Q.(a) Test the consistency of the following system of linear equations by rank method. $x-y+z=-9$ $2x-y+z=4$ $3x-y+z=6$ $4x-y+2z=7$ OR

(a) reduces the augmented matrix of a 4-equation/3-unknown system by row operations and compares rank(A) with rank([A|B]) to test consistency; (b) sets $x,y$ as complex numbers on the unit circle via $2\cos\theta=z+1/z$ and uses De Moivre's theorem on $x^m,y^n$ to prove the two identities.

(a) Consistency by rank method

1. Equations: $x-y+z=-9;\ 2x-y+z=4;\ 3x-y+z=6;\ 4x-y+2z=7$. Augmented matrix $[A|B]=\begin{pmatrix}1&-1&1&|&-9\\2&-1&1&|&4\\3&-1&1&|&6\\4&-1&2&|&7\end{pmatrix}$.
2. $R_2\to R_2-2R_1,\ R_3\to R_3-3R_1,\ R_4\to R_4-4R_1$ gives $\begin{pmatrix}1&-1&1&|&-9\\0&1&-1&|&22\\0&2&-2&|&33\\0&3&-2&|&43\end{pmatrix}$.
3. $R_3\to R_3-2R_2,\ R_4\to R_4-3R_2$ gives $\begin{pmatrix}1&-1&1&|&-9\\0&1&-1&|&22\\0&0&0&|&-11\\0&0&1&|&-23\end{pmatrix}$.
4. Reading the coefficient part alone, the nonzero rows are $(1,-1,1),(0,1,-1),(0,0,1)$, so $\rho(A)=3$. The row $(0\ 0\ 0\,|\,-11)$ is a nonzero row of the augmented matrix (since $0\neq-11$), so $\rho([A|B])=4$.
5. Since $\rho(A)=3\neq\rho([A|B])=4$, the system is inconsistent — it has no solution.

(b) Proving the two identities

1. Since $2\cos\alpha=x+\dfrac1x$, $x$ satisfies $x^2-2x\cos\alpha+1=0$, giving $x=\cos\alpha\pm i\sin\alpha$; take $x=\cos\alpha+i\sin\alpha$, so $\dfrac1x=\cos\alpha-i\sin\alpha$. Likewise take $y=\cos\beta+i\sin\beta$, $\dfrac1y=\cos\beta-i\sin\beta$.
2. By De Moivre's theorem, $x^m=\cos m\alpha+i\sin m\alpha$, $\dfrac{1}{x^m}=\cos m\alpha-i\sin m\alpha$, and $y^n=\cos n\beta+i\sin n\beta$, $\dfrac{1}{y^n}=\cos n\beta-i\sin n\beta$.
3. (i) $\dfrac{x^m}{y^n}=x^m\cdot\dfrac1{y^n}=(\cos m\alpha+i\sin m\alpha)(\cos n\beta-i\sin n\beta)=\cos(m\alpha-n\beta)+i\sin(m\alpha-n\beta)$. …