JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 4)
If $$A = \left( {\matrix{
2 & 2 \cr
9 & 4 \cr
} } \right)$$ and $$I = \left( {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right)$$ then 10A–1 is
equal to :
6I – A
4I – A
A – 6I
A – 4I
Explanation
According to Cayley Hamilton equation
|A – $$\lambda $$I| = 0
$$ \Rightarrow $$ $$\left| {\matrix{ {2 - \lambda } & 2 \cr 9 & {4 - \lambda } \cr } } \right|$$ = 0
$$ \Rightarrow $$ (2 – $$\lambda $$)(4 – $$\lambda $$) – 18 = 0
$$ \Rightarrow $$ 8 – 2$$\lambda $$ – 4$$\lambda $$ + $$\lambda $$2 – 18 = 0
$$ \Rightarrow $$ $$\lambda $$2 – 6$$\lambda $$ – 10 = 0
$$ \therefore $$ A2 – 6A– 10 = 0
$$ \Rightarrow $$ A–1(A2) – 6A–1A – 10A–1 = 0
$$ \Rightarrow $$ A – 6I – 10A–1 = 0
$$ \Rightarrow $$ 10A–1 = A – 6I
|A – $$\lambda $$I| = 0
$$ \Rightarrow $$ $$\left| {\matrix{ {2 - \lambda } & 2 \cr 9 & {4 - \lambda } \cr } } \right|$$ = 0
$$ \Rightarrow $$ (2 – $$\lambda $$)(4 – $$\lambda $$) – 18 = 0
$$ \Rightarrow $$ 8 – 2$$\lambda $$ – 4$$\lambda $$ + $$\lambda $$2 – 18 = 0
$$ \Rightarrow $$ $$\lambda $$2 – 6$$\lambda $$ – 10 = 0
$$ \therefore $$ A2 – 6A– 10 = 0
$$ \Rightarrow $$ A–1(A2) – 6A–1A – 10A–1 = 0
$$ \Rightarrow $$ A – 6I – 10A–1 = 0
$$ \Rightarrow $$ 10A–1 = A – 6I
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