Q.(a) Solve the following system of equations by rank method :
$x+y+z=9,\ 2x+5y+7z=52,\ 2x+y-z=0$

(a) Rank method gives $\rho(A)=\rho([A|B])=3$, so unique solution $x=1,y=3,z=5$. (b) Partial fractions give $A=1,B=2,C=0$, so the integral is $\log|x-1|+\log(x^2+5)+C$.

Part (a) — Rank (echelon) method for $x+y+z=9,\ 2x+5y+7z=52,\ 2x+y-z=0$.

Augmented matrix:

$[A\,|\,B]=\begin{bmatrix}1&1&1&|&9\\2&5&7&|&52\\2&1&-1&|&0\end{bmatrix}.$

$R_2\to R_2-2R_1,\ R_3\to R_3-2R_1$:

$\begin{bmatrix}1&1&1&|&9\\0&3&5&|&34\\0&-1&-3&|&-18\end{bmatrix}.$

$R_3\to 3R_3+R_2$ (to clear the second column):

$\begin{bmatrix}1&1&1&|&9\\0&3&5&|&34\\0&0&-4&|&-20\end{bmatrix}.$

Now $\rho(A)=3$ and $\rho([A|B])=3=$ number of unknowns, so the system is consistent with a unique solution.

Back-substitute: $-4z=-20\Rightarrow z=5$; $3y+5z=34\Rightarrow 3y=34-25=9\Rightarrow y=3$; $x+y+z=9\Rightarrow x=9-3-5=1.$

Part (b) — Integration by partial fractions. Write

$\frac{3x^{2}-2x+5}{(x-1)(x^{2}+5)}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+5}.$

Then $3x^{2}-2x+5=A(x^{2}+5)+(Bx+C)(x-1).$

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