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# Using matrices, solve the following system, of equations : $\large\frac{1}{x}-\frac{1}{y}+\frac{1}{z}$$=4: \: \large\frac{2}{x}+\frac{1}{y}-\frac{3}{z}$$=0: \: \large\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$=2 \qquad x \neq 0, y \neq 0, z \neq 0$

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Toolbox:
• If the value of the determinant of a $3 \times 3$ square matrix is not equal to zero, then it is a non-singular matrix
• If it is a nonsingular matrix, then inverse exists
• $A^{-1}=\frac{1}{|A|} (adjoint A)$
• $A^{-1}B=X$
Step 1:
Given $\large\frac{1}{x}-\frac{1}{y}+\frac{1}{z}=4: \: \frac{2}{x}+\frac{1}{y}-\frac{3}{z}=0: \: \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$
Let $\large\frac{1}{x}=a,\frac{1}{y}=b\;and\; \frac{1}{z}=c$
Therefore the system of equations can be written as
$a-b+c=4$
$2a+b-3c=0$
$a+b+c=2$
This is of the form $AX=B$
$(ie)\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$
where $A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix},\qquad X=\begin{bmatrix} a \\ b \\ c \end{bmatrix}\;and \; B=\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$
Let us first find the determinant value of A
$|A|=1(1+3)+1(2+3)+1(2-1)$
$=4+5+1$
$=10$
$|A| \neq 0$
Hence it is a non singular matrix and $A^{-1}$ exists
Step 2:
Let us now find the cofactors of A
$A_{11}=(-1)^{1+1} \begin{vmatrix} 1 & -3 \\ 1 & 1 \end{vmatrix}=1+3=4$
$A_{12}=(-1)^{1+2} \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix}=-(2+3)=-5$
$A_{13}=(-1)^{1+3} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix}=2-1=1$
$A_{21}=(-1)^{2+1} \begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix}=-(-1-1)=2$
$A_{22}=(-1)^{2+2} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix}=1-1=0$
$A_{23}=(-1)^{2+3} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}=-(1+1)=-2$
$A_{31}=(-1)^{3+1} \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix}=3-1=2$
$A_{32}=(-1)^{3+2} \begin{vmatrix} 1 & 1 \\ 2 & -3 \end{vmatrix}=-(-3-2)=5$
$A_{33}=(-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix}=1+2=3$
Therefore Adj $A=\begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$
$=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$
$A^{-1}=\frac{1}{|A|} adj (A),$
we know $|A|=10$
Hence $A^{-1}=\frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$
Step 3:
$AX=B \qquad => X=A^{-1}B$
Now Substituting for X, $A^{-1}$ and B we get,
$\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$
$\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\frac{1}{10}\begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{bmatrix}=\begin{bmatrix} \frac{20}{10} \\ \frac{-10}{10} \\ \frac{10}{10} \end{bmatrix}$
$\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$
Therefore $a=2,b=-1,and \; c=1$
=> $x=\large\frac{1}{2},y=-1,and \; z=1$