Math, asked by saleemkhan143m, 1 month ago

solve the following equation by reduction method x+y+z=2 x-2y+z=8 3x+y+z=4

Answers

Answered by Mithalesh1602398

Answer:

x+y+z=2 x-2y+z=8 3x+y+z=4 =x=1, y=-2, z=3.

Step-by-step explanation:

Step:1 Given: x+y+z=2

x-2y+z=8

3x+y+z=4

To find:

Solution of the given system by matrix inversion method

Step : 2 Solution:

The given system of equations can be written as

$\left(\begin{array}{ccc}1 & 1 & 1 \\1 & -2 & 1 \\3 & 1 & 1\end{array}\right)\left(\begin{array}{l}x \\y \\z\end{array}\right)=\left(\begin{array}{l}2 \\8 \\4\end{array}\right)$

This can be written as

A X=B

$\Longrightarrow \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}$

$|A|=\left|\begin{array}{ccc}1 & 1 & 1 \\1 & -2 & 1 \\3 & 1 & 1\end{array}\right|$

$\begin{aligned}& |A|=1(-2-1)-1(1-3)+1(1+6) \\& |A|=-3-1(-2)+1(7) \\& |A|=-3+2+7=6\end{aligned}$

$\text { Cofactor matrix of } \mathrm{A}$

$=\left(\begin{array}{ccc}+(-2-1) & -(1-3) & +(1+6) \\-(1-1) & +(1-3) & -(1-3) \\+(1+2) & -(1-1) & +(-2-1)\end{array}\right)$

$=\left(\begin{array}{ccc}-3 & 2 & 7 \\0 & -2 & 2 \\3 & 0 & -3\end{array}\right)$

adj A = $(\text { Cofactor matrix })^{\top}$

adj A = $\left(\begin{array}{ccc}-3 & 0 & 3 \\2 & -2 & 0 \\7 & 2 & -3\end{array}\right)$

$\mathrm{A}^{-1}=\frac{1}{6}\left(\begin{array}{ccc}-3 & 0 & 3 \\2 & -2 & 0 \\7 & 2 & -3\end{array}\right)$

Now,

$X=A^{-1} B$

$\mathrm{X}=\frac{1}{6}\left(\begin{array}{ccc}-3 & 0 & 3 \\2 & -2 & 0 \\7 & 2 & -3\end{array}\right)\left(\begin{array}{l}2 \\8 \\4\end{array}\right)$