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  • Using matrices, solve the following system of linear equations : <img alt="4x-6y+5z= 1" src="https://entrancecorner.codecogs

Using matrices, solve the following system of linear equations :
2x+3y+10z= 4
4x-6y+5z= 1
6x+9y-20z= 2

 

 

 

 
 
 
 
 

Answers (1)

Let  Ax= B \; where \; A= \begin{bmatrix} 2 & 3 &10 \\ 4 & -6 &5 \\ 6& 9 & -20 \end{bmatrix}X= \begin{bmatrix} x\\ y \\ z \end{bmatrix}B= \begin{bmatrix} 4\\ 1 \\ 2 \end{bmatrix}
Here  \left | A \right |= \begin{vmatrix} 2 & 3 &10 \\ 4& -6 &5 \\ 6 & 9 & -20 \end{vmatrix}= 2\left ( 120-45 \right )-3\left ( -80-30 \right )+10\left ( 36+36 \right )
                                           = 1200\neq 0\; \; \therefore A^{-1}exists.
consider Aij be the cofactor of the element aij of A
A_{11}= 75,\; A_{12}= 110,\; A_{13}= 72,A_{21}= 150
A_{22}= -100,\; A_{23}= 0,\; A_{31}= 75,A_{32}= 30,A_{33}= -24
\therefore adj A= \begin{bmatrix} 75 & 150 &75 \\ 110& -100 &30 \\ 72& 0 & -24 \end{bmatrix}
\Rightarrow A^{-1}= \frac{adj\, A}{\left | A \right |}= \frac{1}{1200}\begin{bmatrix} 75 & 150 &75 \\ 110& -100 &30 \\ 72& 0& -24 \end{bmatrix}
Now X= A^{-1}B= \frac{1}{1200}\begin{bmatrix} 75 & 150 &75 \\ 110& -100 &30 \\ 72& 0& -24 \end{bmatrix}\begin{bmatrix} 4\\ 1 \\ 2 \end{bmatrix}
\Rightarrow \begin{bmatrix} x\\ y \\ z \end{bmatrix}= \frac{1}{1200}\begin{bmatrix} 600\\400 \\240 \end{bmatrix}
\Rightarrow \begin{bmatrix} x\\ y \\ z \end{bmatrix}= \begin{bmatrix} \frac{1}{2}\\\frac{1}{3} \\\frac{1}{5} \end{bmatrix}  By equality of matrices, we get x= \frac{1}{2},\; y= \frac{1}{3},\; z= \frac{1}{5}
 

Posted by

Ravindra Pindel

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