Question

x+3y+2z=6, 3x-2y-5z=5, 2x-3y+6z=7 solve them by the method of reduction.​

Answer 1

Answer:

Is this for matrices ??

If yes, then

Step-by-step explanation:

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Answer 2

Answer: x+3y+2z=6, 3x-2y-5z=5, 2x-3y+6z=7

Step-by-step explanation:

The given equations can be written in the matrix form as:

$\left[\begin{array}{ccc}1&3&2\\3&-2&5\\2&-3&6\end{array}\right]$  $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}6\\5\\7\end{array}\right]$

$\left[\begin{array}{ccc}1&3&2\\0&11&1\\7&8&9\end{array}\right]$  $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}6\\13\\-5\end{array}\right]$

$\left[\begin{array}{ccc}1&3&2\\0&2&3\\0&-9&2\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}6\\8\\-5\end{array}\right]$

$\left[\begin{array}{ccc}1&3&2\\0&1&3/2\\0&-9&2\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}6\\4\\-5\end{array}\right]$

$\left[\begin{array}{ccc}1&3&2\\0&1&3/2\\0&0&31/2\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}6\\4\\31\end{array}\right]$

∴$\left[\begin{array}{ccc}x+3y+2z\\0+y+3/2z\\0+0+31/2z\end{array}\right]$ = $\left[\begin{array}{ccc}6\\4\\31\end{array}\right]$

By equality of matrices,  

x + 3y + 2z = 6 …(1)

y + 3/2 z = 4 …(2)

31/2 z= 31  ...(3)

From (3),

z=3

Substituting z=2 in (2), we get,

y+3/2 z= 4

y+ 3/2(2)= 4

y+3= 4

y=4

Substituting y=1, z=2 in (2), we get,

x+3y+2z=6

x+3(1)+2(2)= 6

x+3+4=6

x=-1

Hence,

x=-1

y=1

z=2 is the required solution.

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