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1

MCQ (Single Correct Answer)

Let $$A$$ be $$a\,2 \times 2$$ matrix with real entries. Let $$I$$ be the $$2 \times 2$$ identity matrix. Denote by tr$$(A)$$, the sum of diagonal entries of $$a$$. Assume that $${a^2} = I.$$

**Statement-1 :** If $$A \ne I$$ and $$A \ne - I$$, then det$$(A)=-1$$

**Statement- 2 :** If $$A \ne I$$ and $$A \ne - I$$, then tr $$(A)$$ $$ \ne 0$$.

A

statement - 1 is false, statement -2 is true

B

statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1.

C

statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1.

D

statement - 1 is true, statement - 2 is false.

Let $$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$ $$\,\,\,$$ then $${A^2} = 1$$

$$ \Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0$$

$$ac + cd = 0\,\,\,\,bc + {d^2} = 1$$

From these four relations,

$${a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}$$

and $$\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = - d$$

We can take $$a = 1,b = 0,c = 0,d = - 1$$

as one possible set of values, then

$$A = \left[ {\matrix{ 1 & 0 \cr 0 & { - 1} \cr } } \right]$$

Clearly $$A \ne I\,\,\,$$ and $$\,\,\,\,A \ne - I\,\,$$ and $$\,\,\,A = - 1$$

$$\therefore$$ $$\,\,\,\,\,$$ Statement $$1$$ is true.

Also if $$A \ne I\,\,\,\,\,tr\left( A \right) = 0$$

$$\therefore$$ $$\,\,\,\,\,$$ Statement $$2$$ is false.

$$ \Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0$$

$$ac + cd = 0\,\,\,\,bc + {d^2} = 1$$

From these four relations,

$${a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}$$

and $$\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = - d$$

We can take $$a = 1,b = 0,c = 0,d = - 1$$

as one possible set of values, then

$$A = \left[ {\matrix{ 1 & 0 \cr 0 & { - 1} \cr } } \right]$$

Clearly $$A \ne I\,\,\,$$ and $$\,\,\,\,A \ne - I\,\,$$ and $$\,\,\,A = - 1$$

$$\therefore$$ $$\,\,\,\,\,$$ Statement $$1$$ is true.

Also if $$A \ne I\,\,\,\,\,tr\left( A \right) = 0$$

$$\therefore$$ $$\,\,\,\,\,$$ Statement $$2$$ is false.

2

MCQ (Single Correct Answer)

If$$D = \left| {\matrix{
1 & 1 & 1 \cr
1 & {1 + x} & 1 \cr
1 & 1 & {1 + y} \cr
} } \right|$$ for $$x \ne 0,y \ne 0,$$ then $$D$$ is

A

divisible by $$x$$ but not $$y$$

B

divisible by $$y$$ but not $$x$$

C

divisible by neither $$x$$ nor $$y$$

D

divisible by both $$x$$ and $$y$$

Given, $$D = \left| {\matrix{
1 & 1 & 1 \cr
1 & {1 + x} & 1 \cr
1 & 1 & {1 + y} \cr
} } \right|$$

Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$

and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$

$$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$$

Hence, $$D$$ is divisible by both $$x$$ and $$y$$

Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$

and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$

$$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$$

Hence, $$D$$ is divisible by both $$x$$ and $$y$$

3

MCQ (Single Correct Answer)

Let $$A = \left| {\matrix{
5 & {5\alpha } & \alpha \cr
0 & \alpha & {5\alpha } \cr
0 & 0 & 5 \cr
} } \right|.$$ If $$\,\,\left| {{A^2}} \right| = 25,$$ then $$\,\left| \alpha \right|$$ equals

A

$$1/5$$

B

$$5$$

C

$${5^2}$$

D

$$1$$

$$\left| {{A^2}} \right| = 25 \Rightarrow {\left| A \right|^2} = 25$$

$$ \Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}$$

$$ \Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}$$

4

MCQ (Single Correct Answer)

Let $$A = \left( {\matrix{
1 & 2 \cr
3 & 4 \cr
} } \right)$$ and $$B = \left( {\matrix{
a & 0 \cr
0 & b \cr
} } \right),a,b \in N.$$ Then

A

there cannot exist any $$B$$ such that $$AB=BA$$

B

there exist more then one but finite number of $$B'$$s such that $$AB=BA$$

C

there exists exactly one $$B$$ such that $$AB=BA$$

D

there exist infinitely many $$B'$$s such that $$AB=BA$$

$$A = \left[ {\matrix{
1 & 2 \cr
3 & 4 \cr
} } \right]\,\,\,\,B = \left[ {\matrix{
a & 0 \cr
0 & b \cr
} } \right]$$

$$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$$

$$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$$

Hence, $$AB=BA$$ only when $$a=b$$

$$\therefore$$ There can be infinitely many $$B's$$

for which $$AB=BA$$

$$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$$

$$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$$

Hence, $$AB=BA$$ only when $$a=b$$

$$\therefore$$ There can be infinitely many $$B's$$

for which $$AB=BA$$

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