JEE Main & Advanced Mathematics Matrices Rank of Matrix

Rank of Matrix

Category : JEE Main & Advanced

Definition : Let A be a \[m\times n\] matrix. If we retain any \[r\] rows and \[r\] columns of A we shall have a square sub-matrix of order \[r\]. The determinant of the square sub-matrix of order \[r\] is called a minor of A order \[r\]. Consider any matrix A which is of the order of \[3\times 4\] say, \[A=\left| \begin{matrix} 1 & 3 & 4 & 5  \\ 1 & 2 & 6 & 7  \\ 1 & 5 & 0 & 1  \\ \end{matrix} \right|\]. It is \[3\times 4\] matrix so we can have minors of order 3, 2 or 1. Taking any three rows and three columns minor of order three. Hence minor of order \[3=\left| \,\begin{matrix} 1 & 3 & 4  \\ 1 & 2 & 6  \\ 1 & 5 & 0  \\ \end{matrix}\, \right|=0\]


Making two zeros and expanding above minor is zero. Similarly we can consider any other minor of order 3 and it can be shown to be zero. Minor of order 2 is obtained by taking any two rows and any two columns.


Minor of order \[{{D}_{3}}=\left| \,\begin{matrix} {{a}_{1}} & {{b}_{1}} & {{d}_{1}}  \\ {{a}_{2}} & {{b}_{2}} & {{d}_{2}}  \\ {{a}_{3}} & {{b}_{3}} & {{d}_{3}}  \\ \end{matrix}\, \right|\]. Minor of order 1 is every element of the matrix.


Rank of a matrix: The rank of a given matrix A is said to be \[r\] if  


(1) Every minor of A of order \[r+1\] is zero.


(2) There is at least one minor of A of order \[r\] which does not vanish. Here we can also say that the rank of a matrix A is said to be \[r\], if (i) Every square submatrix of order \[r+1\] is singular.


(ii) There is at least one square submatrix of order \[r\] which is non-singular.


The rank \[r\]  of matrix A is written as \[\rho (A)=r\].

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