JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 11)
If $$A = \left[ {\matrix{
{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr
{{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr
{{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr
} } \right]$$
then A is :
then A is :
invertible for all t$$ \in $$R.
invertible only if t $$=$$ $$\pi $$
not invertible for any t$$ \in $$R
invertible only if t $$=$$ $${\pi \over 2}$$.
Explanation
$$A = \left[ {\matrix{
{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr
{{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr
{{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr
} } \right]$$
$$\left| A \right| = {e^t}.\,{e^{ - t}}.{e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr 1 & {2\sin t} & { - 2\cos t} \cr } } \right|$$
Apply operations R2 < R2 $$-$$R1, R3 < R3 $$-$$ R1, R1 < R1
$$\left| A \right| = {e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr 0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr } } \right|$$
Open the determinant by R1
$$\left| A \right| = 5{e^{ - t}}$$
Invertible for all t $$ \in $$ R
$$\left| A \right| = {e^t}.\,{e^{ - t}}.{e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr 1 & {2\sin t} & { - 2\cos t} \cr } } \right|$$
Apply operations R2 < R2 $$-$$R1, R3 < R3 $$-$$ R1, R1 < R1
$$\left| A \right| = {e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr 0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr } } \right|$$
Open the determinant by R1
$$\left| A \right| = 5{e^{ - t}}$$
Invertible for all t $$ \in $$ R
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