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Mathematics Question on Invertible Matrices

Let A=[132 25t 47−t−6 ]\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

44622

C

2, -3

D

3, -1

Answer

2, -3

Explanation

Solution

We know that inverse of A does not exist
only when |A| = 0
∴\therefore ∣132 25t 47−t−6 ∣\begin{vmatrix} 1 & 3 & 2 \\\ 2 & 5 &t \\\ 4&7-t&-6 \\\ \end{vmatrix}=0
(−30−7t+t2)−3(−12−4t)(-30-7t+t^2)-3(-12-4t)
                  +2(14−2t−20)=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2(14-2t-20)=0
⇒  −30−7t+t2+36+12t−12−4t=0\Rightarrow \ \ -30- 7t + t^2 +36+12t-12-4t=0
⇒  t2+t−6=0⇒  t2+3t −2t−6=0\Rightarrow \ \ t^2+t-6=0 \Rightarrow \ \ t^2 +3t \ -2t-6=0
⇒  t(t+3)−2(t+3)=0\Rightarrow \ \ t(t+3)-2(t+3)=0
⇒    (t+3)(t−2)=0 ⇒ t=2,−3\Rightarrow \ \ \ \ (t+3)(t-2)=0 \ \Rightarrow \ t=2, -3