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**Q.1. If the system of linear equations2x + 2ay + az = 02x + 3by + bz = 02x + 4cy + cz = 0,**

(1) are in A.P.

(2) a, b, c are in G.P.

(3) a + b + c = 0

(4) a, b, c are in A.P.

2x + 2ay + az = 0 (1)

2x + 3by + bz = 0 (2)

2x + 4cy + cz = 0 (3)

The system of linear equations has a non- zero solution, then

(R

â‡’ (3b - 2a) (c -a) - (b-a) (4c - 2a) = 0

â‡’

Hence, are in A.P.

We have b

Now,

x + y + z = 6

x + 2y + 3z = 10

3x + 2y + Î»z = Î¼

The system of linear equations is

x + y + z = 6 (1)

x + 2y + 3z = 10 (2)

3x + 2y + Î»z = Î¼ (3)

The system of equations has more than two solutions, then

â‡’1(2Î» - 6) - 1 (Î» - 9) + 1(2 - 6) = 0

â‡’ 2Î» - 6 - 1 Î» - 9 - 4 = 0 â‡’ Î» = 1

Now,

â‡’ 6(2 - 3) - 1(10 - 3Î¼) + 1(20 - 20Î¼) = 0

â‡’ -6 -10 + 3Î¼ + 20 -2Î¼ = 0 â‡’ Î¼ = 14

Hence, Î¼ - Î»

x + 2y + 3z = 1

3x + 4y + 5z = Î¼

4x + 4y + 4z = Î´

is inconsistent? (2020)

(1) (4, 3)

(2) (4, 6)

(3) (1, 0)

(4) (3, 4)

The system of linear equation is

x + 2y + 3z = 1 (1)

3x + 4y + 5z = Î¼ (2)

4x + 4y + 4z = Î´ (3)

On solving Eqs. (1), (2) and (3), we get

2(3x + 4y + 5z) - 2 (x + 2y + 3z) - 4x + 4y + 4z = 2Î¼ - 2 - Î´

â‡’ 2Î¼ - 2 - Î´ = 0 â‡’ Î´ = 2 (Î¼ - 1)

From the given options only option (4,3) does not satisfies the above equation. So, the system of linear equations is inconsistent for the pair (4,3).

(1) A â€“ 4I

(2) 6I â€“ A

(3) A â€“ 6I

(4) 4I â€“ A

Given,

Now,

Therefore,

Î»x + 2y + 2z = 5

2Î»x + 3y + 5z = 8

4x + Î»y + 6z = 10 has

(1) no solution when Î» = 8

(2) a unique solution when Î» = -8

(3) no solution when Î» = 2

(4) infinitely many solutions when Î» = 2

The system of linear equations is

Î»x + 2y + 2z = 5 (1)

2Î»x + 3y + 5z = 8 (2)

4x + Î»y + 6z = 10 (3)

Therefore,

= Î»

For no solution, D = 0 â‡’ Î» = 2.

Now,

Hence, the system of linear equations has no solution for Î» = 2.

(1) 8

(2) 16

(3) 72

(4) 2

From the properties of determinants of a square matrix of order n, we have

(1)

Now,

Hence, from Eq. (1), we get

7x + 6y - 2z = 0

3x + 4y + 2z = 0

x - 2y - 6z = 0, has

(1) Infinitely many solutions, (x, y, z) satisfying y = 2z.

(2) no solution.

(3) infinitely many solutions, (x, y, z) satisfying x = 2z.

(4) only the trivial solution.

The system of linear equation is

7x + 6y - 2z = 0 (1)

3x + 4y + 2z = 0 (2)

x - 2y - 6z = 0 (3)

Therefore, the coefficient matrix is given by

= 7(-24 + 4) -6(-18-2) -2(-6-4)

= -140 + 120 + 20 = 0

Also, Î”

Hence, for the given system of equation infinite many solutions are possible.

From Eqs. (1) and (3), we get

7x + 6y - 2z + 3(x - 2y - 6z) = 0 â‡’ x = 2z

Given,

R

â‡’ f (x) = {(x+3)

= x

Hence, f (50) = 1

Î± = Ï‰ , Ï‰

Now,

Hence, A

The trace of matrix AA

Hence, the number of such matrices is

2x + 3y + 2z = 5

2x + 3y + (a

x+y + z = 2 ...(1)

2x + 3y + 2z = 5 ...(2)

2x + 3y + (a

Now,

â‡’

(Applying R

= a

When, A = 0 â‡’ a

If a

Hence, the given system of equations is inconsistent.

x - 4y + 7z = g

3y - 5z = h

- 2x + 5y - 9z = k

is consistent, then:

Consider the system of linear equations

x - 4y + 7z = g ...(i)

3y - 5z = h ... (ii)

-2x + 5y - 9z = k ...(iii)

Multiply equation (i) by 2 and add equation (i), equation (ii) and equation (iii)

â‡’ 0 = 2g + h + k. âˆ´ 2g + h + k = 0

then system of equations is

âˆ´ A is invertible.

x + y + z = 5

x + 2y + 3z = 9

x + 3y + Î±z = Î²

(1) 21

(2) 8

(3) 18

(4) 5

Since, the system of equations has infinite many solutions. Hence,

x + 3y + 7z = 0

- x + 4y + 7z = 0

(sin 3Î¸)x + (cos 2Î¸)y + 2z = 0

has a non-trivial solution, is:

(1) three

(2) two

(3) four

(4) one

Since, the system of linear equations has non-trivial solution then determinant of coefficient matrix = 0

sin3Î¸(21 - 28) - cos2Î¸(7 + 7) + 2 (4 + 3) = 0

sin3Î¸ + 2cos2Î¸ - 2 = 0

3sinÎ¸ - 4sin

4sin

sinÎ¸ (4sin

sinÎ¸ (4sin

sinÎ¸ [2sinÎ¸ (2sinÎ¸ - 1) + 3 (2sinÎ¸ - 1)] = 0

sinÎ¸ (2sinÎ¸ - 1) (2sinÎ¸ + 3) = 0

Hence, for

= 2(2b

= 2b

Using A.Mâ‰¥G.M,**Q.19. ****If the system of linear equations (2019)2x + 2y + 3z = a 3x - y + 5z = b x - 3y + 2z = c**

(1) b - c + a = 0

(2) b - c - a = 0

(3) a + b + c = 0

(4) b + c- a = 0

âˆµ System of equations has more than one solution

âˆ´ Î” = Î”

(2) 1/âˆš3

(3) 1/âˆš2

(4) 1/âˆš6

(1) abc

(2) -(a + b + c)

(3) 2(a + b + c)

(4) -2(a + b + c)

(1) 1/4

(2) 1

(3) 1/16

(4) 16

ax + (1 + Î²)y + z = 3

Î±x+ Î²y+ 2z = 2

has a unique solution, is : (2019)

(2) (-3, 1)

(3) (-4, 2)

(4) (1, -3)

Solution.

(1) is a singleton

(2) contains exactly two elements

(3) is an empty set

(4) contains more than two elements

Consider the given system of linear equations

Now, for a non-trivial solution, the determinant of coefficient matrix is zero.

â‡’ (1 - Î»)

Î» = 1

x - cy - cz = 0

cx - y + cz = 0

cx + cy - z = 0

(1) -1

(2) 1/2

(3) 2

(4) 0

If the system of equations has non-trivial solutions, then the determinant of coefficient matrix is zero

Hence, the greatest value of c is 1/2 for which the system of linear equations has non-trivial solution.

(3)

2x + y + z = 2

3x - y - kz = 3

(1) 3x - 4y - 1 = 0

(2) 4x - 3y - 4 = 0

(3) 4x - 3y - 1 = 0

(4) 3x - 4y - 4 = 0

Given system of linear equations,

âˆ´ System of equation has infinite many solutions.

(2) y(y

(3) y

(4) y

Let Î± = Ï‰ and Î² = Ï‰

(1) 2

(2) 3

(3) 6

(4) 4

Number of combinations of (x, y) = 2 x 2 = 4**Q.34. If the system of equations 2x + 3y - z = 0, x + ky - 2z - 0 and 2x - y + z = 0 has a non-trivial solution (x, y, z), then (2019)****(1) 3/4****(2) 1/2****(4) -4****Ans. **(2)**Solution.**

Given system of equations has a non-trivial solution.

âˆ´ equations are 2x + 3y - z = 0 ...(i)

2x - y + z = 0 ... (ii)

2x + 9y - 4z = 0 ...(iii)**Q.35. **

** (2019)****(1) Î” _{1} - Î”_{2} = -2x^{3}**

(2) Î”_{1} - Î”_{2} = x(cos2Î¸ - cos4Î¸)

(3) Î”_{1} x Î”_{2} = -2(x^{3} + x - 1)

(4) Î”_{1} + Î”_{2} = -2x

Solution.

= x (- x

= - x

= -x

Similarly, Î”

x + y + z = 5

x + 2y + 2z = 6

(1) 12

(2) 9

(3) 7

(4) 10

Given system of linear equations: x + y + z = 5; x + 2y + 2z = 6 and x + 3y + Î»z = Î¼ have infinite solution

âˆ´ for

On expanding,

x (- 3x

â‡’ x (- 5x

â‡’ x

âˆµ all the roots are real.

âˆ´ sum of real roots = 0/1 = 0

On comparing each term,

(1) (-4, -5)

(2) (-4, 3)

(3) (-4, 5)

(4) (4, 5)

B = 5

x + ky + 3z = 0

3x + ky - 2z = 0

2x + 4y - 3z = 0

has a non-zero solution (x, y, z), then xz/y

(1) -10

(2) 10

(3) -30

(4) 30

Ans:

Solution.

hence equations are x + 11y + 3z = 0

3x + 11y - 2z = 0

and 2x + 4y - 3z = 0

let z = t

x+ y + z = 2

2x + y - z = 3

3x + 2y + kz = 4

has a unique solution.

(1) S equal to {0}

(2) equal to R-{0}

(3) an empty set

(4) equal to R

Therefore, set S = equal to R-{0}

(2)

(3)

(4)

Ans:

Solution:

âˆ´ c = 0, 2a + 3b = 0, a = 2c + 3d a = 3d

âˆ´a

|3A| = 108

Hence, option 3 is the answer.

(1) 210

(2) 211

(3) 251

(4) 231

Ans:

Solution:

Sum of the elements of first column =

x +y+ z = 1

x +ay+ z = 1

ax + by+ z = 0

has no solution, then S is

(1) A singleton

(2) An empty set

(3) An infinite set

(4) A finite set containing two or more elements

Ans.

Solution

â‡’ â€“(1 â€“ a)2 = 0

â‡’ a = 1

For a = 1

Eq. (1) & (2) are identical i.e.,x + y + z = 1

To have no solution with x + by + z = 0

(1)

(2)

(3)

(4)

Ans.

Solution.

= (2 â€“ 2Î»- Î» + Î»

f (Î»)= Î»

âˆµ A satisfies f (Î»)

âˆ´ A

A

3A

3A

(2) 1

(3) 2

(4) 0

It will give only one real value of Î»

(1) 10A + 5B = 3I

(3) 5A + 10B = 2I

(4) B + 2A = I

**Q.50. If A = and A adj A = A A ^{T}, then 5a + b is equal to: (2016)(1) -1(2) 5(3) 4(4) 13Ans.** (2)

Equate, 10a + 3b = 25a

& 10a + 3b = 13

& 15a - 2b = 0

a/2 = b/15 = k (let)

Solving a = 2/5, b = 3

So, 5a + b = 5 x 2/5 + 3 = 5

x + Î»y - z = 0

Î»x - y - z = 0

x + y - Î»z = 0

has a non-trivial solution for: (2016)

(1) infinitely many values of Î»

(2) exactly one value of Î»

(3) exactly two values of Î»

(4) exactly three values of Î»

Ans.

For trivial solution,

(1) 4

(2) 1

(3) 2

(4) 3

Ans.

tanx = 1 â‡’ x = Ï€/4

(1)

(2)

(3)

(4)

Ans.

â‡’ A

â‡’

Statement - I: A

Statement - II : The polynomial A

Then (2016)

(1) Statement-I is false, but Statement-II is true.

(2) Both the statements are false.

(3) Both the statements are true.

(4) Statement-I is true, but Statement-II is false.

Ans.

Hence, statement 1 is true

Now A

Statement 2 is also correct

(1) 2014

(2) 2016

(3) -175

(4) -25

Ans.

â‡’ A

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