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Q)

Find the rank of the following matrix :$\begin{bmatrix} 6 & 12 & 6 \\1 & 2 & 1 \\4 & 8 & 4 \end{bmatrix}$

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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} 6 & 12 & 6 \\ 1 & 2 & 1 \\ 4 & 8 & 4 \end{bmatrix}$ is a 3 x 3 matrix
$ \therefore \rho(A) \leq 3$
$ |A| = \begin{vmatrix} 6 & 12 & 6 \\ 1 & 2 & 1 \\ 4 & 8 & 4 \end{vmatrix} = 6(8-8)-12(4-4)+6(8-8) = 0$
$ \rho(A) \neq 3$
$ \therefore \rho(A) \leq 2$
All 2nd order minors are also 0.
$ \begin{vmatrix} 6 & 12 \\ 1 & 2 \end{vmatrix} = \begin{vmatrix}2 & 1 \\ 8 & 4 \end{vmatrix} = \begin{vmatrix}12 & 6 \\ 8 & 4 \end{vmatrix} = 0 $ etc..
$ \therefore \rho(A) \neq 2$
$ \rho(A) = 1$ because there are nonzero minors of order 1.
Method 2
$ A = \begin{bmatrix} 6 & 12 & 6 \\ 1 & 2 & 1 \\ 4 & 8 & 4 \end{bmatrix}\: \: R_3 \rightarrow R_3-4R_2$
$ \sim \begin{bmatrix} 6 & 12 & 6 \\ 1 & 2 & 1 \\ 0 & 0 & 0 \end{bmatrix} R_2 \rightarrow R_2 - \large\frac{1}{6}R_1$
$ \sim \begin{bmatrix} 6 & 12 & 6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
The last equivalent matrix is in echeron form and there is one nonzero row.
$ \therefore \rho(A) = 1$

 

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