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Find the rank of the following matrix :$\begin{bmatrix} 0 & 1 & 2 &1 \\2 & -3 & 0 &-1 \\1 & 1 & -1 & 0 \end{bmatrix}$

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  • The rank of a matrix A is equal to $r$ if (i) A has atleast one minor of order r that does not vanish (ii) every minor of order $r+1$ or higher vanishes. $ \rho (A)=r.$
  • By elementary transformations, a matrix can by reduced to echelon ( or triangular form ) so that (i) Every row of A with all its entries being 0 occurs below every row with non-zero entries. (ii) The first non-zero entry in each non-zero row is 1. (iii) The number of zeros before the first non-zero element in a row is less than the number of zeros in the next row. The rank of the matrix in echelon form is equal to the number of non-zero rows.
Method 1
$ A = \begin{bmatrix} 0 & 1 & 2 & 1 \\ 2 & -3 & 0 & 1 \\ 1 & 1 & -1 & 0 \end{bmatrix}$ is a $3$ x $4$ matrix. $ \rho(A) \leq 3$
A has 4 3rd order minors.
Consider the minor $\begin{vmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & 1 & -1 \end{vmatrix} = 0-1(-2-0)+2(2+3)$
$ = 2+10=12 \neq 0$
A has atleast one 3rd order minor $ \neq 0$
$ \therefore \rho(A) = 3$
Method 2
$ A = \begin{bmatrix} 0 & 1 & 2 & 1 \\ 2 & -3 & 0 & 1 \\ 1 & 1 & -1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 2 & 0 \\ 1 & -3 & 0 & 2 \\ 0 & 1 & -1 & 1 \end{bmatrix} (C_1 \leftrightarrow C_4)$
$ \sim \begin{bmatrix} 1 & 1 & 2 & 0 \\ 0 & -2 & -2 & 2 \\ 0 & 1 & -1 & 1 \end{bmatrix} R_2 \rightarrow R_2-R_1 \sim \begin{bmatrix} 1 & 1 & 2 & 0 \\ 0 & -2 & -2 & 2 \\ 0 & 0 & -4 & 4 \end{bmatrix} R_3 \rightarrow 2R_3+R_2$
The last equivalent matrix is in echelon form with 3 nonzero rows. $ \therefore \rho(A) = 3$

 

answered May 26, 2013 by thanvigandhi_1
edited Jun 3, 2013 by thanvigandhi_1
 

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