JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 7)
If $$B = \left[ {\matrix{
5 & {2\alpha } & 1 \cr
0 & 2 & 1 \cr
\alpha & 3 & { - 1} \cr
} } \right]$$ is the inverse of a 3 × 3 matrix A, then the sum of all values of $$\alpha $$ for which
det(A) + 1 = 0, is :
2
- 1
0
1
Explanation
Given |A| + 1 = 0
$$ \Rightarrow $$ |A| = -1
$$\left| B \right| = \left| {{A^{ - 1}}} \right| = {1 \over {\left| A \right|}} = - 1$$
$$\left| {\matrix{ 5 & {2\alpha } & 1 \cr 0 & 2 & 1 \cr \alpha & 3 & { - 1} \cr } } \right| $$ = -1
$$ \Rightarrow $$ $$ 5( - 2 - 3) + 2\alpha (\alpha ) + 1( - 2\alpha ) = - 1$$
$$ \Rightarrow $$ $$2{\alpha ^2} - 2\alpha - 24 = 0$$
$$ \therefore $$ sum of value of $$\alpha $$ = $${{ - ( - 2)} \over 2} = 1$$
$$ \Rightarrow $$ |A| = -1
$$\left| B \right| = \left| {{A^{ - 1}}} \right| = {1 \over {\left| A \right|}} = - 1$$
$$\left| {\matrix{ 5 & {2\alpha } & 1 \cr 0 & 2 & 1 \cr \alpha & 3 & { - 1} \cr } } \right| $$ = -1
$$ \Rightarrow $$ $$ 5( - 2 - 3) + 2\alpha (\alpha ) + 1( - 2\alpha ) = - 1$$
$$ \Rightarrow $$ $$2{\alpha ^2} - 2\alpha - 24 = 0$$
$$ \therefore $$ sum of value of $$\alpha $$ = $${{ - ( - 2)} \over 2} = 1$$
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