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Q)

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

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A)
Toolbox:
  • 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} 3 & 1 & 2 & 0 \\ 1 & 0 & -1 & 0 \\ 2 & 1 & 3 & 0 \end{bmatrix} $ is of type $3$ x $4$
$ \therefore \rho(A) \leq 3$
Now A is 4 3rd order minors of which three minors with $C_3$ being zero, will vanish.
Consider the minor $ \begin{vmatrix} 3 & 1 & 2 \\ 1 & 0 & -1 \\ 2 & 1 & 3 \end{vmatrix} = 3(0+1)-1(3+2)+2(1-0)$
$ = 3-5+2=0$
All 3rd order minors vanish $ \therefore \rho(A) \neq 3$
$ \rho(A) \leq 2$
Consider the minor $\begin{vmatrix} 3 & 1 \\ 1 & 0 \end{vmatrix} = 0-1=-1 \neq 0$
Atleast one second order minor is nonzero.
$ \therefore \rho(A) = 2$
Method 2
$A = \begin{bmatrix} 3 & 1 & 2 & 0 \\ 1 & 0 & -1 & 0 \\ 2 & 1 & 3 & 0 \end{bmatrix} $
$ \sim \begin{bmatrix} 1 & 3 & 2 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 2 & 3 & 0 \end{bmatrix} C_1 \leftrightarrow C_2 $
$ \sim \begin{bmatrix} 1 & 3 & 2 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & -1 & 1 & 0 \end{bmatrix} R_3 \leftrightarrow R_3-R_1$
$ \sim \begin{bmatrix} 1 & 3 & 2 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} R_3 \leftrightarrow R_3+R_2$
The equivalent matrix is in echelon form with 2 non-zero rows. $ \therefore \rho(A) = 2$

 

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