First slide

Applications of determinant

Question

Let $A_{t} = (\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7 - t & - 6 \end{array})$ then the value(s) of t for which inverse of $A_{t}$ does not exist.

Moderate

A

– 2, 1
B

$3, 2$
C

$2, - 3$
D

$3, - 1$

Solution

since ${A_{t}}^{- 1}$ does not exist, $|A_{t}| = 0$
$\Rightarrow (- 30 - t (7 - t)) - 3 (- 12 - 4 t) + 2 (14 - 2 t - 20) = 0$
$\Rightarrow t^{2} + t - 6 = 0 \Rightarrow t = - 3, 2$

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