X+y+z=1, 2x+3y+2z=2 and ax+ay+2az=4 solve the equation by the method o

1.	X+y+z=1, 2x+3y+2z=2 and ax+ay+2az=4 solve the equation by the method of inversion
Answer» $\textbf{Given:}$ $<klux>X</klux>+y+z=1,\;2x+<klux>3Y</klux>+2z=2,\;ax+ay+2az=4$ $\textbf{To find:}$ $\text{The values of x,y and z by matrix inversion <klux>METHOD</klux>}$ $\textbf{Solution:}$ $\text{The given system of equations can be written as}$ $\left[\begin{array}{ccc}1&1&1\\2&3&2\\a&a&2a\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\2\\4\end{array}\right]$ $\text{This is of the form}\;AX=B$ $\text{Here}$ $A=\left[\begin{array}{ccc}1&1&1\\2&3&2\\a&a&2a\end{array}\right]$ $\|A\|=1(6a-2a)-1(4a-2a)+1(2a-3a)$ $\|A\|=4a-2a-a$ $\|A\|=a$ $\text{Cofactor matrix of A is}$ $\left[\begin{array}{ccc}4a&-2a&-a\\-a&a&0\\-1&1&0\end{array}\right]$ $\text{Adjoint of matrix A = Transpose of cofactor matrix of A}$ $\text{Then}$ $adj\,A=\left[\begin{array}{ccc}4a&-a&-1\\-2a&a&0\\-a&0&1\end{array}\right]$ $A^{-1}=\frac{1}{\|A\|}(adjA)$ $A^{-1}=\frac{1}{a}\left[\begin{array}{ccc}4a&-a&-1\\-2a&a&0\\-a&0&1\end{array}\right]$ $\text{Now,}$ $X=A^{-1}B$ $X=\frac{1}{a}\left[\begin{array}{ccc}4a&-a&-1\\-2a&a&0\\-a&0&1\end{array}\right]\left[\begin{array}{c}1\\2\\4\end{array}\right]$ $X=\frac{1}{a}\left[\begin{array}{c}4a-2a-4\\-2a+2a+0\\-a+0+4\end{array}\right]$ $X=\frac{1}{a}\left[\begin{array}{c}2a-4\\0\\4-a\end{array}\right]$ $X=\left[\begin{array}{c}\frac{2a-4}{a}\\0\\\frac{4-a}{a}\end{array}\right]$ $\therefore\textbf{The solution is}$ $\bf\,x=\frac{2a-4}{a}$ $\bf\,y=0$ $\bf\,z=\frac{4-a}{a}$ Find more: SOLVE the following system of linear equations using Inverse Matrix Method. x + 6Y – z = 10 2x + 3y + 3z = 17 3x - 3y – 2z = -9 brainly.in/question/16281483

X+y+z=1, 2x+3y+2z=2 and ax+ay+2az=4 solve the equation by the method of inversion

Answer»

$\textbf{Given:}$

$<klux>X</klux>+y+z=1,\;2x+<klux>3Y</klux>+2z=2,\;ax+ay+2az=4$

$\textbf{To find:}$

$\text{The values of x,y and z by matrix inversion <klux>METHOD</klux>}$

$\textbf{Solution:}$

$\text{The given system of equations can be written as}$

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