Test: System of Linear Equations - SAT MCQ

Concept:

Consider the system of m linear equations

a₁₁ x₁ + a₁₂ x₂ + … + a_1n x_n = b₁

a₂₁ x₁ + a₂₂ x₂ + … + a_2n x_n = b₂

…

a_m1 x₁ + a_m2 x₂ + … + a_mn x_n = b_m

The above equations containing the n unknowns x₁, x₂, …, x_n. 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

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

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.

Concept:

Consider the system of m linear equations

a₁₁ x₁ + a₁₂ x₂ + … + a_1n x_n = 0

a₂₁ x₁ + a₂₂ x₂ + … + a_2n x_n = 0

…

a_m1 x₁ + a_m2 x₂ + … + a_mn x_n = 0

The above equations containing the n unknowns x₁, x₂, …, x_n. 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

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).

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

Concept:

Consider the system of m linear equations

a₁₁ x₁ + a₁₂ x₂ + … + a_1n x_n = 0

a₂₁ x₁ + a₂₂ x₂ + … + a_2n x_n = 0

a_m1 x₁ + a_m2 x₂ + … + a_mn x_n = 0

• The above equations containing the n unknowns x₁, x₂, …, x_n. 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, A_x, A_y, A_z 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

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

K² (K – 2(K – 1) = 0

K² (K – 2K + 2) = 0

K² (-K + 2) = 0

K = 0, 0, 2

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

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).

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

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

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 A^T is 4 × 2

The order of A^TA is 4 × 4

Rank of (A) ≤ min (2, 4) = 2

Rank of (A^T) ≤ min (2, 4) = 2

Rank (A^TA) ≤ min (2, 2) = 2

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

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

(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

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

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