A rectangular array of m x n numbers in the form of m rows and n columns, is called a matrix of order m by n, written as m x n matrix. This article gives an idea of the rank of a matrix and some special matrices.
If A=(aij)m×n is a matrix, and B is its sub-matrix of order r, then∣β∣ the determinant is called r-rowed minor of A.
Definition:
Let A=(aij)m×n be a matrix. A positive integer r is said to be a rank of A if
A possesses at least one r-rowed minor which is different from zero; and
Every (r + 1) rowed minor of A is zero.
From (ii), it automatically follows that all minors of higher order are zeros. We denote rank of A by ρ(A)
Note:
The rank of a matrix does not change when the following elementary row operations are applied to the matrix:
(a) Two rows are interchanged (Ri↔Rj);
(b) A row is multiplied by a non-zero constant, (Ri→kRi,withk=0);
(c) A constant multiple of another row is added to a given row (Ri→Ri,+kRj) where i=j.
Note: The arrow → means “replaced by”.
Note that the application of these elementary row operations does not change a singular matrix to a non-singular matrix nor does a non-singular matrix change to a singular matrix. Therefore, the order of the largest non-singular square sub-matrix is not affected by the application of any of the elementary row operations. Thus, the rank of a matrix does not change by the application of any of the elementary row operations. A matrix obtained from a given matrix by applying any of the elementary row operations is said to be equivalent to it. If A and B are two equivalent matrices, we write A ~ B. Note that if A ~ B, then ρ(A) = ρ(B)
By using the elementary row operations, we shall try to transform the given matrix in the following
Where * stands for zero or non-zero element. That is, we shall try to make aii as 1 and all the elements below aij as zero.
Illustration: For what values of x does the matrix ⎣⎢⎡3+x1257+x5263+x⎦⎥⎤ have the rank 2?
Solution:
The given matrix has only one 3rd-order minor. In order that the rank arrive at 2, we must bring about its determinant to zero. Hence, by applying the invariance method we can obtain values of x.
∣∣∣∣∣∣∣3+x1257+x5263+x∣∣∣∣∣∣∣=0
Now, using R1→R1−R3∣∣∣∣∣∣∣3+x1257+x5263+x∣∣∣∣∣∣∣=∣∣∣∣∣∣∣1+x1207+x5−1−x63+x∣∣∣∣∣∣∣;usingC3→C3+C1=∣∣∣∣∣∣∣1+x1207+x5075+x∣∣∣∣∣∣∣=(1+x)∣∣∣∣∣7+x575+x∣∣∣∣∣=(1+x)[(7+x(5+x)−35]=(1+x)(x2+12x)=x(1+x)(x+12)
If f(x)=a0xn+a1xn+a2xn−2+…….+anx0, then we define a matrix polynomial a, b f(A)=a0An+a1An−1+a2An−2+……+anIn where A is the given square matrix. If f(A) is a null matrix, then A is called the zero or root of the matrix polynomial f(A).
Illustration: Let A=⎣⎢⎡22−101−1130⎦⎥⎤ and f(x)=x2−5x+6I3. Find f (A).
Solution:
By using methods of multiplication and addition of matrices we will obtain the required result. Here f(A)=A2−5A+5I3=⎣⎢⎡22−101−1130⎦⎥⎤2−5⎣⎢⎡22−101−1130⎦⎥⎤+6⎣⎢⎡100010001⎦⎥⎤=⎣⎢⎡22−101−1130⎦⎥⎤×⎣⎢⎡22−101−1130⎦⎥⎤−⎣⎢⎡1010−505−55150⎦⎥⎤+⎣⎢⎡600060006⎦⎥⎤=⎣⎢⎡2×2+0×2+1(−1)2×2+1×2+3(−1)(−1)2+(−1)2+0(−1)2×0+0×1+1(−1)2×0+1×1+3(−1)(−1)0+(−1)1+0(−1)2×1+0×3+1×02×1+1×3+3×0(−1)1+(−1)3+0×0⎦⎥⎤+⎣⎢⎡6−100−100−(−5)0−06−50−(−15)0−50−156−0⎦⎥⎤=⎣⎢⎡33−4−1−2−125−4⎦⎥⎤+⎣⎢⎡−4−105015−5−156⎦⎥⎤=⎣⎢⎡3−43−10−4+5−1+0−2+1−1+52−55−15−4+6⎦⎥⎤=⎣⎢⎡−1−71−1−14−3−102⎦⎥⎤
Illustration: Let A=[acbd] be such that
then A3=0,butA3=0
Solution:
(a) As A3=0,weget∣∣∣A3∣∣∣=0;∣∣∣A3∣∣∣=0⇒∣A∣=0⇒ad−bc=0
In this problem, A3=0 means | A | also is equal to 0. Therefore, by calculating A2 we can obtain the result.