Using matrices, solve the following system of linear equations:
          x+2y-3z= -4
         2x+3y+2z= 2
         3x-3y-4z= 11

 

 

 

 
 
 
 
 

Answers (1)

we have x+2y-3z= -4,\; 2x+3y+2z= 2,\; 3x-3y-4z= 11
Let  A= \begin{bmatrix} 1 & 2 & -3\\ 2&3 &2 \\ 3& -3 & -4 \end{bmatrix},\; B= \begin{bmatrix} -4\\ 2 \\ 11 \end{bmatrix},\; x= \begin{bmatrix} x\\ y \\ z \end{bmatrix}
so that
          Ax= B     i.e             x= A^{-1}B---(i)
        \because \left | A \right |= \begin{vmatrix} 1 & 2 &-3 \\ 2& 3 &2 \\ 3&-3 & -4 \end{vmatrix}
= 1\left ( -6 \right )-2\left ( -14 \right )-3\left ( -15 \right )= -6+28+45= 67\not\equiv 0
                           \therefore A^{-1}   exists
consider Aij  be the cofactor of element aij of matrix A
A_{11}= -6\;, A_{12}= 14\;, A_{13}= -15\;
A_{21}= 17\;, A_{22}= 5\;, A_{23}= 9\;
A_{31}= 13\;, A_{32}= -8\;, A_{33}= -1\;
\therefore adj\, A = \begin{bmatrix} -6 & 17 &13 \\ 14& 5&-8 \\ -15&9 & -1 \end{bmatrix}
A^{-1} =\frac{1}{67} \begin{bmatrix} -6 & 17 &13 \\ 14& 5&-8 \\ -15&9 & -1 \end{bmatrix}
by (i)   x =\frac{1}{67} \begin{bmatrix} -6 & 17 &13 \\ 14& 5&-8 \\ -15&9 & -1 \end{bmatrix}\begin{bmatrix} -4\\ 2 \\ 11 \end{bmatrix}
        x = \frac{1}{67}\begin{bmatrix} 201\\ -134 \\ 67 \end{bmatrix}
           \Rightarrow \begin{bmatrix} x\\ y \\z \end{bmatrix}= \begin{bmatrix} 3\\ -2 \\1 \end{bmatrix}   

By equality of matrices, we get
\therefore x= 3,\; y= -2,z= 1

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