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  • x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2

x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2

Answers (1)

given that

\\ x+y+2z=1\\ 3x+2y+z=7\\ 2x+y+3z=2\\

find: solutions

Ans:

this system can be written as

\begin{bmatrix} 1 &1 &2 \\ 3 &2 &1 \\2 &1 &3 \end{bmatrix} \begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 1\\7 \\2 \end{bmatrix} or\ BX=C

As\ B^{-1}\ \text{exists the given system has unique solutions}

X=B^{-1}C

we can solve and find the inverse of B

B^{-1}=\frac{1}{4}\begin{bmatrix} -5 &1 &3 \\7 &1 &-5 \\ 1& -1 &1 \end{bmatrix}

X=\begin{bmatrix} x\\y \\z \end{bmatrix}= \frac{1}{4}\begin{bmatrix} -5 &1 &3 \\7 &1 &-5 \\ 1& -1 &1 \end{bmatrix}\begin{bmatrix} 1\\7 \\2 \end{bmatrix}

\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 2\\ 1 \\ -1 \end{bmatrix}

Hence, the solution is x=2, y=1, z=-1

 

Posted by

Ramraj Saini

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