Explanation: 1. The Geometric Series Formula for Matrices: For a square matrix , the sum of the infinite series is given by: This formula is valid if the series converges (i.e., all eigenvalues of have a magnitude less than ). Note: In competitive exam contexts, if the sum is asked, we proceed with calculating . 2. Calculate : Given and : 3. Find the Inverse : The inverse of a matrix is given by: Step A: Calculate the determinant : Step B: Calculate the Adjoint: Swap diagonal elements and change the signs of off-diagonal elements: Step C: Calculate the Inverse: Divide each element by : Final Answer: The sum of the series is . The correct option is (c) .

How do I solve NIMCET 2009 Mathematics questions?

Practice NIMCET 2009 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 2009 — Mathematics PYQ

NIMCET | Mathematics | 2009

If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ , then $I + A + A^2 + \dots \infty$ equals to:

Choose the correct answer:

Explanation

1. The Geometric Series Formula for Matrices:

For a square matrix $A$ , the sum of the infinite series $S = I + A + A^2 + A^3 + \dots$ is given by:

$S = (I - A)^{-1}$

This formula is valid if the series converges (i.e., all eigenvalues of $A$ have a magnitude less than $1$ ). Note: In competitive exam contexts, if the sum is asked, we proceed with calculating $(I - A)^{-1}$ .

2. Calculate $(I - A)$ :

Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ :

$I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

$I - A = \begin{bmatrix} 1-1 & 0-2 \\ 0-3 & 1-4 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ -3 & -3 \end{bmatrix}$

3. Find the Inverse $(I - A)^{-1}$ :

The inverse of a $2 \times 2$ matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by:

$M^{-1} = \frac{1}{|M|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Step A: Calculate the determinant $|I - A|$ :

$|I - A| = (0)(-3) - (-2)(-3)$

$|I - A| = 0 - 6 = -6$

Step B: Calculate the Adjoint:

Swap diagonal elements and change the signs of off-diagonal elements:

$\text{adj}(I - A) = \begin{bmatrix} -3 & 2 \\ 3 & 0 \end{bmatrix}$

Step C: Calculate the Inverse:

$(I - A)^{-1} = \frac{1}{-6} \begin{bmatrix} -3 & 2 \\ 3 & 0 \end{bmatrix}$

Divide each element by $-6$ :

$(I - A)^{-1} = \begin{bmatrix} \frac{-3}{-6} & \frac{2}{-6} \\ \frac{3}{-6} & \frac{0}{-6} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{3} \\ -\frac{1}{2} & 0 \end{bmatrix}$

Final Answer:

The sum of the series is $\begin{bmatrix} \frac{1}{2} & -\frac{1}{3} \\ -\frac{1}{2} & 0 \end{bmatrix}$ . The correct option is (c).

(Correct Answer)