Question

Let A=$\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 &t \\ 4&7-t&-6 \\ \end{bmatrix}$, then the values of t for which inverse of A does not exist

• A. -2, 1
• B. 3, 2
• C. 2, -3
• D. 3, -1

Answer 1

Answer: (C)

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