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Test: System of Linear Equations - Class 10 MCQ


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15 Questions MCQ Test The Complete SAT Course - Test: System of Linear Equations

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Test: System of Linear Equations - Question 1

If the system

2x – y + 3z = 2

x + y + 2z = 2

5x – y + az = b

Has infinitely many solutions, then the values of a and b, respectively, are

Detailed Solution for Test: System of Linear Equations - Question 1

Concept:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = b1

a21 x1 + a22 x2 + … + a2n xn = b2

am1 x1 + am2 x2 + … + amn xn = bm

The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of following matrices.

A is the coefficient matrix and [A|B] is called as augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

  • If the rank of matrix A is equal to the rank of augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution, i.e.
    Rank of A = Rank of augmented matrix = n
  • If the rank of matrix A is equal to the rank of augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.
    Rank of A = Rank of augmented matrix < n
  • If the rank of matrix A is not equal to the rank of the augmented matrix, then the system is inconsistent, and it has no solution.
    Rank of A ≠ Rank of augmented matrix

Calculation:

Given linear system is

2x – y + 3z = 2

x + y + 2z = 2

5x – y + az = b

Then augmented matrix form is written below;

For rank (A) < n = 3

‘a’ must be = 8

For rank [A|B] < 3, b = 6

Therefore a = 8 & b = 6

Test: System of Linear Equations - Question 2

For what value of μ do the simultaneous equations 5x + 7y = 2, 15x + 21y = μ have no solution?

Detailed Solution for Test: System of Linear Equations - Question 2

Concept:

System of equations

a1x + b1y = c1

a2x + b2y = c2

For unique solution

For Infinite solution

For no solution

Calculation:

Given:

5x + 7y = 2, 15x + 21y = μ

Here For no solution

Hence for no solution μ ≠ 6

*Answer can only contain numeric values
Test: System of Linear Equations - Question 3

Gauss-Seidel method is used to solve the following equations (as per the given order):

x1 + 2x2 + 3x3 = 5

2x1 + 3x2 + x3 = 1

3x1 + 2x2 + x3 = 3

Assuming initial guess as x1 = x2 = x3 = 0, the value of x3 after the first iteration is ________


Detailed Solution for Test: System of Linear Equations - Question 3

Gauss Seidel Method:

In Gauss Seidel method, the value of x calculated is used in next calculation putting other variable as 0.

x1 + 2x2 + 3x3 = 5

Putting x2 = 0, x3 = 0 ⇒ x1 = 5

2x1 + 3x2 + x3 = 1

Putting x1 = 5, x3 = 0 ⇒ x2 = -3

3x1 + 2x2 + x3 = 3

Putting x1 = 5, x2 = -3 ⇒ x3 = 3 – 3(5) – 2 (-3)

x3 = 3 – 15 + 6

x3 = -6

Mistake Point: Don’t arrange them diagonally because It is given in question solve as per given order.

Test: System of Linear Equations - Question 4

The value of k, for which the following system of linear equations has a non-trivial solution.

x + 2y - 3z = 0

2x + y + z = 0

x - y + kz = 0

Detailed Solution for Test: System of Linear Equations - Question 4

Concept:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = 0

a21 x1 + a22 x2 + … + a2n xn = 0

am1 x1 + am2 x2 + … + amn xn = 0

The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrix.

A is the coefficient matrix of the given system of equations.

  • The system of homogeneous equations has a unique solution (trivial solution) if and only if the determinant of A is non-zero.
  • The system of homogeneous equations has a Non - trivial solution if and only if the determinant of A is zero.

Calculation:

Given:

x + 2y – 3z = 0

2x + y + z = 0

x – y + kz = 0

For non-trivial solution the determinant should be zero

∴ 1(k + 1) – 2(2k - 1) – 3(-2 - 1) = 0

∴ k + 1 – 4k + 2 + 9 = 0

∴ 12 = 3k

∴ k = 4

Test: System of Linear Equations - Question 5

For what value of λ, do the simultaneous equation 2x + 3y = 1, 4x + 6y = λ have infinite solutions?

Detailed Solution for Test: System of Linear Equations - Question 5

Concept:

Non-homogeneous equation of type AX = B has infinite solutions;

if ρ(A | B) = ρ(A) < Number of unknowns

Calculation:

Given:

2x + 3y = 1

4x + 6y = λ

The augmented matrix is given by:

For the system to have infinite solutions, the last row must be a fully zero row.

So if λ = 2 then the system of equations has infinitely many solutions.

Key Points:

Remember the system of equations

AX = B have

1. Unique solution, if ρ(A : B) = ρ(A) = Number of unknowns.

2. Infinite many solutions, if ρ(A : B) = ρ(A) < Number of solutions

3. No solution, if ρ(A : B) ≠ ρ(A).

Test: System of Linear Equations - Question 6

For what value of k, the system linear equation has no solution

(3k + 1)x + 3y - 2 = 0

(k2 + 1)x + (k - 2)y - 5 = 0

Detailed Solution for Test: System of Linear Equations - Question 6

Given:

a1 = 3k + 1

b1 = 3

c1 = -2

a2 = k2 + 1

b2 = k - 2

c2 = -5

Formula Used:

Calculation: 

By cross multiplication

⇒ (3k + 1)(k - 2) = 3(k2 + 1)

⇒ 3k2 - 6k + k - 2 = 3k2 + 3

⇒ -5k - 2 = 3

⇒ -5k = 5

∴ k = -1 

The correct option is 2 i.e. -1

Test: System of Linear Equations - Question 7

The value of k for which the system of equations x + ky + 3z = 0, 4x + 3y + kz = 0, 2x + y + 2z = 0 has non-trivial solution is

Detailed Solution for Test: System of Linear Equations - Question 7

Concept:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = 0

a21 x1 + a22 x2 + … + a2n xn = 0

am1 x1 + am2 x2 + … + amn xn = 0

  • The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrix.

  • A is the coefficient matrix of the given system of equations.

  • Where, Ax, Ay, Az is the coefficient matrix of the given system of equations after replacing the first, second, and third columns from the constant term column which will be having all the entries as 0.
  • In the case of homogeneous equations, the determinants of, Ax, Ay, Az will be 0 definitely.
  • So, for the system of homogeneous equations having the the-trivial solution, the determinant of A should be zero.
  • The system of homogeneous equations has a unique solution (trivial solution) if and only if the determinant of A is non-zero.

Calculation:
For non - trivial solution, the |A| = 0

⇒ 1(6 - K) - K(8 - 2K) + 3(4 - 6) = 0

⇒ 9K -  2K2 = 0

⇒ k = 0 or 9/2

*Answer can only contain numeric values
Test: System of Linear Equations - Question 8

Consider matrix  The number of distinct real values of k for which the equation Ax = 0 has infinitely many solution is________


Detailed Solution for Test: System of Linear Equations - Question 8

Concept:

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to the rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

The rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to the rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

The rank of A = Rank of augmented matrix < n

Then |A| = 0

(iii) If the rank of matrix A is not equal to the rank of the augmented matrix, then the system is inconsistent, and it has no solution.

The rank of A ≠ Rank of an augmented matrix

Application:

A system to have infinitely many solutions must satisfy:

|A| = 0

K(K – 2(K – 1) = 0

K(K – 2K + 2) = 0

K(-K + 2) = 0

K = 0, 0, 2

Hence, there are 3 eigen values, and two distinct eigen value and 1 repeated eigen value.

Test: System of Linear Equations - Question 9

The set of equations

x + y + z = 1

ax – ay + 3z = 5

5x – 3y + az = 6

has infinite solutions, if a =

Detailed Solution for Test: System of Linear Equations - Question 9

Concept:

Non-homogeneous equation of type AX = B has infinite solutions if ρ(A : B) = ρ(A) < Number of unknowns

Calculation:

Given set of equations

x + y + z = 1

ax – ay + 3z = 5

5x – 3y + az = 6

a2 – a – 12 = 0

a2 – 4a + 3a – 12 = 0

a(a - 4) + 3(a - 4) = 0

(a - 4)(a + 3) = 0

A = 4, -3

When a = 4, then ρ(A : B) = ρ(A) = 2 < 3

Hence, given system of equations have infinite solutions when a = 4.

Note: here a = -3 we cannot consider because for a = -3  ρ(A : B) ≠  ρ(A) 

Key Points:

Remember the system of equations

AX = B have

1. Unique solution, if ρ(A : B) = ρ(A) = Number of unknowns.

2. Infinite many solutions, if ρ(A : B) = ρ(A) <  Number of unknowns

3. No solution, if ρ(A : B) ≠ ρ(A).

*Answer can only contain numeric values
Test: System of Linear Equations - Question 10

Consider the system of equations  The value of x3 (round off to the nearest integer), is ______.


Detailed Solution for Test: System of Linear Equations - Question 10

The given system has 4 equations and 3 unknowns.

Hence it is a over-determined system of equations.

The equations are;

x1 + 3x2 + 2x3 = 1       --(1)

2x1 + 2x2 - 3x3 = 1     ---(2)

4x1 + 4x2 - 6x3 = 2     ---(3)

2x1 + 5x2 + 2x3 = 1     ---(4)

Note that equation (2) and (3) are linearly dependent on each other and equation 3 is twice that of equation (2).

Hence considering equation 1, 2 and 4 -


⇒ x3 = 3

Important Points:

  • Under-determined system: Number of equations < Number of unknowns
  • Over-Determined system: Number of equation > Number of unknowns
  • Equally Determined system: Number of equation = Number of unknowns
Test: System of Linear Equations - Question 11

The approximate solution of the system of simultaneous equations

2x - 5y + 3z = 7

x + 4y - 2z = 3

2x + 3y + z = 2

by applying Gauss-Seidel method one time (using initial approximation as x - 0, y - 0, z - 0) will be:

Detailed Solution for Test: System of Linear Equations - Question 11

Gauss Seidel Method:

In Gauss Seidel method, the value of x calculated is used in next calculation putting other variable as 0.

2x - 5y + 3z = 7

Putting y = 0, z = 0 ⇒ x = 3.5

x + 4y - 2z = 3

Putting x = 3.5, z = 0 ⇒ y = - 0.125

2x + 3y + z = 2

Putting x = 3.5, y = - 0.125 ⇒ z = 2 – 3(-0.125) – 2(3.5)

z = - 4.625

Test: System of Linear Equations - Question 12

A set of linear equations is given in the form Ax = b, where A is a 2 × 4 matrix with real number entries and b ≠ 0. Will it be possible to solve for x and obtain a unique solution by multiplying both left and right sides of the equation by AT (the super script T denotes the transpose) and inverting the matrix AT A?

Detailed Solution for Test: System of Linear Equations - Question 12

Concept:

From the properties of a matrix,

The rank of m × n matrix is always ≤ min {m, n}

If the rank of matrix A is ρ(A) and rank of matrix B is ρ(B), then the rank of matrix AB is given by

ρ(AB) ≤ min {ρ(A), ρ(B)}

If n × n matrix is singular, the rank will be less than ≤ n

Calculation:

Given:

AX = B

Where A is 2 × 4 matrices and b ≠ 0

The order of AT is 4 × 2

The order of ATA is 4 × 4

Rank of (A) ≤ min (2, 4) = 2

Rank of (AT) ≤ min (2, 4) = 2

Rank (ATA) ≤ min (2, 2) = 2

As the matrix ATA is of order 4 × 4, to have a unique solution the rank of ATA should be 4.

Therefore, the unique solution of this equation is not possible.

Test: System of Linear Equations - Question 13

If  then which one of the following is correct?

Detailed Solution for Test: System of Linear Equations - Question 13

(1 - λ) (-3λ + λ2 + 2) - 3(6 - 2λ + 1) + 2(4 + λ) = 0

(1 - λ) (λ2 - 3λ + 2) - 3(7 - 2λ) + 2(4 + λ) = 0

λ3 - 4λ2 - 3λ + 11 = 0

By C-H theorem replace λ by A

A3 - 4A2 - 3A + 11I = 0

Test: System of Linear Equations - Question 14

Consider a matrix  
The matrix A satisfies the equation 6A-1 = A2 + cA + dI, where c and d are scalars and I is the identity matrix. Then (c + d) is equal to

Detailed Solution for Test: System of Linear Equations - Question 14

Concept:

for the given square matrix, the characteristic equation will be

|B - AI| = 0

B = Given matrix

I = Unit matrix

A = Characteristic roots

Calculation:

|B - AI| = 0

Take the determinant of matrix, then 

(1 - A) [(4 - A) (1 - A) + 2] = 0

(1 - A) [4 - 4A - A + A2 + 2] = 0

(1 - A) [4 - 5A + A2 + 2] = 0

(1 - A) [A2 - 5A + 6] = 0

A2 - 5A + 6 - A3 + 5A2 - 6A = 0

-A3 + 6A2 - 11A + 6 = 0

A3 - 6A2 + 11A = 6

A2 - 6A + 11 = 6A-1       ........(1)

Given 6A-1 = A2 + cA + dI     .........(2)

Compare 1 and 2

c = -6, d = +11

c + d = +5

Test: System of Linear Equations - Question 15

The system of linear equations 

-y + z = 0

(4d - 1) x + y + Z = 0

(4d - 1) z = 0 

has a non-trivial solution, if d equals 

Detailed Solution for Test: System of Linear Equations - Question 15

Concept

For a homogeneous system of linear equations:

Having non-trivial solution:

The rank of the matrix should be less than the number of variables.

Or determinant of the matrix should be equal to zero.

Calculation:

Given:

-y + z = 0

(4d - 1) x + y + Z = 0

(4d - 1) z = 0

For non-trivial solution:

det. A = 0

⇒ |A| = 0

⇒ 0 × [(4d - 1) - 0] + 1 × [(4d - 1)2 - 0] + 1(0 - 0) = 0

⇒ (4d - 1)2 = 0

∴ The system of linear equations has a non-trivial solution if d equals to 1/4

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