Using matrix method,Solve the following system of equation:
3x-2y+3z= 8
2x+y-z= 1
4x-3y+2z= 4

 

 

 

 
 
 
 
 

Answers (1)

3x-2y+3z= 8
2x+y-z= 1          \left ( given \right )
4x-2y+2z= 4
These equation can be written in the form Ax= 13
where
A= \begin{bmatrix} 3 & -2& 3\\ 2 &1 &-1 \\ 4& -3 &2 \end{bmatrix}\: x= \begin{bmatrix} x\\ y \\ z \end{bmatrix}\: and\: B= \begin{bmatrix} 8\\ 1 \\4 \end{bmatrix}
Now
\left | A \right |= 3\left ( 2-3 \right )+2\left ( 4+4 \right )+3\left ( -6-4 \right )
       = -17\neq 0
Hence,A is non-singular and so its inverse exists cofactors of A are,
A_{11}= -1 \: \:\; A_{12}= -8\: \:\; A_{13}= -10
A_{21}= -5 \: \:\; A_{22}= -6\: \:\; A_{23}= 1
A_{31}= -1 \: \:\; A_{32}= 9\: \:\; A_{33}= 7
\therefore adj\: A= \begin{bmatrix} -1 & -8& -10\\ -5&-6 & 1\\ -1& 9 & 7 \end{bmatrix}
                   = \begin{bmatrix} -1 & -5& -1\\ -8&-6 & 9\\ 10& 1 & 7 \end{bmatrix}  
Now A^{-1}= \frac{1}{\left | A \right |}\: adj\: A
                 = \frac{1}{-17}\begin{bmatrix} -1 & -5& -1\\ -8& -6 &9 \\ -10& 1 &7 \end{bmatrix}
                \Rightarrow x= A^{-1}B
                   =\frac{-1}{17}\begin{bmatrix} -1 & -5 &-1 \\ -8& -6&9 \\ -10& 1 & 7 \end{bmatrix}\begin{bmatrix} 8\\ 1 \\4 \end{bmatrix}
           \begin{bmatrix} x\\ y \\z \end{bmatrix}= \frac{-1}{17}\begin{bmatrix} -17\\-34 \\ -51 \end{bmatrix}= \begin{bmatrix} 1\\ 2 \\ 3 \end{bmatrix}   
Hence,x= 1,y= 2,z= 3

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