JEE MAIN - Mathematics (2018 - 15th April Evening Slot - No. 5)
Suppose A is any 3$$ \times $$ 3 non-singular matrix and ( A $$-$$ 3I) (A $$-$$ 5I) = O where I = I3 and O = O3. If $$\alpha $$A + $$\beta $$A-1 = 4I, then $$\alpha $$ + $$\beta $$ is equal to :
8
7
13
12
Explanation
Given,
( A $$-$$ 3I) (A $$-$$ 5I) = O
$$ \Rightarrow $$ A2 - 8A + 15I = O
Multiplying both sides by A- 1, we get,
A- 1A.A - 8A- 1A + 15A- 1I = A- 1O
$$ \Rightarrow $$ A - 8I + 15A- 1 = O
$$ \Rightarrow $$ A + 15A- 1 = 8I
$$ \Rightarrow $$$${A \over 2} + {{15{A^{ - 1}}} \over 2} = 4I$$
Comparing with the equation $$\alpha $$A + $$\beta $$A-1 = 4I, we get
$$\alpha $$ = $${1 \over 2}$$ and $$\beta $$ = $${15 \over 2}$$
$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $${1 \over 2}$$ + $${15 \over 2}$$ = $${16 \over 2}$$ = 8
( A $$-$$ 3I) (A $$-$$ 5I) = O
$$ \Rightarrow $$ A2 - 8A + 15I = O
Multiplying both sides by A- 1, we get,
A- 1A.A - 8A- 1A + 15A- 1I = A- 1O
$$ \Rightarrow $$ A - 8I + 15A- 1 = O
$$ \Rightarrow $$ A + 15A- 1 = 8I
$$ \Rightarrow $$$${A \over 2} + {{15{A^{ - 1}}} \over 2} = 4I$$
Comparing with the equation $$\alpha $$A + $$\beta $$A-1 = 4I, we get
$$\alpha $$ = $${1 \over 2}$$ and $$\beta $$ = $${15 \over 2}$$
$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $${1 \over 2}$$ + $${15 \over 2}$$ = $${16 \over 2}$$ = 8
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