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

OR

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

Q: How many values of x satisfy part (b)?

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

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): $x = 0, y = -5, z = -3$

For $A = \begin{bmatrix}0 & 2 & 1\\ -2 & -1 & -2\\ 1 & -1 & 0\end{bmatrix}$ , $|A| = -1 \ne 0$ , so $A^{-1}$ exists.
$A^{-1} = \begin{bmatrix}2 & 1 & 3\\ 2 & 1 & 2\\ -3 & -2 & -4\end{bmatrix}$ .
The system's coefficient matrix is $\begin{bmatrix}0 & -2 & 1\\ 2 & -1 & -1\\ 1 & -2 & 0\end{bmatrix} = A^T$ , so the system is $A^TX = B$ with $B = [7, 8, 10]^T$ .
Then $X = (A^T)^{-1}B = (A^{-1})^TB = \begin{bmatrix}2 & 2 & -3\\ 1 & 1 & -2\\ 3 & 2 & -4\end{bmatrix} \cdot [7, 8, 10]^T = [0, -5, -3]^T$ .
Hence $x = 0, y = -5, z = -3$ .

Part (b): $x = 0$ or $x = \pi/6$

A matrix is singular when its determinant is 0. Expanding gives $10 - 10\cos 2x - 5\sin 3x = 0$ , i.e. $2 - 2\cos 2x - \sin 3x = 0$ .
Using $2 - 2\cos 2x = 4\sin^2x$ and $\sin 3x = 3\sin x - 4\sin^3x$ : $4\sin^2x - 3\sin x + 4\sin^3x = 0 \implies \sin x(4\sin^2x + 4\sin x - 3) = 0$ .
$\sin x = 0 \implies x = 0$ ; $4\sin^2x + 4\sin x - 3 = 0 \implies \sin x = 1/2$ or $\sin x = -3/2$ (rejected).
$\sin x = 1/2 \implies x = \pi/6$ . Both $x = 0$ and $x = \pi/6$ lie in $[0, \pi/2]$ , so $x = 0$ or $x = \pi/6$ .

inverse of a matrixadjoint methodsingular matrixsystem of linear equationscofactordeterminantmatrix method

Marking Scheme

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

Hint

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

Quick Oral Answer

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

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

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

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

Common Mistakes

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

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 ( $O(n!)$ 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 ( $|A|=0$ ) 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.

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Frequently Asked Questions

How do you know if $A^{-1}$ exists?

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

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