Question

Consider the following statements.
P : If a system of linear equations Ax = b has a unique solution, where A is an m × n matrix and b is an m × 1 matrix, then m = n.
Q : For a subspace W of a nonzero vector space V, whenever u ∈ V \ W and v ∈ V \ W, then u + v ∈ V \ W.
Which one of the following holds?

  1. Both P and Q are true
  2. P is true but Q is false
  3. P is false but Q is true
  4. Both P and Q are false

Answer: Both P and Q are false

Solution :-

Statement P :

Statement ‘P’ is not always true. Lets consider an example of system of linear equations

x1−2x2=1,2x1−5x2=3,−3x1+6x2=−3

Now, lets solve this system of linear equations. Here,

A=[1−22−5−36]3×2,x=[x1x2]2×1andb=[13−3]3×1⋅

Here, Matrix A is 3 × 2 and b is 3 × 1.

Hence, augmented matrix is

[1−212−53−36−3]

Applying R2→R2+(−2)R1 and R3→R3+3R1 , then we get

[1−210−11000]

Applying R2→−R2 , then we get

[1−2101−1000]

Hence, x1−2x2=1⋯⁢⋯⁢(1) and x2=−1
put the value of x2=−1 in eq(1), we get x1=−1

So, x1=−1 and x2=−1 is a solution of the system of linear equations and we can see that this solution is unique but 3≠2 . Here, 3 is the number of rows of A and 2 is the number of columns of A.

Therefore, If a system of linear equations Ax = b has a unique solution, where A is an m × n matrix and b is an m × 1 matrix, then m = n is not always true, So Statement ‘P’ is false.

Statement Q :

Statement ‘Q’ is also not always true. lets us consider vector space V and subspace W.

V=ℝ3(ℝ)={(x,y,z):x,y,z∈ℝ}

W={(x,y,0):x,y∈ℝ} , Here we can see that W is subspace of V.

Now, V\W={(x,y,z):x,y,z∈ℝ;z≠0} . Now, we take two vector from V\W , u=(x,y,z)∈V\W,v=(x,y,−z)∈V\W.
Now, u+v=(x,y,z)+(x,y,−z)=(2⁢x,2⁢y,0)∈W , not in V\W .

Hence, Statement ‘Q’ For a subspace W of a nonzero vector space V, whenever u ∈ V \ W and v ∈ V \ W, then u + v ∈ V \ W is false. Here, ‘whenever’ word is important.

So, the correct answer is option-4, → Both P and Q are false