Browse Questions

# Show that $A=\begin{bmatrix} xc_{r} & xc_{r+1} & xc_{r+2} \\ yc_r & yc_{r+1} & yc_{r+2} \\ zc_r & zc_{r+1} & zc_{r+2} \end{bmatrix}=\begin{bmatrix} xc_{r} & x+1c_{r+1} & x+2c_{r+2} \\ yc_r & y+1c_{r+1} & y+2c_{r+2} \\ zc_r & z+1c_{r+1} & z+2c_{r+2} \end{bmatrix}$

Toolbox:
• (i) Elementary transformation can be made by
• a)interchanging two columns or rows
• b)By adding or subtracting two or more rows or columns.
• $nc_r+nc_{r+1}=n+1c_{r+1}$
Step 1:
Let us consider the $LHS$
$\Delta =\begin{bmatrix} xc_{r} & xc_{r+1} & xc_{r+2} \\ yc_r & yc_{r+1} & yc_{r+2} \\ zc_r & zc_{r+1} & zc_{r+2} \end{bmatrix}$
Apply $c_3 \to c_2 +c_3$
$\Delta=\begin{bmatrix} xc_{r} & xc_{r+1} & xc_{r+1}+xc_{r+2} \\ yc_r & yc_{r+1} & yc_{r+1}+yc_{r+2} \\ zc_r & zc_{r+1} & zc_{r+1}+zc_{r+2} \end{bmatrix}$
$\large \frac{n}{2}+\frac{n}{3}=\frac{n+1}{2}$
$\Delta=\begin{bmatrix} xc_{r} & xc_{r+1} & x+1c_{r+2} \\ yc_r & yc_{r+1} & y+1c_{r+2} \\ zc_r & zc_{r+1} & z+1c_{r+2} \end{bmatrix}$
Step 2:
Now apply $c_2 \to c_1+c_2$
$\Delta=\begin{bmatrix} xc_{r} & xc_r+xc_{r+1} & x+1c_{r+2} \\ yc_r & yc_r+yc_{r+1} & y+1c_{r+2} \\ zc_r & zc_r+zc_{r+1} & z+1c_{r+2} \end{bmatrix}$
$nc_r+nc_{r+1}=n+1c_{r+1}$
$\Delta=\begin{bmatrix} xc_{r} & x+1c_{r+1} & x+1c_{r+2} \\ yc_r & y+1c_{r+1} & y+1c_{r+2} \\ zc_r & z+1zc_{r+1} & z+1c_{r+2} \end{bmatrix}$
Now apply $c_3 \to c_2+c_3$
Step 3:
$\Delta=\begin{bmatrix} xc_{r} & x+1c_{r+1} & x+1c_{r+1}+x+1c_{r+2} \\ yc_r & y+1c_{r+1} & y+1c_{r+2}+y+1c_{r+2} \\ zc_r & z+1c_{r+1} & z+1c_{r+1}+y+1c_{r+2} \end{bmatrix}$
$n+1c_{r+1}+n+1 c_{r+2}=n+2c_{r+2}$
Therefore $\Delta=\begin{bmatrix} xc_{r} & x+1c_{r+1} & x+2c_{r+2} \\ yc_r & y+1c_{r+1} & y+2c_{r+2} \\ zc_r & z+1c_{r+1} & z+2c_{r+2} \end{bmatrix}$
$=R.H.S$
Solution : Hence proved