Rank of Matrix by Minor Method :
Here we are going to see some example problems to know the method of finding rank of a 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 In 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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