Joint Entrance Examination

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1

MCQ (Single Correct Answer)

Let $$A = \left( {\matrix{
1 & { - 1} & 1 \cr
2 & 1 & { - 3} \cr
1 & 1 & 1 \cr
} } \right).$$ and $$10$$ $$B = \left( {\matrix{
4 & 2 & 2 \cr
{ - 5} & 0 & \alpha \cr
1 & { - 2} & 3 \cr
} } \right)$$. if $$B$$ is

the inverse of matrix $$A$$, then $$\alpha $$ is

A

$$5$$

B

$$-1$$

C

$$2$$

D

$$-2$$

Given that $$10B$$ $$\,\,\, = \left[ {\matrix{
4 & 2 & 2 \cr
{ - 5} & 0 & \alpha \cr
1 & { - 2} & 3 \cr
} } \right]$$

$$ \Rightarrow B = {1 \over {10}}\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$$

Also since, $$B = {A^{ - 1}} \Rightarrow AB = I$$

$$ \Rightarrow {1 \over {10}}\left[ {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$

$$ \Rightarrow {1 \over {10}}\left[ {\matrix{ {10} & 0 & {5 - 2} \cr 0 & {10} & { - 5 + \alpha } \cr 0 & 0 & {5 + \alpha } \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$

$$ \Rightarrow {{5 - \alpha } \over {10}} = 0$$

$$ \Rightarrow \alpha = 5$$

$$ \Rightarrow B = {1 \over {10}}\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$$

Also since, $$B = {A^{ - 1}} \Rightarrow AB = I$$

$$ \Rightarrow {1 \over {10}}\left[ {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$

$$ \Rightarrow {1 \over {10}}\left[ {\matrix{ {10} & 0 & {5 - 2} \cr 0 & {10} & { - 5 + \alpha } \cr 0 & 0 & {5 + \alpha } \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$

$$ \Rightarrow {{5 - \alpha } \over {10}} = 0$$

$$ \Rightarrow \alpha = 5$$

2

MCQ (Single Correct Answer)

If the system of linear equations

$$x + 2ay + az = 0;$$ $$x + 3by + bz = 0;\,\,x + 4cy + cz = 0;$$

has a non - zero solution, then $$a, b, c$$.

$$x + 2ay + az = 0;$$ $$x + 3by + bz = 0;\,\,x + 4cy + cz = 0;$$

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.

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.

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

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

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

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

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

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

3

MCQ (Single Correct Answer)

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

$$\omega $$

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

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

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

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

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

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

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

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

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

4

MCQ (Single Correct Answer)

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

$$\alpha = {a^2} + {b^2},\,\beta = {a^2} - {b^2}$$

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

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

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

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

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

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Complex Numbers

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