Q46 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics

Let $\mathrm{A}$ be a matrix such that $\mathrm{AB}$ is a scalar matrix where $B=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$ and $\operatorname{det}(3 A)=27$. Then $3 A^{-1}+A^2=$

  • A
    $\left[\begin{array}{cc}4 & -6 \\ 0 & 2\end{array}\right]$
  • B
    $\left[\begin{array}{cc}9 & -4 \\ 0 & 3\end{array}\right]$
  • C
    $\left[\begin{array}{cc}10 & -6 \\ 0 & 2\end{array}\right]$
  • Correct answer
    D
    $\left[\begin{array}{cc}10 & -6 \\ 0 & 4\end{array}\right]$

Solution

$\begin{aligned} & A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \\ & A B=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]=\left[\begin{array}{ll}a & 2 a+3 b \\ c & 2 c+3 d\end{array}\right]=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right] \\ & \therefore \quad c=0, a=k \\ & \Rightarrow 2 a+3 b=0 \\ & 2 c+3 d=k \\ & \therefore 3 d=k=a \\ & \therefore 6 d+3 b=0 \Rightarrow b+2 d=0 \\ & \text { Also }|3 A|=27 \Rightarrow 9|A|=27 \Rightarrow|A|=3 \\ & a b-b c=3 \\ & \therefore \quad a d=3 \\ & 3 d \cdot d=3 \Rightarrow d=1 \\ & \therefore \quad b=-2, a=3 \\ & \therefore \quad A=\left[\begin{array}{cc}3 & -2 \\ 0 & 1\end{array}\right] \\ & A^2=\left[\begin{array}{cc}3 & -2 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & -2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}9 & -8 \\ 0 & 1\end{array}\right] \\ & A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj} A=\frac{1}{3}\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right] \\ & \therefore \quad 3 A^{-1}+A^2=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]+\left[\begin{array}{cc}9 & -8 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}10 & -6 \\ 0 & 4\end{array}\right] \\ & \end{aligned}$

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

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