Question

Using theorem, solve the equation
i. x + y + z = 9
ii. 2x + 5y +7z = 52
iii. 2x + y – z =0

Answer 1

difficult sum but okay I will send you answer

Answer 2

$x=1,y=3,z=5$ is the solution of the given equation.

Step-by-step explanation:

The given equations are $\[x+y+z=9,2x+5y+7z=52,2x+y-z=0\]$

The matrix equation is written as AX=B

$\left( \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \\\end{matrix} \right)\left( \begin{matrix} x \\ y \\ z \\\end{matrix} \right)=\left( \begin{matrix} 9 \\ 52\\ 0 \\\end{matrix} \right)$

Augmented matrix,

$\[\left[ A,B \right]\]=\left( \begin{matrix} 1 & 1 & 1 & 9 \\ 2 & 5 & 7 &52 \\ 2 & 1 & -1&0 \\\end{matrix} \right)$

to simplify the matrix,

$\[ & {{R}_{2}}\to {{R}_{2}}-2{{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-2{{R}_{1}} \\ \end{align}\]$

$\[\left[ A,B \right]\]=\left( \begin{matrix} 1 & 1 & 1 & 9 \\ 0 & 3 & 5 &34 \\ 0 &- 1 & -3&-18 \\\end{matrix} \right)$

to simplify the matrix,

$\[ & {{R}_{3}}\to {{3R}_{3}}+{{R}_{2}} \\ \end{align}\]$

$\[\left[ A,B \right]\]\sim \]\left( \begin{matrix} 1 & 1 & 1 & 9 \\ 0 & 3 & 5 &34 \\ 0 &0 & -4&-20 \\\end{matrix} \right)$

Now,

$A \[\sim \]\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 3 & 5 \\ 0 &0 & -4\\\end{matrix} \right)$

$P(A)=3$

$\[\left[ A,B \right]\]\sim \]\left( \begin{matrix} 1 & 1 & 1 & 9 \\ 0 & 3 & 5 &34 \\ 0 &0 & -4&-20 \\\end{matrix} \right)$

has the three non-zero rows,

$P(\[\left[ A,B \right]\])=3$

That is,$P(A)=P(\[\left[ A,B \right]\])=3=$number of unknowns.

So, the given system is consistent and has unique solution.

To find the solution, we rewrite the echelon form into the matrix form.

$\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 3 & 5 \\ 0 & 0 & -4 \\\end{matrix} \right)\left( \begin{matrix} x \\ y \\ z \\\end{matrix} \right)=\left( \begin{matrix} 9 \\ 34\\ -20 \\\end{matrix} \right)$

$x+y+z=9$          (1)

$2x+5y+7z=52$   (2)

$2x+y-z=0$         (3)

(3) $\[\Rightarrow \]$ $z=5$

(2) $\[\Rightarrow \]$ $3y=34-25=9$

$y=3$

(1) $\[\Rightarrow \]$

$x=9-3-5\\\\x=1$

$x=1,y=3,z=5$ is the unique solution of the given equation.