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Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Let $$A$$ be a square matrix all of whose entries are integers.

Then which one of the following is true?

Then which one of the following is true?

A

If det $$A = \pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are not necessarily integers

B

If det $$A \ne \pm 1,$$ then $${A^{ - 1}}$$ exists and all its entries are non integers

C

If det $$A = \pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are integers

D

If det $$A = \pm 1,$$ then $${A^{ - 1}}$$ need not exists

As all entries of square matrix $$A$$ are integers, therefore all co-factors should also be integers.

If det $$A = \pm 1\,\,$$ then $${A^{ - 1}}\,\,$$ exists. Also all entries of $${A^{ - 1}}$$ are integers.

If det $$A = \pm 1\,\,$$ then $${A^{ - 1}}\,\,$$ exists. Also all entries of $${A^{ - 1}}$$ are integers.

2

MCQ (Single Correct Answer)

Let $$a, b, c$$ be any real numbers. Suppose that there are real numbers $$x, y, z$$ not all zero such that $$x=cy+bz,$$ $$y=az+cx,$$ and $$z=bx+ay.$$ Then $${a^2} + {b^2} + {c^2} + 2abc$$ is equal to

A

$$2$$

B

$$-1$$

C

$$0$$

D

$$1$$

The given equations are

$$\matrix{ { - x + cy + bz = 0} \cr {cx - y + az = 0} \cr {bx + ay - z = 0} \cr } $$

As $$x,y,z$$ are not all zero

$$\therefore$$ The above system should not have unique (zero) solution

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

$$ \Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$$

$$ \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$$

$$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$$

$$\matrix{ { - x + cy + bz = 0} \cr {cx - y + az = 0} \cr {bx + ay - z = 0} \cr } $$

As $$x,y,z$$ are not all zero

$$\therefore$$ The above system should not have unique (zero) solution

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

$$ \Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$$

$$ \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$$

$$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$$

3

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.

4

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 $$y$$

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Complex Numbers

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Definite Integrals and Applications of Integrals

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