RANK OF MATRIX BY MINOR METHOD

The rank of a matrix A is defined as the order of a highest order non-vanishing minor of the matrix A. It is denoted by the symbol ρ (A).The rank of a zero matrix is defined to be 0.

Note

(i) If a matrix contains at-least one non-zero element, then ρ (A) ≥ 1

(ii) The rank of the identity matrix I_n is n.

(iii) If the rank of a matrix A is r, then there exists at-least one minor of A of order r which does not vanish and every minor of A of order r + 1 and higher order (if any) vanishes.

(iv) If A is an m × n matrix, then ρ (A) ≤ min {m, n} = minimum of m, n.

(v) A square matrix A of order n has inverse if and only if ρ (A) = n.

Question 1 :

Solution :

Then A is a matrix of order 2×2. So ρ (A)  min {2, 2} = 2. The highest order of minors of A is 2 . There is only one third order minor of A .

  =  4 - 4 

|A|  =  0

The rank of the given matrix will be less than 2.

Hence the rank of the given matrix is 1.

Question 2 :

Solution :

Then A is a matrix of order 3 × 2. So ρ (A)  min {3, 2} = 2. The highest order of minors of A is 2 . 

There are four 2 x 2 minor matrices in the above matrix. By finding the determinants, we get

Since the minor of 2 x 2 matrix is not equal to zero, the rank of the given matrix is 2.

Question 3 :

Solution :

Then A is a matrix of order 2 × 4. So ρ (A)  min {2, 4} = 2. The highest order of minors of A is 2 . 

There are four 2 x 2 minor matrices in the above matrix.

Rank of the given matrix is 2.

Question 4 :

Solution :

Then A is a matrix of order 3 × 3. So ρ (A)  min {3, 3} = 3. The highest order of minors of A is 3 . 

By finding determinant of given matrix, we get 

  =  1(-4 + 6) + 2(-2 + 30) + 3(2 - 20)

  =  1(2) + 2(28) + 3(-18)

  =  2 + 56 - 54

  =   58 - 54

|A|  =  4 ≠ 0

Hence the rank of the given matrix is 3.

Question 5 :

Solution :

Then A is a matrix of order 3 × 4. So ρ (A)  min {3, 4} = 3. The highest order of minors of A is 3 . 

By finding determinant of given matrix, we get 

	0(0 - 4) - 1(0-32) + 2(0-16)   =  0 - 1(-32) + 2(-16)   =  32 - 32   =  0
	=  1(8-0) - 2(4-3) + 1(0-4)   =  8 - 2(1) + 1(-4)   =  8 - 2 - 4   =  8 - 6   =  2 ≠ 0

Hence the rank of the given matrix is 3.

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