Q47 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

If $A=\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]$ and $f(x)=x+x^2+x^3+\ldots . .+x^{2023}$ then $\mathrm{f}(\mathrm{A})+\mathrm{I}=$

  • A
    $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
  • B
    $\left[\begin{array}{ll}1 & 3 \\ 0 & 0\end{array}\right]$
  • Correct answer
    C
    $\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$
  • D
    $\left[\begin{array}{ll}1 & 3 \\ 1 & 1\end{array}\right]$

Solution

$A=\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]$ $\begin{aligned} & \mathrm{A}^2=\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \\ & \therefore \mathrm{A}^n=0, n \geq 2 \\ & f(\mathrm{~A})=\mathrm{A}+\mathrm{A}^2+\mathrm{A}^3+\ldots .+\mathrm{A}^{2023} \\ & f(\mathrm{~A})=\mathrm{A} \\ & f(\mathrm{~A})+\mathrm{I}=\mathrm{A}+\mathrm{I} \\ & =\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]\end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions