Q46 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics

If $\mathrm{A}=\left[\begin{array}{ccc}k & 5 & 2 \\ 2 & -k & 5 \\ 5 & 2 & -k\end{array}\right]$ and $\operatorname{det} A=190$ then $\operatorname{Adj} A=$

  • Correct answer
    A
    $\left[\begin{array}{ccc}-1 & 19 & 31 \\ 31 & -19 & -11 \\ 19 & 19 & -19\end{array}\right]$
  • B
    $\left[\begin{array}{ccc}-1 & 31 & 19 \\ 19 & -19 & 19 \\ 31 & -11 & -19\end{array}\right]$
  • C
    $\left[\begin{array}{ccc}-1 & 19 & 31 \\ -31 & -19 & -11 \\ 19 & 19 & -19\end{array}\right]$
  • D
    $\left[\begin{array}{ccc}-1 & -31 & 19 \\ 19 & -19 & 19 \\ 31 & -11 & -19\end{array}\right]$

Solution

$A=\left[\begin{array}{ccc}K & 5 & 2 \\ 2 & -K & 5 \\ 5 & 2 & -K\end{array}\right]$ $\begin{array}{ll}\therefore & |A|=K\left(K^2-10\right)-5(-2 K-25)+2(4+5 K)=190 \\ \Rightarrow & K^3-10 K+10 K+125+8+10 K=190 \\ \Rightarrow & K^3+10 K-57=0 \Rightarrow(K-3)\left(K^2+3 K+19\right)=0 \\ \Rightarrow & K=3 \\ \therefore & A=\left[\begin{array}{ccc}3 & 5 & 2 \\ 2 & -3 & 5 \\ 5 & 2 & -3\end{array}\right]\end{array}$ $\begin{array}{lll}C_{11}=-1 & C_{21}=19 & C_{31}=31 \\ C_{12}=31 & C_{22}=-19 & C_{32}=-11 \\ C_{13}=19 & C_{23}=19 & C_{33}=-19 \\ \therefore & \text { adj } A=\left[\begin{array}{ccc}-1 & 19 & 31 \\ 31 & -19 & -11 \\ 19 & 19 & -19\end{array}\right]\end{array}$

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

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