Consider the matrix . Then is:

None of these Explanation: Solving: 1. Find the Characteristic Equation: The characteristic equation is given by : Expanding along the second row: 2. Applying Cayley-Hamilton Theorem: Every square matrix satisfies its own characteristic equation. Replace with : (Note: The question asks for based on the expression provided in the image context). Let's find the value of : From , we get: Subtract from both sides: 3. Direct Calculation of : 4. Finding the resulting matrix: Since this resulting matrix is not the Identity, not all entries are 10, and it is not diagonal: Correct Option: (d) None of these

How do I solve NIMCET 2010 Mathematics questions?

Practice NIMCET 2010 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Tip:A–D to answerE for explanationV for videoS to reveal answer

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

Consider the matrix $A = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & 0 \\ 1 & 1 & 3 \end{bmatrix}$ . Then $A^3 - 9A$ is:

Choose the correct answer:

Explanation

Solving:

1. Find the Characteristic Equation:

The characteristic equation is given by $|A - \lambda I| = 0$ :

\begin{vmatrix} 1-\lambda & -2 & 2 \\ 0 & 2-\lambda & 0 \\ 1 & 1 & 3-\lambda \end{vmatrix} = 0

Expanding along the second row:

(2-\lambda) \begin{vmatrix} 1-\lambda & 2 \\ 1 & 3-\lambda \end{vmatrix} = 0

(2-\lambda) [(1-\lambda)(3-\lambda) - 2] = 0

(2-\lambda) [3 - \lambda - 3\lambda + \lambda^2 - 2] = 0

(2-\lambda) [\lambda^2 - 4\lambda + 1] = 0

2\lambda^2 - 8\lambda + 2 - \lambda^3 + 4\lambda^2 - \lambda = 0

-\lambda^3 + 6\lambda^2 - 9\lambda + 2 = 0 \implies \lambda^3 - 6\lambda^2 + 9\lambda - 2 = 0

2. Applying Cayley-Hamilton Theorem:

Every square matrix satisfies its own characteristic equation. Replace $\lambda$ with $A$ :

A^3 - 6A^2 + 9A - 2I = 0

A^3 + 9A = 6A^2 + 2I

(Note: The question asks for $A^3 - 9A$ based on the expression provided in the image context).

Let's find the value of $A^3 - 9A$ :

From $A^3 - 6A^2 + 9A - 2I = 0$ , we get:

A^3 + 9A = 6A^2 + 2I

Subtract $18A$ from both sides:

A^3 - 9A = 6A^2 - 18A + 2I

3. Direct Calculation of $A^2$ :

A^2 = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & 0 \\ 1 & 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & 0 \\ 1 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 8 \\ 0 & 4 & 0 \\ 4 & 3 & 11 \end{bmatrix}

4. Finding the resulting matrix:

A^3 - 9A = 6A^2 - 18A + 2I

= 6\begin{bmatrix} 3 & -4 & 8 \\ 0 & 4 & 0 \\ 4 & 3 & 11 \end{bmatrix} - 18\begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & 0 \\ 1 & 1 & 3 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}

= \begin{bmatrix} 18 & -24 & 48 \\ 0 & 24 & 0 \\ 24 & 18 & 66 \end{bmatrix} - \begin{bmatrix} 18 & -36 & 36 \\ 0 & 36 & 0 \\ 18 & 18 & 54 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}

= \begin{bmatrix} 2 & 12 & 12 \\ 0 & -10 & 0 \\ 6 & 0 & 14 \end{bmatrix}

Since this resulting matrix is not the Identity, not all entries are 10, and it is not diagonal:

Correct Option:

(d) None of these