Question

Define rank and give one example.

Answer 1

In this question, we are going to give a definition for the rank, rank is a term associated with the matrices and signifies the dimension of the vector space that is generated by the columns of a matrix.

Formula used:
The rank of a matrix is always less than or equal to the order of the $m\times m$ matrix and is less than or equal to $m$ in an $m\times n$ matrix if $m **Complete step by step solution:** Definition: The rank of a matrix is defined as the dimension of a vector space which is generated by the columns of the matrix. In other words, the rank is defined as the maximum number of linearly independent column vectors in the matrix or the maximum number of the row vectors of a matrix that are linearly independent. For an \[m\times n$matrix, if $mExample of rank: Let us consider a \[3\times 3$ matrix, the largest order of the submatrix , to be got is $3\times 3$ only
the rank can be found by the normal form.
Firstly, convert the matrix to identity matrix, for this applying row to row operations on first column,

& \left( \begin{matrix} 3 & 1 & 2 \\\ 2 & 0 & 5 \\\ 1 & 2 & 3 \\\ \end{matrix} \right) \\\ & R1\to R1-R2 \\\ & \left( \begin{matrix} 1 & 1 & -3 \\\ 2 & 0 & 5 \\\ 1 & 2 & 3 \\\ \end{matrix} \right) \\\ & R2\to R2-2R1;R3\to R1-R3 \\\ & \left( \begin{matrix} 1 & 1 & -3 \\\ 0 & -2 & 11 \\\ 0 & -1 & -6 \\\ \end{matrix} \right) \\\ \end{aligned}$$ Now, applying row to row operations on the column number $$2$$ $$\begin{aligned} & R2\to -R2+R3 \\\ & \left( \begin{matrix} 1 & 1 & -3 \\\ 0 & 1 & -17 \\\ 0 & -1 & -6 \\\ \end{matrix} \right) \\\ & R1\to R1-R2;R3\to R2+R3 \\\ & \left( \begin{matrix} 1 & 0 & 14 \\\ 0 & 1 & -17 \\\ 0 & 0 & -23 \\\ \end{matrix} \right) \\\ \end{aligned}$$ Now, operating the third column $$\begin{aligned} & R3\to \dfrac{R3}{-23} \\\ & \left( \begin{matrix} 1 & 0 & 14 \\\ 0 & 1 & -17 \\\ 0 & 0 & 1 \\\ \end{matrix} \right) \\\ & R1\to R1-14R3;R2\to R2+17R3 \\\ & \left( \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right) \\\ \end{aligned}$$ Thus, we obtained an identity matrix of order 3 so, $$rank\left( A \right)=3$$ **Note:** Alternatively, $$A=\left( \begin{matrix} 3 & 1 & 2 \\\ 2 & 0 & 5 \\\ 1 & 2 & 3 \\\ \end{matrix} \right)$$ Rank of $$A\le 3$$ Now, $$\begin{aligned} & \det \left( A \right)=\left| \begin{matrix} 3 & 1 & 2 \\\ 2 & 0 & 5 \\\ 1 & 2 & 3 \\\ \end{matrix} \right| \\\ & \Rightarrow \det \left( A \right)=-23\ne 0 \\\ \end{aligned}$$ As the determinant is not zero, so the rank equals the order of the matrix which is the largest non-zero matrix. Therefore, $$rank\left( A \right)=3$$