91Ó°ÊÓ

For finding the value of a, b, c, d, e in the given matrix we will follow the following steps:

For the value of a:

$A=\left[\begin{array}{ccccc}0& \mathrm{a}& 2& 1& \mathrm{b}\\ 0& 0& 0& \mathrm{c}& \mathrm{d}\\ 0& 0& \mathrm{e}& 0& 0\end{array}\right]$

Since, 1^st column has all zero entry and a lies in the in second column.

Thus, according to rule a must be a pivot. Hence, the value of a is 1.

Since, e is in the third column which do not have any pivot entry.

Hence, e can have any value. Let the value of e be 1.

Since, c is in the Fourth column which does not has any pivot entry.

Hence, c must be equal to 0.

Since, d is in the Fifth column which not has any pivot entry. If we make d as the pivot it should be 1or value of d can be 0 too.

Since, b is in the Fifth column which not has any pivot entry. If we make d as the pivot then the value of b is 0 or if we make value of d as 0 then b can have any value.

Values if given arbitrary value for the matrix to be in reduced echelon form are

$a=1,e=0,c=0,d=1,b=0$

Matrix will be

$\left[\begin{array}{ccccc}0& 1& 2& 1& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 1& 0& 0\end{array}\right]$