Question

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

Answer 1

$\textbf{Given:}$

$x+y+z=1,\;2x+3y+2z=2,\;ax+ay+2az=4$

$\textbf{To find:}$

$\text{The values of x,y and z by matrix inversion method}$

$\textbf{Solution:}$

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

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

$\text{This is of the form}\;AX=B$

$\text{Here}$

$A=\left[\begin{array}{ccc}1&1&1\\2&3&2\\a&a&2a\end{array}\right]$

$|A|=1(6a-2a)-1(4a-2a)+1(2a-3a)$

$|A|=4a-2a-a$

$|A|=a$

$\text{Cofactor matrix of A is}$

$\left[\begin{array}{ccc}4a&-2a&-a\\-a&a&0\\-1&1&0\end{array}\right]$

$\text{Adjoint of matrix A = Transpose of cofactor matrix of A}$

$\text{Then}$

$adj\,A=\left[\begin{array}{ccc}4a&-a&-1\\-2a&a&0\\-a&0&1\end{array}\right]$

$A^{-1}=\frac{1}{|A|}(adjA)$

$A^{-1}=\frac{1}{a}\left[\begin{array}{ccc}4a&-a&-1\\-2a&a&0\\-a&0&1\end{array}\right]$

$\text{Now,}$

$X=A^{-1}B$

$X=\frac{1}{a}\left[\begin{array}{ccc}4a&-a&-1\\-2a&a&0\\-a&0&1\end{array}\right]\left[\begin{array}{c}1\\2\\4\end{array}\right]$

$X=\frac{1}{a}\left[\begin{array}{c}4a-2a-4\\-2a+2a+0\\-a+0+4\end{array}\right]$

$X=\frac{1}{a}\left[\begin{array}{c}2a-4\\0\\4-a\end{array}\right]$

$X=\left[\begin{array}{c}\frac{2a-4}{a}\\0\\\frac{4-a}{a}\end{array}\right]$

$\therefore\textbf{The solution is}$

$\bf\,x=\frac{2a-4}{a}$

$\bf\,y=0$

$\bf\,z=\frac{4-a}{a}$

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

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