NIMCET 2009 — Mathematics PYQ

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

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 $A=\left[\begin{array}{cc}1& amp;2\\ 3& amp;4\end{array}\right]$ and $I=\left[\begin{array}{cc}1& amp;0\\ 0& amp;1\end{array}\right]$:

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

The inverse of a $2\times 2$ matrix $M=\left[\begin{array}{cc}a& amp;b\\ c& amp;d\end{array}\right]$ is given by:

• 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:

  $$\text{adj}(I-A)=\left[\begin{array}{cc}-3& amp;2\\ 3& amp;0\end{array}\right]$$

• Step C: Calculate the Inverse:

  $$(I-A{)}^{-1}=\frac{1}{-6}\left[\begin{array}{cc}-3& amp;2\\ 3& amp;0\end{array}\right]$$

  Divide each element by $-6$:

  $$(I-A{)}^{-1}=\left[\begin{array}{cc}\frac{-3}{-6}& amp;\frac{2}{-6}\\ \frac{3}{-6}& amp;\frac{0}{-6}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& amp;-\frac{1}{3}\\ -\frac{1}{2}& amp;0\end{array}\right]$$

Final Answer:

The sum of the series is $\left[\begin{array}{cc}\frac{1}{2}& amp;-\frac{1}{3}\\ -\frac{1}{2}& amp;0\end{array}\right]$. The correct option is (c).