JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 1)
Let A and B be 3 $$\times$$ 3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations (A2B2 $$-$$ B2A2) X = O, where X is a 3 $$\times$$ 1 column matrix of unknown variables and O is a 3 $$\times$$ 1 null matrix, has :
no solution
exactly two solutions
infinitely many solutions
a unique solution
Explanation
AT = A, BT = $$-$$B
Let A2B2 $$-$$ B2A2 = P
PT = (A2B2 $$-$$ B2A2)T = (A2B2)T $$-$$ (B2A2)T
= (B2)T (A2)T $$-$$ (A2)T (B2)T
= B2A2 $$-$$ A2B2
$$ \Rightarrow $$ P is skew-symmetric matrix
$$\left[ {\matrix{ 0 & a & b \cr { - a} & 0 & c \cr { - b} & { - c} & 0 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$
$$ \therefore $$ ay + bz = 0 ..... (1)
$$-$$ax + cz = 0 .... (2)
$$-$$bx $$-$$cy = 0 ..... (3)
From equation 1, 2, 3
$$\Delta$$ = 0 & $$\Delta$$1 = $$\Delta$$2 = $$\Delta$$3 = 0
$$ \therefore $$ equation have infinite number of solution
Let A2B2 $$-$$ B2A2 = P
PT = (A2B2 $$-$$ B2A2)T = (A2B2)T $$-$$ (B2A2)T
= (B2)T (A2)T $$-$$ (A2)T (B2)T
= B2A2 $$-$$ A2B2
$$ \Rightarrow $$ P is skew-symmetric matrix
$$\left[ {\matrix{ 0 & a & b \cr { - a} & 0 & c \cr { - b} & { - c} & 0 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$
$$ \therefore $$ ay + bz = 0 ..... (1)
$$-$$ax + cz = 0 .... (2)
$$-$$bx $$-$$cy = 0 ..... (3)
From equation 1, 2, 3
$$\Delta$$ = 0 & $$\Delta$$1 = $$\Delta$$2 = $$\Delta$$3 = 0
$$ \therefore $$ equation have infinite number of solution
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