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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices1 Mark
MCQ
If $A=\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]$, then the value of $I-A+A^2-A^3+\ldots$ is :
  • ✓
    $\left[\begin{array}{cc}-1 & -1 \\ 4 & 3\end{array}\right]$
  • B
    $\left[\begin{array}{cc}3 & 1 \\ -4 & -1\end{array}\right]$
  • C
    $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
  • D
    $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
✓

Answer

Correct option: A.
$\left[\begin{array}{cc}-1 & -1 \\ 4 & 3\end{array}\right]$
Given $A=\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]$, then
$A^2=\left[\begin{array}{cc} 2 & 1 \\ -4 & -2 \end{array}\right]\left[\begin{array}{cc} 2 & 1 \\ -4 & -2 \end{array}\right]$
$=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0 $
$\Rightarrow A^n=0 \forall n \geq 2$
$\therefore I-A+A^2-A^3+\ldots=I-A$
$=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} 2 & 1 \\ -4 & -2 \end{array}\right]$
$=\left[\begin{array}{cc} -1 & -1 \\ 4 & 3 \end{array}\right]$

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