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1

AIEEE 2003

If the system of linear equations
$$x + 2ay + az = 0;$$ $$x + 3by + bz = 0;\,\,x + 4cy + cz = 0;$$
has a non - zero solution, then $$a, b, c$$.
A
satisfy $$a+2b+3c=0$$
B
are in A.P
C
are in G.P
D
are in H.P.

Explanation

For homogeneous system of equations to have non zero solution, $$\Delta = 0$$

$$\left| {\matrix{ 1 & {2a} & a \cr 1 & {3b} & b \cr 1 & {4c} & c \cr } } \right| = 0\,{C_2} \to {C_2} - 2{C_3}$$

$$\left| {\matrix{ 1 & 0 & a \cr 1 & b & b \cr 1 & {2c} & c \cr } } \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}$$

$$\left| {\matrix{ 1 & 0 & a \cr 0 & b & {b - a} \cr 0 & {2c - b} & {c - b} \cr } } \right| = 0$$

$$b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0$$

On simplification, $${2 \over b} = {1 \over a} + {1 \over c}$$

$$\therefore$$ $$a,b,c$$ are in Harmonic Progression.
2

AIEEE 2003

If $$1,$$ $$\omega ,{\omega ^2}$$ are the cube roots of unity, then

$$\Delta = \left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr } } \right|$$ is equal to

A
$${\omega ^2}$$
B
$$0$$
C
$$1$$
D
$$\omega$$

Explanation

$$\Delta = \left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr } } \right|$$

$$= 1\left( {{\omega ^{3n}} - 1} \right) - {\omega ^n}\left( {{\omega ^{2n}} - {\omega ^{2n}}} \right) +$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^{2n}}\left( {{\omega ^n} - {\omega ^{4n}}} \right)$$

$$= {\omega ^{3n}} - 1 - 0 + {\omega ^{3n}} - {\omega ^{6n}}$$

$$= 1 - 1 + 1 - 1 = 0$$ $$\left[ {} \right.$$ as $$\,\,\,\,\,$$ $${\omega ^{3n}} = 1$$ $$\left. {} \right]$$
3

AIEEE 2003

If $$A = \left[ {\matrix{ a & b \cr b & a \cr } } \right]$$ and $${A^2} = \left[ {\matrix{ \alpha & \beta \cr \beta & \alpha \cr } } \right]$$, then
A
$$\alpha = 2ab,\,\beta = {a^2} + {b^2}$$
B
$$\alpha = {a^2} + {b^2},\,\beta = ab$$
C
$$\alpha = {a^2} + {b^2},\,\beta = 2ab$$
D
$$\alpha = {a^2} + {b^2},\,\beta = {a^2} - {b^2}$$

Explanation

$${A^2} = \left[ {\matrix{ \alpha & \beta \cr \beta & \alpha \cr } } \right] = \left[ {\matrix{ a & b \cr b & a \cr } } \right]\left[ {\matrix{ a & b \cr b & a \cr } } \right]$$

$$= \left[ {\matrix{ {{a^2} + {b^2}} & {2ab} \cr {2ab} & {{a^2} + {b^2}} \cr } } \right]$$

$$\alpha = {a^2} + {b^2};\,\,\beta = 2ab$$
4

AIEEE 2002

If $$a>0$$ and discriminant of $$\,a{x^2} + 2bx + c$$ is $$-ve$$, then
$$\left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr {ax + b} & {bx + c} & 0 \cr } } \right|$$ is equal to
A
$$+ve$$
B
$$\left( {ac - {b^2}} \right)\left( {a{x^2} + 2bx + c} \right)$$
C
$$-ve$$
D
$$0$$

Explanation

We have $$\left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr {ax + b} & {bx + c} & 0 \cr } } \right|$$

By $$\,\,\,{R_3} \to {R_3} - \left( {x{R_1} + {R_2}} \right);$$

$$= \left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr 0 & 0 & { - \left( {a{x^2} + 2bx + C} \right)} \cr } } \right|$$

$$= \left( {a{x^2} + 2bx + c} \right)\left( {{b^2} - ac} \right)$$

$$= \left( + \right)\left( - \right) = - ve.$$

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