Question 335 marksLong AnswerSection D

(a) If A=[021−2−1−21−10]A = \begin{bmatrix}0 & 2 & 1\\ -2 & -1 & -2\\ 1 & -1 & 0\end{bmatrix}, find A−1A^{-1} and use it to solve the following system of equations: −2y+z=7-2y + z = 7, 2x−y−z=82x - y - z = 8, x−2y=10x - 2y = 10.


OR


(b) If [3−1sin⁡3x−74cos⁡2x−1172]\begin{bmatrix}3 & -1 & \sin 3x\\ -7 & 4 & \cos 2x\\ -11 & 7 & 2\end{bmatrix} is a singular matrix, then find all values of x, where x∈[0,π/2]x \in [0, \pi/2].

Determinants
Inverse of a Matrix and Singular Matrices

Official answer

This is an OR question — attempt either Part (a) or Part (b) fully.


Part (a): x=0,y=−5,z=−3x = 0, y = -5, z = -3


  • For A=[021−2−1−21−10]A = \begin{bmatrix}0 & 2 & 1\\ -2 & -1 & -2\\ 1 & -1 & 0\end{bmatrix}, ∣A∣=−1≠0|A| = -1 \ne 0, so A−1A^{-1} exists.
  • A−1=[213212−3−2−4]A^{-1} = \begin{bmatrix}2 & 1 & 3\\ 2 & 1 & 2\\ -3 & -2 & -4\end{bmatrix}.
  • The system's coefficient matrix is [0−212−1−11−20]=AT\begin{bmatrix}0 & -2 & 1\\ 2 & -1 & -1\\ 1 & -2 & 0\end{bmatrix} = A^T, so the system is ATX=BA^TX = B with B=[7,8,10]TB = [7, 8, 10]^T.
  • Then X=(AT)−1B=(A−1)TB=[22−311−232−4]⋅[7,8,10]T=[0,−5,−3]TX = (A^T)^{-1}B = (A^{-1})^TB = \begin{bmatrix}2 & 2 & -3\\ 1 & 1 & -2\\ 3 & 2 & -4\end{bmatrix} \cdot [7, 8, 10]^T = [0, -5, -3]^T.
  • Hence x=0,y=−5,z=−3x = 0, y = -5, z = -3.

Part (b): x=0x = 0 or x=π/6x = \pi/6


  • A matrix is singular when its determinant is 0. Expanding gives 10−10cos⁡2x−5sin⁡3x=010 - 10\cos 2x - 5\sin 3x = 0, i.e. 2−2cos⁡2x−sin⁡3x=02 - 2\cos 2x - \sin 3x = 0.
  • Using 2−2cos⁡2x=4sin⁡2x2 - 2\cos 2x = 4\sin^2x and sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3x: 4sin⁡2x−3sin⁡x+4sin⁡3x=0  ⟹  sin⁡x(4sin⁡2x+4sin⁡x−3)=04\sin^2x - 3\sin x + 4\sin^3x = 0 \implies \sin x(4\sin^2x + 4\sin x - 3) = 0.
  • sin⁡x=0  ⟹  x=0\sin x = 0 \implies x = 0; 4sin⁡2x+4sin⁡x−3=0  ⟹  sin⁡x=1/24\sin^2x + 4\sin x - 3 = 0 \implies \sin x = 1/2 or sin⁡x=−3/2\sin x = -3/2 (rejected).
  • sin⁡x=1/2  ⟹  x=π/6\sin x = 1/2 \implies x = \pi/6. Both x=0x = 0 and x=π/6x = \pi/6 lie in [0,π/2][0, \pi/2], so x=0x = 0 or x=π/6x = \pi/6.
Answer keywords:inverse of a matrixadjoint methodsingular matrixsystem of linear equationscofactordeterminantmatrix method

Marking scheme

  1. 1.Part (a) — 1 mark: correct value of ∣A∣=−1|A| = -1 and conclusion that A⁻¹ exists.
  2. 2.Part (a) — 2 marks: correct cofactors/adjoint and A−1=[213212−3−2−4]A^{-1} = \begin{bmatrix}2 & 1 & 3\\ 2 & 1 & 2\\ -3 & -2 & -4\end{bmatrix}.
  3. 3.Part (a) — 2 marks: recognising the coefficient matrix is ATA^T, computing X=(A−1)TBX = (A^{-1})^TB and obtaining x=0,y=−5,z=−3x = 0, y = -5, z = -3.
  4. 4.Part (b) — 2 marks: correctly expanding the 3×33\times 3 determinant to 10−10cos⁡2x−5sin⁡3x=010 - 10\cos 2x - 5\sin 3x = 0.
  5. 5.Part (b) — 2 marks: correct simplification to sin⁡x(4sin⁡2x+4sin⁡x−3)=0\sin x(4\sin^2x + 4\sin x - 3) = 0 using trigonometric identities.
  6. 6.Part (b) — 1 mark: obtaining both x=0x = 0 and x=π/6x = \pi/6 and rejecting sin⁡x=−3/2\sin x = -3/2.

Hint

For (a), compute ∣A∣|A| and A−1=adj(A)/∣A∣A^{-1} = \text{adj}(A)/|A|, then note the system's coefficient matrix is ATA^T, so X=(A−1)TBX = (A^{-1})^TB. For (b), expand the 3×33\times 3 determinant, set it to 0, and simplify using 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2\sin^2x and sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3x.

Quick oral answer

For (a), ∣A∣=−1|A| = -1 so A−1A^{-1} exists; since the coefficient matrix is ATA^T, X=(A−1)TBX = (A^{-1})^TB gives x=0,y=−5,z=−3x = 0, y = -5, z = -3. For (b), setting the determinant to 0 yields sin⁡x=0\sin x = 0 or 1/21/2, so x=0x = 0 or π/6\pi/6 in [0,π/2][0, \pi/2].

Common mistakes

  1. 1.Sign errors while computing cofactors, leading to a wrong adjoint/A⁻¹ — always verify with AA−1=IAA^{-1} = I.
  2. 2.In part (a), using A−1BA^{-1}B directly without noticing the coefficient matrix is ATA^T (the solution needs (A−1)TB(A^{-1})^TB).
  3. 3.In part (b), dropping the valid root sin⁡x=1/2\sin x = 1/2 (x=π/6x = \pi/6) and reporting only x=0x = 0, or keeping the invalid root sin⁡x=−3/2\sin x = -3/2.

Analysis and explanation

This 5-mark question tests the matrix (inverse) method for a linear system and the determinant condition for a singular matrix — both core applications from the Determinants chapter.


Concept (a): A−1=adj(A)/∣A∣A^{-1} = \text{adj}(A)/|A|. A key feature here is that the system's coefficient matrix equals Aᵀ, not A, so the system is ATX=BA^TX = B and the solution is X=(A−1)TBX = (A^{-1})^TB — students must notice this transpose relationship rather than blindly using A⁻¹B.


Concept (b): A matrix is singular exactly when ∣A∣=0|A| = 0. Expanding the determinant converts the condition into a trigonometric equation, factorised using 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2\sin^2x and sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3x.


Exam trap: In (a), sign errors in the cofactors are common — verify with AA−1=IAA^{-1} = I, or substitute the found (x, y, z) back into all three equations. In (b), after factoring sin⁡x(4sin⁡2x+4sin⁡x−3)=0\sin x(4\sin^2x + 4\sin x - 3) = 0, students must keep both valid roots sin⁡x=0\sin x = 0 and sin⁡x=1/2\sin x = 1/2 (giving x=0x = 0 and x=π/6x = \pi/6) and reject only sin⁡x=−3/2\sin x = -3/2; dropping the sin x = 1/2 root is the most common error.


Real-world relevance: Solving AX = B by matrix inversion underlies computer-graphics transformations, structural load analysis, and Leontief input–output economic models.

Interesting facts

The adjoint (adjugate) method for matrix inversion was formalised by Arthur Cayley in his 1858 paper 'A Memoir on the Theory of Matrices', which also introduced the term 'matrix' itself.

Matrix inversion via the determinant/adjoint method is computationally expensive for large matrices (O(n!)O(n!) via cofactors), which is why real-world software uses Gaussian elimination or LU decomposition instead — but the adjoint method remains the standard hand-calculation technique for CBSE board exams.

The singular matrix condition (∣A∣=0|A|=0) is exactly the condition under which a system of linear equations either has no solution or infinitely many — a concept central to consistency analysis in linear algebra.

Question details

Q33 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section D
Marks
5
Type
Long Answer
Chapter
Determinants
Topic
Inverse of a Matrix and Singular Matrices
Bloom's level
apply
Syllabus topic
Unit 3: Matrices and Determinants — Inverse of a Matrix by Adjoint Method, Solving a System of Linear Equations, Condition for a Singular Matrix
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How do you know if A−1A^{-1} exists?

A⁻¹ exists if and only if ∣A∣≠0|A| \ne 0 (A is non-singular). If ∣A∣=0|A| = 0, the matrix is singular and A⁻¹ does not exist.

Why use 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2\sin^2x in part (b)?

This double-angle identity converts the mixed cos 2x and sin 3x terms into expressions of sin x alone, allowing the determinant equation to be factorised and solved for x.

How many values of x satisfy part (b)?

Two: x=0x = 0 and x=π/6x = \pi/6, both lying in the closed interval [0,π/2][0, \pi/2]. They come from sin⁡x=0\sin x = 0 and sin⁡x=1/2\sin x = 1/2; the third algebraic root sin⁡x=−3/2\sin x = -3/2 is rejected as it is outside [−1,1][-1, 1].