Q62 / 150 · MHT CET 2008 · Mathematics

The inverse matrix of $\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$ is

  • Correct answer
    A
    $\left[\begin{array}{rrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}\right]$
  • B
    $\left[\begin{array}{rrr}\frac{1}{2} & -4 & \frac{5}{2} \\ 1 & -6 & 3 \\ 1 & 2 & -1\end{array}\right]$
  • C
    $\frac{1}{2}\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 2 & 1 \\ 4 & 2 & 3\end{array}\right]$
  • D
    $\frac{1}{2}\left[\begin{array}{rrr}1 & -1 & -1 \\ -8 & 6 & -2 \\ 5 & -3 & 1\end{array}\right]$

Solution

Let $\begin{aligned} A &=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right] \\ \therefore \begin{aligned}|A| &=0-1(1-9)+2(1-6) \\ &=8-10=-2 \neq 0 \\ \operatorname{adj}(A) &=\left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right] \\ \therefore A^{-1} &=\frac{\operatorname{adj}(A)}{|A|} \\ &=\frac{1}{-2}\left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right] \\ &=\left[\begin{array}{rrrr}1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right] \end{aligned} \end{aligned}$

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

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