a. Solve by inverse matrix or row equivalent method: x y = 0 , 2x y + 4z = 18 , 3x + z + 2 = 0 b. Find the equation whose roots are reciprocal of the roots of x^2 x + 1 = 0

a. Solve the system of equations using the inverse matrix or row equivalent method

Given system of equations: $$\{\begin{array}{l}x-y=0\text{(1)}\\ 2x-y+4z=18\text{(2)}\\ -3x+z+2=0\text{(3)}\end{array}$$

Step 1: Write the system in matrix form $AX=B$

The augmented matrix $[A|B]$ is: $$\left[\begin{array}{ccccc}1& -1& 0& |& 0\\ 2& -1& 4& |& 18\\ -3& 0& 1& |& -2\end{array}\right]$$

Step 2: Solve using the row equivalent method (Gaussian elimination)

We perform row operations to obtain the row echelon form.

Row Operation 1: $R_2\to R_2-2R_1$ $$\left[\begin{array}{ccccc}1& -1& 0& |& 0\\ 0& 1& 4& |& 18\\ -3& 0& 1& |& -2\end{array}\right]$$

Row Operation 2: $R_3\to R_3+3R_1$ $$\left[\begin{array}{ccccc}1& -1& 0& |& 0\\ 0& 1& 4& |& 18\\ 0& -3& 1& |& -2\end{array}\right]$$

Row Operation 3: $R_3\to R_3+3R_2$ $$\left[\begin{array}{ccccc}1& -1& 0& |& 0\\ 0& 1& 4& |& 18\\ 0& 0& 13& |& 52\end{array}\right]$$

Row Operation 4: $R_2\to R_2-4R_3$ $$\left[\begin{array}{ccccc}1& -1& 0& |& 0\\ 0& 1& 0& |& -10\\ 0& 0& 13& |& 52\end{array}\right]$$

Row Operation 5: $R_1\to R_1+R_2$ $$\left[\begin{array}{ccccc}1& 0& 0& |& -10\\ 0& 1& 0& |& -10\\ 0& 0& 1& |& 4\end{array}\right]$$

Step 3: Back substitution

From the final matrix, we get: $$x=-10,y=-10,z=4$$

Final solution: $$\boxed{x=-10,y=-10,z=4}$$

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b. Find the equation whose roots are the reciprocals of the roots of $x^2-x+1=0$

Step 1: Let the original equation be $f(x)=0$

Given: $$f(x)=x^2-x+1=0$$ Let the roots of $f(x)=0$ be $\alpha$ and $\beta$.

Step 2: Find the sum and product of the roots

For a quadratic equation $a{x}^2+bx+c=0$, the sum and product of the roots are: $$\alpha +\beta =-\frac{b}{a}=1,\alpha \beta =\frac{c}{a}=1$$

Step 3: Find the sum and product of the reciprocal roots

Let the new roots be $\frac{1}{\alpha }$ and $\frac{1}{\beta }$.

The sum of the new roots: $$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{1}{1}=1$$

The product of the new roots: $$\frac{1}{\alpha }\cdot \frac{1}{\beta }=\frac{1}{\alpha \beta }=\frac{1}{1}=1$$

Step 4: Form the new quadratic equation

The quadratic equation with roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ is: $$x^2-(\text{sum of roots})x+(\text{product of roots})=0$$ Substituting the values: $$x^2-(1)x+(1)=0$$ $$x^2-x+1=0$$

Wait! This is the same as the original equation. This suggests that the original equation is self-reciprocal (i.e., its roots' reciprocals are the same as the original roots).

However, let us verify this by solving the original equation and checking the reciprocals.

Verification:

Solve $x^2-x+1=0$: $$x=\frac{1\pm \sqrt{1-4}}{2}=\frac{1\pm \sqrt{-3}}{2}$$ The roots are complex: $\alpha =\frac{1+i\sqrt{3}}{2}$ and $\beta =\frac{1-i\sqrt{3}}{2}$.

Now, compute the reciprocals: $$\frac{1}{\alpha }=\frac{2}{1+i\sqrt{3}}=\frac{2(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}=\frac{2(1-i\sqrt{3})}{1+3}=\frac{1-i\sqrt{3}}{2}=\beta$$ Similarly, $\frac{1}{\beta }=\alpha$.

Thus, the roots are swapped, but the equation remains the same.

Conclusion: The equation whose roots are the reciprocals of the roots of $x^2-x+1=0$ is the same as the original equation.

However, if the question intended a different interpretation (e.g., forming a new equation with roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$), the answer remains: $$\boxed{x^2-x+1=0}$$

Alternative Approach (General Method): If we want to construct a new equation with roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$, we can use the substitution $y=\frac{1}{x}$:

Given $x^2-x+1=0$, substitute $x=\frac{1}{y}$: $${\left(\frac{1}{y}\right)}^2-\left(\frac{1}{y}\right)+1=0$$ Multiply through by $y^2$: $$1-y+y^2=0$$ $$y^2-y+1=0$$

Thus, the equation with roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ is: $$\boxed{x^2-x+1=0}$$