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Find the rank of the following matrix :$\begin{bmatrix} 1 & 2 & -1 &3 \\2 & 4 & 1 &-2 \\3 & 6 & 3 &-7 \end{bmatrix}$

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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} 1 & 2 & -1 & 3 \\ 2 & 4 & 1 & -2 \\ 3 & 6 & 3 & -7 \end{bmatrix} $ is a $3$ x $4$ matrix.
$ \therefore \rho(A) \leq 3$
The two order minors of A with $C_1$ and $C_2$ will vanish since the elements in $C_1$ and $C_2$ are proportional.
Consider the 3rd order minor
$ \begin{vmatrix} 1 & -1 & 3 \\ 2 & 1 & -2 \\ 3 & 3 & -7 \end{vmatrix} = 1(-7+6)+1(-14+6)+3(6-3)$
$ = -1-8+9=0$
Consider the 3rd order minor
$ \begin{vmatrix} 2 & -1 & 3 \\ 4 & 1 & -2 \\ 6 & 3 & -7 \end{vmatrix} = 2(-7+6)+1(-28+12)+3(12-6)$
$ -2-16+18=0$
All the 3rd order minors of A are zero.
$ \therefore \rho(A) \leq 2$
Consider the 2nd order minor $ \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} = 2+4=6 \neq 0$
Since at least one 2nd order minor $ \neq 0, \rho(A) = 2$
Method 2
$A=\begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & 1 & -2 \\ 3 & 6 & 3 & -7 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -8 \\ 0 & 0 & 6 & -16 \end{bmatrix} R_2 \rightarrow R_2-2R_1$
$ R_3 \rightarrow R_3-3R_1$
$ \sim \begin{vmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -8 \\ 0 & 0 & 0 & 0 \end{vmatrix} R_3 \rightarrow R_3 - 2R_2$
The last equivalent matrix is in echelon form with 2 nonzero rows.
$ \therefore \rho(A) = 2$

 

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

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