JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 5)
Let $$A = \left[ {\matrix{
2 & 3 \cr
a & 0 \cr
} } \right]$$, a$$\in$$R be written as P + Q where P is a symmetric matrix and Q is skew symmetric matrix. If det(Q) = 9, then the modulus of the sum of all possible values of determinant of P is equal to :
36
24
45
18
Explanation
$$A = \left[ {\matrix{
2 & 3 \cr
a & 0 \cr
} } \right]$$, $${A^T} = \left[ {\matrix{
2 & a \cr
3 & 0 \cr
} } \right]$$
$$A = {{A + {A^T}} \over 2} + {{A - {A^T}} \over 2}$$
Let $$P = {{A + {A^T}} \over 2}$$ and $$Q = {{A - {A^T}} \over 2}$$
$$Q = \left( {\matrix{ 0 & {{{3 - a} \over 2}} \cr {{{a - 3} \over 2}} & 0 \cr } } \right)$$
Det (Q) = 9
$$0 - \left( {{{3 - a} \over 2}} \right)\left( {{{a - 3} \over 2}} \right) = 9$$
$$ \Rightarrow {\left( {{{a - 3} \over 2}} \right)^2} = 9 \Rightarrow {(a - 3)^2} = 36$$
$$a - 3 = \pm \,6 \Rightarrow a = 9, - 3$$
$$P = \left[ {\matrix{ 2 & {{{a + 3} \over 2}} \cr {{{a + 3} \over 2}} & 0 \cr } } \right]$$
$$P = \left[ {\matrix{ 2 & 6 \cr 6 & 0 \cr } } \right]$$ or $$\left[ {\matrix{ 2 & 0 \cr 0 & 0 \cr } } \right]$$
| P | = - 36 or 0
$$\therefore$$ | $$-$$36 + 0 | = 36
$$A = {{A + {A^T}} \over 2} + {{A - {A^T}} \over 2}$$
Let $$P = {{A + {A^T}} \over 2}$$ and $$Q = {{A - {A^T}} \over 2}$$
$$Q = \left( {\matrix{ 0 & {{{3 - a} \over 2}} \cr {{{a - 3} \over 2}} & 0 \cr } } \right)$$
Det (Q) = 9
$$0 - \left( {{{3 - a} \over 2}} \right)\left( {{{a - 3} \over 2}} \right) = 9$$
$$ \Rightarrow {\left( {{{a - 3} \over 2}} \right)^2} = 9 \Rightarrow {(a - 3)^2} = 36$$
$$a - 3 = \pm \,6 \Rightarrow a = 9, - 3$$
$$P = \left[ {\matrix{ 2 & {{{a + 3} \over 2}} \cr {{{a + 3} \over 2}} & 0 \cr } } \right]$$
$$P = \left[ {\matrix{ 2 & 6 \cr 6 & 0 \cr } } \right]$$ or $$\left[ {\matrix{ 2 & 0 \cr 0 & 0 \cr } } \right]$$
| P | = - 36 or 0
$$\therefore$$ | $$-$$36 + 0 | = 36
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