) = −adj(A) = [[2,1,3],[2,1,2],[−3,−2,−4]]. The system −2y+z=7, 2x−y−z=8, x−2y=10 has coefficient matrix [[0,−2,1],[2,−1,−1],[1,−2,0]], which is exactly Aᵀ. So the system is AᵀX = B with X=(x,y,z)ᵀ, B=(7,8,10)ᵀ. X = (Aᵀ)⁻¹B = (A⁻¹)ᵀB. (A⁻¹)ᵀ = [[2,2,−3],[1,1,−2],[3,2,−4]]. X = [[2,2,−3],[1,1,−2],[3,2,−4]]·[7,8,10]ᵀ = [14+16−30, 7+8−20, 21+16−40] = [0,−5,−3].
- A⁻¹ = [[2,1,3],[2,1,2],[−3,−2,−4]]; x=0, y=−5, z=−3.
Alternative 2 5 Marks
If [[3,-1,sin3x],[-7,4,cos2x],[-11,7,2]] is a singular matrix, then find all values of x where x ∈ [0, π/2].
Answer: x = 0, π/6
- Set determinant = 0: 3(8−7cos2x) + 1(−14+11cos2x) + sin3x(−49+44) = 0.
- Simplify: 24−21cos2x−14+11cos2x−5sin3x = 0 ⇒ 10−10cos2x−5sin3x=0 ⇒ 2cos2x+sin3x=2. Using cos2x=1−2sin²x and sin3x=3sinx−4sin³x: 2(1−2sin²x)+(3sinx−4sin³x)=2 ⇒ −4sin²x+3sinx−4sin³x=0. Factor: sinx(4sin²x+4sinx−3)=0 ⇒ sinx(2sinx−1)(2sinx+3)=0. sinx=0 or sinx=1/2 or sinx=−3/2 (rejected, out of range). In [0, π/2]: x=0 or x=π/6.
- x = 0, π/6