Q.(a) Solve the following system of equations, using matrix inversion method. $2x_1+3x_2+3x_3=5$ $x_1-2x_2+x_3=-4$ $3x_1-x_2-2x_3=3$ OR

(a) Inverts the coefficient matrix via its adjoint and multiplies by the constant vector to solve the linear system; (b) converts to polar form to show the equation forces $r=0$ or a fixed nonzero modulus with four distinct arguments, five roots total. Both alternatives answered below.

(a) Solve by matrix inversion

1. Write in matrix form $AX=B$.

$A=\begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix},\quad X=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix},\quad B=\begin{bmatrix}5\\-4\\3\end{bmatrix}$

2. Find $|A|$.

$|A|=2[(-2)(-2)-(1)(-1)]-3[(1)(-2)-(1)(3)]+3[(1)(-1)-(-2)(3)]$

$=2(4+1)-3(-2-3)+3(-1+6)=2(5)-3(-5)+3(5)=10+15+15=40 e0$

So $A^{-1}$ exists.

3. Cofactors of $A$.

$C_{11}=5,\ C_{12}=5,\ C_{13}=5,\ C_{21}=3,\ C_{22}=-13,\ C_{23}=11,\ C_{31}=9,\ C_{32}=1,\ C_{33}=-7$

4. Adjoint (transpose of the cofactor matrix):

$\text{adj}(A)=\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}$

5. $A^{-1}=\dfrac1{40}\,\text{adj}(A)$, then $X=A^{-1}B$:

$x_1=\dfrac{5(5)+3(-4)+9(3)}{40}=\dfrac{25-12+27}{40}=\dfrac{40}{40}=1$

$x_2=\dfrac{5(5)+(-13)(-4)+1(3)}{40}=\dfrac{25+52+3}{40}=\dfrac{80}{40}=2$

$x_3=\dfrac{5(5)+11(-4)+(-7)(3)}{40}=\dfrac{25-44-21}{40}=\dfrac{-40}{40}=-1$

6. Verify in the original equations: $2(1)+3(2)+3(-1)=5$✓; $1-2(2)+(-1)=-4$✓; $3(1)-2-2(-1)=3$✓.

(b) Show $z^3+2\bar z=0$ has five solutions

1. Polar form. Let $z=re^{i\theta}$, $r\ge0$, so $\bar z=re^{-i\theta}$. The equation becomes

$r^3e^{3i\theta}+2re^{-i\theta}=0$

2. Case $r=0$. This satisfies the equation, giving $z=0$ — one solution.

3. Case $r>0$: divide by $r$.

$r^2e^{3i\theta}+2e^{-i\theta}=0$

4. Multiply through by $e^{i\theta}$. …