# find the rank of 4 1 3 3

find the rank of 4 1 3 3 3

7 -1 6 0 6

2 3 8 2 8

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Haidar Ali

The rank of a matrix is the number of non-zero rows in the echelon form of the matrix. The echelon form can be obtained by row operations such that the leading entry of each row must have zeros below it.

The given matrix is:

$\left[\begin{array}{ccccc}4& 1& 3& 3& 3\\ 7& -1& 6& 0& 6\\ 2& 3& 8& 2& 8\end{array}\right]$

Divide ${R}_{3}$ by 2:

$\left[\begin{array}{ccccc}4& 1& 3& 3& 3\\ 7& -1& 6& 0& 6\\ 1& \frac{3}{2}& 4& 1& 4\end{array}\right]$

Replace ${R}_{1}$ by ${R}_{1}-4{R}_{3}$:

$\left[\begin{array}{ccccc}4-4& 1-4\left(\frac{3}{2}\right)& 3-4\left(4\right)& 3-4\left(1\right)& 3-4\left(4\right)\\ 7& -1& 6& 0& 6\\ 1& \frac{3}{2}& 4& 1& 4\end{array}\right]$

Rewrite the above matrix:

$\left[\begin{array}{ccccc}0& -5& -13& -1& -13\\ 7& -1& 6& 0& 6\\ 1& \frac{3}{2}& 4& 1& 4\end{array}\right]$

Interchange  ${R}_{1}$ with ${R}_{3}$:

$\left[\begin{array}{ccccc}1& \frac{3}{2}& 4& 1& 4\\ 7& -1& 6& 0& 6\\ 0& -5& -13& -1& -13\end{array}\right]$

Replace ${R}_{2}$ by ${R}_{2}-7{R}_{1}$:

$\left[\begin{array}{ccccc}1& \frac{3}{2}& 4& 1& 4\\ 7-7\left(1\right)& -1-7\left(\frac{3}{2}\right)& 6-7\left(4\right)& 0-7\left(1\right)& 6-7\left(4\right)\\ 0& -5& -13& -1& -13\end{array}\right]$

Rewrite the above matrix:

$\left[\begin{array}{ccccc}1& \frac{3}{2}& 4& 1& 4\\ 0& -\frac{23}{2}& -22& -7& -22\\ 0& -5& -13& -1& -13\end{array}\right]$

Multiply  ${R}_{2}$ by -2, we get:

$\left[\begin{array}{ccccc}1& \frac{3}{2}& 4& 1& 4\\ 0& 23& 44& 14& 44\\ 0& -5& -13& -1& -13\end{array}\right]$

Replace ${R}_{3}$ by ${R}_{3}+\frac{5}{23}{R}_{2}$:

$\left[\begin{array}{ccccc}1& \frac{3}{2}& 4& 1& 4\\ 0& 23& 44& 14& 44\\ 0& -5+\frac{5}{23}\left(23\right)& -13+\frac{5}{23}\left(44\right)& -1+\frac{5}{23}\left(14\right)& -13+\frac{5}{23}\left(44\right)\end{array}\right]$

Rewrite the above matrix:

$\left[\begin{array}{ccccc}1& \frac{3}{2}& 4& 1& 4\\ 0& 23& 44& 14& 44\\ 0& 0& -\frac{79}{23}& \frac{47}{23}& -\frac{79}{23}\end{array}\right]$

Multiply  ${R}_{3}$ by -23:

$\left[\begin{array}{ccccc}1& \frac{3}{2}& 4& 1& 4\\ 0& 23& 44& 14& 44\\ 0& 0& 79& -47& 79\end{array}\right]$

The above matrix is the echelon form of the given matrix. As the number of non-zero rows in the echelon form of the given matrix is 3, it follows that the rank of the given matrix is 3.