Q61 / 180 · KCET 2010 · Mathematics

If $A=\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]$, then $A^{2}+x A+y I=0$ for $(x, y)$ is

  • Correct answer
    A
    $(-4,1)$
  • B
    $(-1,3)$
  • C
    $(4,-1)$
  • D
    $(1,3)$

Solution

$$ A=\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right] $$ The characteristic equation of ' $A$ ' is $$ \begin{array}{r} |\mathrm{A}-\lambda \mathrm{I}|=0 \\ \left|\begin{array}{cc} 3-\lambda & 2 \\ 1 & 1-\lambda \end{array}\right|=0 \\ (3-\lambda)(1-\lambda)-2=0 \\ 3-\lambda-3 \lambda+\lambda^{2}-2=0 \\ \left(\lambda^{2}-4 \lambda+1\right)=0 \quad \text{...(i)} \end{array} $$ By Caylay-Hamilton theorem : Every square matrix satisfied its characteristic equation, then put $(\lambda=\mathrm{A})$ is in Eq. (i) $$ \begin{aligned} &\mathrm{A}^{2}-4 \mathrm{~A}+\mathrm{I}=0 \\ &\text { On comparing with } \mathrm{A}^{2}+\mathrm{xA}+\mathrm{yI}=0 \\ &\Rightarrow \quad \mathrm{x}=-4, \mathrm{y}=1 \end{aligned} $$

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

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