Q44 / 120 · VITEEE 2006 · Mathematics

Let $\mathrm{A}=\left(\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & \mathrm{t} \\ 4 & 7-\mathrm{t} & -6\end{array}\right)$, then the values of $\mathrm{t}$ for which inverse of A does not exist

  • A
    $-2,1$
  • B
    3,2
  • Correct answer
    C
    $2,-3$
  • D
    $3,-1$

Solution

We know that inverse of A does not exist only when $|\mathrm{A}|=0$ $\therefore\left|\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6\end{array}\right|=0$ $\Rightarrow\left(-30-7 t+t^{2}\right)-3(-12-4 t)$ $+2(14-2 t-20)=0$ $\Rightarrow-30-7 t+t^{2}+36+12 t-12-4 t=0$ $\Rightarrow t^{2}+t-6=0 \Rightarrow t^{2}+3 t-2 t-6=0$ $\Rightarrow t(t+3)-2(t+3)=0$ $\Rightarrow(t+3)(t-2)=0 \Rightarrow t=2,-3$

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

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