91Ó°ÊÓ

The system of equation is given as,

$\left[\begin{array}{cccc}2\mathrm{x}& +\mathrm{y}& & =\mathrm{C}\\ & 3\mathrm{y}& +\mathrm{z}& =\mathrm{C}\\ \mathrm{x}& & +4\mathrm{z}& =\mathrm{C}\end{array}\right]$

It is given that C is a constant and x, y, z are integers.

In case of Echelon matrix zeros lies at the bottom.Echelon matrix is used to determine the rank of the matrix. The number of zeros to the left of first non zero element in each row increase from top to bottom.

The Given system of equation can be written in the form of $\mathrm{AX}=\mathrm{B}$,

$\left[\begin{array}{ccc}2& 1& 0\\ 0& 3& 1\\ 1& 0& 4\end{array}\right]\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}\mathrm{C}\\ \mathrm{C}\\ \mathrm{C}\end{array}\right]$

The Augmented matrix corresponding to the above equation is,

$\left[\mathrm{A}|\mathrm{B}\right]=\left[\begin{array}{cccc}2& 1& 0& |\mathrm{C}\\ 0& 3& 1& |\mathrm{C}\\ 1& 0& 4& |\mathrm{C}\end{array}\right]$

Now we can convert the above augmented matrix into the reduced row echelon matrix,

Use operation data-custom-editor="chemistry" ${\mathrm{R}}_1→\frac{{\mathrm{R}}_1}{2}$

[A| B] =$\left[\begin{array}{cccc}1& \frac{1}{2}& 0& |\frac{\mathrm{C}}{2}\\ 0& 3& 1& |\mathrm{C}\\ 1& 0& 4& |\mathrm{C}\end{array}\right]$

Use operation data-custom-editor="chemistry" ${\mathrm{R}}_3→{\mathrm{R}}_3-{\mathrm{R}}_1$

[A| B] =$\left[\begin{array}{cccc}1& \frac{1}{2}& 0& |\frac{\mathrm{C}}{2}\\ 0& 3& 1& |\mathrm{C}\\ 0& -\frac{1}{2}& 4& |\frac{\mathrm{C}}{2}\end{array}\right]$

Use operation data-custom-editor="chemistry" ${\mathrm{R}}_2→\frac{1}{3}{\mathrm{R}}_2$

[A| B] =$\left[\begin{array}{cccc}1& \frac{1}{2}& 0& |\frac{C}{2}\\ 0& 1& \frac{1}{3}& |\frac{C}{3}\\ 0& -\frac{1}{2}& 4& |\frac{C}{2}\end{array}\right]$

Use operation ${\mathrm{R}}_1→{\mathrm{R}}_1-\frac{1}{2}{\mathrm{R}}_2$

data-custom-editor="chemistry" $\left[\mathrm{A}|\mathrm{B}\right]=\left[\begin{array}{cccc}1& 0& -\frac{1}{6}& |\frac{\mathrm{C}}{3}\\ 0& 1& \frac{1}{3}& |\frac{\mathrm{C}}{3}\\ 0& -\frac{1}{2}& 4& |\frac{\mathrm{C}}{2}\end{array}\right]$

Use operation data-custom-editor="chemistry" ${\mathrm{R}}_3→{\mathrm{R}}_3+\frac{1}{2}{\mathrm{R}}_2$

data-custom-editor="chemistry" $\left[\mathrm{A}|\mathrm{B}\right]=\left[\begin{array}{cccc}1& 0& -\frac{1}{6}& |\frac{\mathrm{C}}{3}\\ 0& 1& \frac{1}{3}& |\frac{\mathrm{C}}{3}\\ 0& 0& \frac{25}{6}& |\frac{4\mathrm{C}}{6}\end{array}\right]$

Use operation data-custom-editor="chemistry" ${\mathrm{R}}_3→\frac{6}{25}{\mathrm{R}}_3$

data-custom-editor="chemistry" $\left[\mathrm{A}|\mathrm{B}\right]=\left[\begin{array}{cccc}1& 0& -\frac{1}{6}& |\frac{\mathrm{C}}{3}\\ 0& 1& \frac{1}{3}& |\frac{\mathrm{C}}{3}\\ 0& 0& 1& |\frac{4\mathrm{C}}{25}\end{array}\right]$

Use operationdata-custom-editor="chemistry" ${\mathrm{R}}_1→{\mathrm{R}}_1+\frac{1}{6}{\mathrm{R}}_3$ and data-custom-editor="chemistry" ${\mathrm{R}}_2→{\mathrm{R}}_2-\frac{1}{3}{\mathrm{R}}_3$

$\left[\mathrm{A}|\mathrm{B}\right]=\left[\begin{array}{cccc}1& 0& 0& |\frac{9\mathrm{C}}{25}\\ 0& 1& 0& |\frac{7\mathrm{C}}{25}\\ 0& 0& 1& |\frac{4\mathrm{C}}{25}\end{array}\right]$

From the above matrix we can write in the equation form,

data-custom-editor="chemistry" $\mathrm{x}=\frac{9\mathrm{C}}{25}$,data-custom-editor="chemistry" $\mathrm{y}=\frac{7\mathrm{C}}{25}$ and data-custom-editor="chemistry" $\mathrm{z}=\frac{4\mathrm{C}}{25}$.

Now it is given that the x, y, z are integers. So should be a positive multiple of 25.

Therefore the least positive value of C is 25.