Q51 / 150 · MHT CET 2024 (16 May Shift 1) · Mathematics

Let $\mathrm{A}=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right], x \in \mathbb{R}^{+}$and $\mathrm{A}^4=\left[\mathrm{a}_{\mathrm{ij}}\right]_2$. If $\mathrm{a}_{11}=109$, then $\left(\mathrm{A}^4\right)^{-1}=$

  • A
    $\left[\begin{array}{ll}109 & 33 \\ 33 & 10\end{array}\right]$.
  • B
    $\left[\begin{array}{ll}10 & 33 \\ 33 & 10\end{array}\right]$
  • C
    $\left[\begin{array}{cc}10 & 33 \\ 33 & 109\end{array}\right]$
  • Correct answer
    D
    $\left[\begin{array}{cc}10 & -33 \\ -33 & 109\end{array}\right]$

Solution

$\begin{aligned} & A=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right] \\ & A^2=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}x^2+1 & x \\ x & 1\end{array}\right] \\ \therefore \quad & A^4=A^2 \cdot A^2\end{aligned}$ $\begin{aligned} & =\left[\begin{array}{cc} x^2+1 & x \\ x & 1 \end{array}\right]\left[\begin{array}{cc} x^2+1 & x \\ x & 1 \end{array}\right] \\ & =\left[\begin{array}{ll} \left(x^2+1\right)^2+x^2 & x\left(x^2+1+1\right) \\ x\left(x^2+1+1\right) & \left(x^2+1\right) \end{array}\right] \\ & =\left[\begin{array}{cc} \left(x^2+1\right)^2+x^2 & x\left(x^2+2\right) \\ x\left(x^2+2\right) & x^2+1 \end{array}\right] \\ \therefore \quad \begin{array}{l} \dot{A}^4 \end{array} & =\left[\mathrm{a}_{\mathrm{ij}}\right] \text { and } \mathrm{a}_{11}=109 \\ \mathrm{a}_{11} & =109 \\ \Rightarrow & \Rightarrow\left(x^2+1\right)^2+x^2=109 \end{aligned}$ ...[Given] i.e. $\begin{aligned} \left(x^2+1\right)^2+x^2 & =100+9 \\ & =(10)^2+3^2=\left(3^2+1\right)^2+3^2 \end{aligned}$
Comparing we get, $x^2=9 \Rightarrow x=3$ $\mathrm{a}_{12}=x\left(x^2+2\right)=3(9+2)=33$ $\begin{aligned} & \mathrm{a}_{21}=x\left(x^2+2\right)=3(9+2)=33 \\ & \mathrm{a}_{22}=x^2+1=9+1=10 \end{aligned}$ $\begin{array}{ll} \therefore & A^4=\left[\begin{array}{cc} 109 & 33 \\ 33 & 10 \end{array}\right] \\ \therefore & \left|A^4\right|=1 \neq 0 \end{array}$
If $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ and $a d-b c \neq 0$, then $\begin{array}{ll} & A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \\ \therefore & \left(A^4\right)^{-1}=\left[\begin{array}{cc} 10 & -33 \\ -33 & 109 \end{array}\right] \end{array}$

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

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