JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 14)
Let M be a 2 $$\times$$ 2 symmetric matrix with integer entries. Then, M is invertible, if
the first column of M is the transpose of the second row of M
the second row of M is the transpose of the first column of M
M is a diagonal matrix with non-zero entries in the main diagonal
the product of entries in the main diagonal of M is not the square of an integer
Explanation
Note : A square matrix M is invertible if det(M) or | M | $$\ne$$ 0.
Let $$M = \left[ {\matrix{ a & b \cr b & c \cr } } \right]$$
(a) Given that $$\left[ {\matrix{ a \cr b \cr } } \right] = \left[ {\matrix{ b \cr c \cr } } \right]$$
$$ \Rightarrow a = b = c = \alpha $$ (let)
$$ \Rightarrow M = \left[ {\matrix{ \alpha & \alpha \cr \alpha & \alpha \cr } } \right]$$
$$\Rightarrow$$ | M | = 0
$$\Rightarrow$$ M is non-invertible.
(b) Given that [bc] = [ab]
$$ \Rightarrow a = b = c = \alpha $$ (let)
Again | M | = 0
$$\Rightarrow$$ M is non-invertible.
(c) As given $$M = \left[ {\matrix{ a & 0 \cr 0 & c \cr } } \right]$$
$$ \Rightarrow \left| M \right| = ac \ne 0$$ ($$\because$$ a and c are non-zero)
$$\Rightarrow$$ M is invertible.
(d) $$M = \left[ {\matrix{ a & b \cr b & c \cr } } \right] \Rightarrow \left| M \right| = ac - {b^2} \ne 0$$
$$\because$$ ac is not equal to square of an integer.
$$\therefore$$ M is invertible.
Let $$M = \left[ {\matrix{ a & b \cr b & c \cr } } \right]$$
(a) Given that $$\left[ {\matrix{ a \cr b \cr } } \right] = \left[ {\matrix{ b \cr c \cr } } \right]$$
$$ \Rightarrow a = b = c = \alpha $$ (let)
$$ \Rightarrow M = \left[ {\matrix{ \alpha & \alpha \cr \alpha & \alpha \cr } } \right]$$
$$\Rightarrow$$ | M | = 0
$$\Rightarrow$$ M is non-invertible.
(b) Given that [bc] = [ab]
$$ \Rightarrow a = b = c = \alpha $$ (let)
Again | M | = 0
$$\Rightarrow$$ M is non-invertible.
(c) As given $$M = \left[ {\matrix{ a & 0 \cr 0 & c \cr } } \right]$$
$$ \Rightarrow \left| M \right| = ac \ne 0$$ ($$\because$$ a and c are non-zero)
$$\Rightarrow$$ M is invertible.
(d) $$M = \left[ {\matrix{ a & b \cr b & c \cr } } \right] \Rightarrow \left| M \right| = ac - {b^2} \ne 0$$
$$\because$$ ac is not equal to square of an integer.
$$\therefore$$ M is invertible.
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