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1

### IIT-JEE 2011 Paper 2 Offline

Let $$\omega$$ $$\ne$$ 1 be a cube root of unity and S be the set of all non-singular matrices of the form $$\left[ {\matrix{ 1 & a & b \cr \omega & 1 & c \cr {{\omega ^2}} & \omega & 1 \cr } } \right]$$, where each of a, b, and c is either $$\omega$$ or $$\omega$$2. Then the number of distinct matrices in the set S is

A
2
B
6
C
4
D
8

## Explanation

$$\left| A \right| \ne 0$$, as non-singular.

$$\therefore$$ $$\left| {\matrix{ 1 & a & b \cr \omega & 1 & c \cr {{\omega ^2}} & \omega & 1 \cr } } \right| \ne 0$$

$$\Rightarrow 1(1 - c\omega ) - a(\omega - c{\omega ^2}) + b({\omega ^2} - {\omega ^2}) \ne 0$$

$$\Rightarrow 1 - c\omega - a\omega + ac{\omega ^2} \ne 0$$

$$\Rightarrow (1 - c\omega )(1 - a\omega ) \ne 0$$

$$\Rightarrow a \ne {1 \over \omega },c \ne {1 \over \omega } \Rightarrow a = \omega ,c = \omega$$

and $$b \in \{ \omega ,{\omega ^2}\} \Rightarrow 2$$ solutions

2

### IIT-JEE 2011 Paper 1 Offline

Let a, b and c be three real numbers satisfying

$$[\matrix{ a & b & c \cr } ]\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right] = [\matrix{ 0 & 0 & 0 \cr } ]$$ ........ (E)

Let b = 6, with a and c satisfying (E). If $$\alpha$$ and $$\beta$$ are the roots of the quadratic equation ax2 + bx + c = 0, then $$\sum\limits_{n = 0}^\infty {{{\left( {{1 \over \alpha } + {1 \over \beta }} \right)}^n}}$$ is

A
6
B
7
C
$${6 \over 7}$$
D
$$\infty$$

## Explanation

Given, $${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$$

$$\Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

$$\Rightarrow a + 8b + 7c = 0$$ ..... (i)

$$\Rightarrow 9a + 2b + 3c = 0$$ .... (ii)

$$\Rightarrow a + b + c = 0$$ .... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$$7a + c = 0$$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$$6a - b = 0$$ ..... (v)

$$\therefore$$ b = 6a and c = $$-$$ 7a

If b = 6, a = 1 and c = $$-$$7

$$\therefore$$ $$a{x^2} + bx + c = 0$$

$$\Rightarrow {x^2} + 6x - 7 = 0$$

$$\Rightarrow (x + 7)(x - 1) = 0$$

$$\therefore$$ $$x = 1,7$$

$$\Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {{1 \over 1} - {1 \over 7}} \right)}^n} \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {{6 \over 7}} \right)}^n}} }$$

$$\Rightarrow 1 + {6 \over 7} + {\left( {{6 \over 7}} \right)^n} + ....\infty$$

$$= {1 \over {1 - {6 \over 7}}} = {1 \over {1/7}} = 7$$

3

### IIT-JEE 2011 Paper 1 Offline

Let a, b and c be three real numbers satisfying

$$[\matrix{ a & b & c \cr } ]\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right] = [\matrix{ 0 & 0 & 0 \cr } ]$$ ........(E)

Let $$\omega$$ be a solution of $${x^3} - 1 = 0$$ with $${\mathop{\rm Im}\nolimits} (\omega ) > 0$$. If a = 2 with b and c satisfying (E), then the value of $${3 \over {{\omega ^a}}} + {1 \over {{\omega ^b}}} + {3 \over {{\omega ^c}}}$$ is equal to

A
$$-$$2
B
2
C
3
D
$$-$$3

## Explanation

Given, $${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$$

$$\Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

$$\Rightarrow a + 8b + 7c = 0$$ ..... (i)

$$\Rightarrow 9a + 2b + 3c = 0$$ .... (ii)

$$\Rightarrow a + b + c = 0$$ .... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$$7a + c = 0$$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$$6a - b = 0$$ ..... (v)

$$\therefore$$ b = 6a and c = $$-$$ 7a

If a = 2, b = 12 and c = $$-$$14

$$\therefore$$ $${3 \over {{\omega ^a}}} + {1 \over {{\omega ^b}}} + {3 \over {{\omega ^c}}}$$

$$\Rightarrow {3 \over {{\omega ^2}}} + {1 \over {{\omega ^{12}}}} + {3 \over {{\omega ^{ - 14}}}} = {3 \over {{\omega ^2}}} + 1 + 3{\omega ^2}$$

$$= 3\omega + 1 + 3{\omega ^2}$$

$$= 1 + 3(\omega + {\omega ^2})$$

$$= 1 - 3 = - 2$$

4

### IIT-JEE 2011 Paper 1 Offline

Let a, b and c be three real numbers satisfying

$$[\matrix{ a & b & c \cr } ]\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right] = [\matrix{ 0 & 0 & 0 \cr } ]$$ .......(E)

If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is

A
0
B
12
C
7
D
6

## Explanation

Given, $${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$$

$$\Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

$$\Rightarrow a + 8b + 7c = 0$$ ..... (i)

$$\Rightarrow 9a + 2b + 3c = 0$$ .... (ii)

$$\Rightarrow a + b + c = 0$$ ....... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$$7a + c = 0$$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$$6a - b = 0$$ ..... (v)

$$\therefore$$ $$b = 6a$$ and $$c = - 7a$$

As (a, b, c) lies on $$2x + y + z = 1$$

$$\Rightarrow 2a + b + c = 1$$

$$\Rightarrow 2a + 6a - 7a = 1$$

$$\Rightarrow$$ a = 1, b = 6 and c = $$-$$7

$$\therefore$$ $$7a + b + c = 7 + 6 - 7 = 6$$

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