Q43 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

If $A=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}8 & 0 \\ 7 & 1\end{array}\right]$ and $A^3=B$, then $x=$

  • A
    -2 or 3
  • B
    $-2$
  • C
    2 or -3
  • Correct answer
    D
    $2$

Solution

$\mathrm{A} \cdot \mathrm{A}=\mathrm{A}^2=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$ $=\left[\begin{array}{cc}x^2 & 0 \\ x+1 & 1\end{array}\right]$ $A^3=A^2 \cdot A=\left[\begin{array}{cc}x^2 & 0 \\ x+1 & 1\end{array}\right]\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$ $=\left[\begin{array}{cc}x^3 & 0 \\ x(x+1)+1 & 1\end{array}\right]$ $\begin{aligned} & \mathrm{A}^3=\mathrm{B} \\ & \Rightarrow x^3=8 \\ & \therefore x=2\end{aligned}$

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

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