Test: System of Linear Equations (May 6) - JEE MCQ

Let's take the example of it
The nine numbers x₁, x₂, x₃ ... x₉, are in ascending order
Let  the nine number is 1,1,1,1,1,1,1,1,10
Their Avg will be = 18/9 = 2
Also, this satisfies that their average m is strictly greater than all the first eight numbers. 
now varifing
Average (x₁, x₂ ... x₉, m) = m
Avg[1,1,1,1,1,1,1,1,10,2] will be 20/10 = 2
Which satisfies the given statement
Now checking Average (x₂, x₃ ... x₉) > m
Average (1,1,1,1,1,1,1,10) will 17/8 = 2.12
Which is greater then 2
So it is also satisfied  Average (x₂, x₃ ... x₉) > m
So, Average (x1, x2 ... x9, m) = m and Average (x2, x3 ... x9) > m is the true statement.

When a system of simultaneous equations has infinite solutions, it's called a "dependent" system of equations.
In such a scenario, the equations are representing the same line or are multiples of each other, leading to an infinite number of solutions where the equations overlap entirely.
Therefore, the correct term for a system with infinite solutions is "dependent."

v1, . . . , vn be the column vectors of a non-zero n × n real matrix A. Let a₁, . . . , an ∈ ℝ, not all zero, be such that 
that means v₁, . . . , v_n are Linearly independent.
In the system of equations when the system is L.I. then rank (A) = number of unknowns then it has a unique solution. 
if at least one ��_�� not equal to zero vector(s) are Linearly dependent. 
In the system of equations when rank of A < n then it has infinitely many solution(s) 
Here, given that ,  where all ��_�� are not zero Hence vectors are linearly dependent 
When vectors are Linearly independent then in system  rank of A is less than 9 (number of column vectors )
Hence, it has infinitely many solutions.
Therefore, the Correct Option is Option (4).

Let consider a standard pair of linear equation such that,

For the pair of linear equations, the condition for the unique solution is,

Given a pair of linear equation is,
kx - y = 2
6x - 2y = 3
These equations can be written as,
⇒ kx - y - 2 = 0
⇒ 6x - 2y - 3 = 0
On comparing it with standard pair of linear equation, we get, 
⇒ a₁ = k, b₁ = -1, c₁ = -2
⇒ a₂ = 6, b₂ = -2, c₂ = -3
Using the condition for unique solution,

So, the value of k for which the system of equations has unique solution,
k ≠ 3
Hence, the correct option is 2.

Given:
a₁ = 3k + 1
b₁ = 3
c₁ = -2
a₂ = k² + 1
b₂ = k - 2
c₂ = -5
Formula Used:


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.

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
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.

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.

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

Non-homogeneous equation of type AX = B has infinite solutions if ρ(A : B) = ρ(A) < Number of unknowns
Given set of equations
x + y + z = 1
ax – ay + 3z = 5
5x – 3y + az = 6

R₂ → R₂ – aR₁ and R₃ → R₃ – 5R₁


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) 
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).

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

CAYLEY-HAMILTON THEOREM:
Statement: Every square matrix satisfies its own characteristic equation.
The Cayley–Hamilton theorem states that substituting the matrix A for x in polynomial, p(x) = det(xI_n – A), results in the zero matrices, such as:
p(A) = 0
It states that a ‘n x n’ matrix A is demolished by its characteristic polynomial det(tI – A), which is monic polynomial of degree n.
Uses of Cayley-Hamilton theorem:
(1) To calculate the positive integral powers of A
(2) To calculate the inverse of a square matrix A