JEE Advance - Mathematics (2021 - Paper 1 Online - No. 7)
Let $$\alpha$$, $$\beta$$ and $$\gamma$$ be real numbers such that the system of linear equations
x + 2y + 3z = $$\alpha$$
4x + 5y + 6z = $$\beta$$
7x + 8y + 9z = $$\gamma $$ $$-$$ 1
is consistent. Let | M | represent the determinant of the matrix
$$M = \left[ {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right]$$
Let P be the plane containing all those ($$\alpha$$, $$\beta$$, $$\gamma$$) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P.
The value of | M | is _________.
x + 2y + 3z = $$\alpha$$
4x + 5y + 6z = $$\beta$$
7x + 8y + 9z = $$\gamma $$ $$-$$ 1
is consistent. Let | M | represent the determinant of the matrix
$$M = \left[ {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right]$$
Let P be the plane containing all those ($$\alpha$$, $$\beta$$, $$\gamma$$) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P.
The value of | M | is _________.
Answer
1
Explanation
$$7x + 8y + 9z - (\gamma - 1) = A(4x + 5y + 6z - \beta ) + B(x + 2y + 3z - \alpha )$$
On equating the coefficients,
4A + B = 7 .... (i)
5A + 2B = 8 .... (ii)
and $$-$$ ($$\gamma$$ $$-$$ 1) = $$-$$ A$$\beta$$ $$-$$ $$\alpha$$B ..... (iii)
On solving Eqs. (i) and (ii), we get A = 2 and B = $$-$$1
From Eq. (iii), we get
$$-$$ $$\gamma$$ + 1 = $$-$$ 2$$\beta$$ $$-$$ $$\alpha$$($$-$$1)
$$\Rightarrow$$ $$\alpha$$ $$-$$ 2$$\beta$$ + $$\gamma$$ = 1 ..... (iv)
Now, determinant of
$$M = \left| M \right| = \left| {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right| = \alpha - 2\beta + \gamma = 1$$ [from Eq. (iv)]
On equating the coefficients,
4A + B = 7 .... (i)
5A + 2B = 8 .... (ii)
and $$-$$ ($$\gamma$$ $$-$$ 1) = $$-$$ A$$\beta$$ $$-$$ $$\alpha$$B ..... (iii)
On solving Eqs. (i) and (ii), we get A = 2 and B = $$-$$1
From Eq. (iii), we get
$$-$$ $$\gamma$$ + 1 = $$-$$ 2$$\beta$$ $$-$$ $$\alpha$$($$-$$1)
$$\Rightarrow$$ $$\alpha$$ $$-$$ 2$$\beta$$ + $$\gamma$$ = 1 ..... (iv)
Now, determinant of
$$M = \left| M \right| = \left| {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right| = \alpha - 2\beta + \gamma = 1$$ [from Eq. (iv)]
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