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Problem 36: Define Rank of matrix and provide examples

Problem 36: Define Rank of matrix and provide examples

Solution:

A number r is said to be the rank of a matrix A if it possesses the following two properties:

i. There is at least one square sub matrrix of A of order r whose determinant is not equal to zero.

ii. If the matrix A contains any square submatrix of order r+1, then the determinant of every square submatrix of A of order should be zero.

In short the rank of a matrix is the order of any highest order non-vanising minor of the matrix.

Examples:

(a.) Let \begin{align*} A =\left[\begin{array}{ccc} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ \end{array}\right]\\ \end{align*} We have $|A|=1$. Therefore A is a non-singular matrix. Hence rank A = 3.

(b.) Let \begin{align*} A =\left[\begin{array}{ccc} 0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\\ \end{array}\right]\\ \end{align*} Since A is the null matrix. Hence rank A = 0.

(c.) Let \begin{align*} A =\left[\begin{array}{ccc} 1& 2& 3\\ 2& 3& 4\\ 0& 2& 2\\ \end{array}\right]\\ \end{align*} We have $|A|=2 \ne 0$. Therefore A is a non-ingular matrix. Hence rank A = 3.

(d.) Let \begin{align*} A =\left[\begin{array}{ccc} 1& 2& 3\\ 3& 4& 5\\ 4& 5& 6\\ \end{array}\right]\\ \end{align*} We have $|A|=1(24-25)-2(18-20)+3(15-16)=0$. Therefore the rank of A is less than 3. Now there is atlease one minor of A of order which is not equal to zero. Henxe rank A =2.

(e.) Let \begin{align*} A =\left[\begin{array}{ccc} 3& 1& 2\\ 6& 2& 4\\ 3& 1& 2\\ \end{array}\right]\\ \end{align*} We have $|A|=0$. Also each 2-rowed minot of A is equal to zero. But A is not a null matrix. Henxe rank A =2.

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