Group A

$3\times4=12$

1. Answer any four questions

(a) If $a$ be a non-zero complex number and $z$ be any complex number, then define $a^z$. What is the principle value of $a^z$?

(b) Apply Descarte's rule of sign to find the nature of the roots of the equation:

$$x^4+2x^2+3x-1=0$$

Given equation:

$$f(x)=x^4+2x^2+3x-1=0$$

The above equation is a polynomial of degree $4$. Therefore, there are $4$ roots to the above polynomial.

In order to find the number of positive roots, we have to count the number of sign changes in $f(x)$.

$$f(x)=x^4+2x^2+3x-1\\$$

The only sign change is between, $3x$ and $-1$. Therefore, the number of positive roots is $1$.

In order to find the number of negative roots, we have to count the number of sign changes in $f(-x)$.

$$f(-x)=x^4+2x^2-3x-1$$

The only sign change is between $2x^2$ and $-3x$. Therefore, the number of negative roots is $1$.

The remaining roots are imaginary roots. Therefore, imaginary roots is $4-1-1=2$.

(c) Prove that zero is an eigenvalue of a square matrix.

As per Cayley-Hamilton theorem, any square matrix satisfies its characteristic equation $\psi_A(x)$ as follows:

$$\begin{align*} \psi_A(x)=c_0x^n+c_1x^{n-1}+\ldots+c_{n-1}x+c_n=0&&&&&&&&&&(1) \end{align*}$$

But for a square matrix $A$, its determinant is: $\det A=0$. And in the characteristic equation $(1)$:

$$c_n=\det A=0$$

Therefore, rewriting the characteristic equation, we get:

$$\psi_A(x)=c_0x^n+c_1x^{n-1}+\ldots+c_{n-1}x=0$$

In the above equation, $0$ satisfies $\psi_A(x)$, therby making it an eigenvalue.

Hence proved: Zero is an eigenvalue of a square matrix.

(d) Define consistent system of equations. Show that the following system of linear equations has infinite number of solutions:

$$x_1+2x_2=674\\ 3x_1+6x_2=2022$$

Consistent system of equations: A system of equations for which a solution exists is said to be consistent system of equations. The solution could either be unique, or there might exist an infinite number of solutions, based on the following conditions:

1. $p(A)=p(A|B)=n$
2. $p(A)=p(A|B)<n$

Where,

• $p(A)$ is the rank of coefficient matrix,
• $p(A|B)$ is the rank of augmented matrix, and
• $n$ is the number of unknowns in the system of equations.

Given equations:

$$x_1+2x_2=674\\ 3x_1+6x_2=2022$$

Creating the augmented matrix:

$$\begin{align*} [A|B]&=\left[\begin{matrix} 1&2&:&674\\ 3&6&:&2022 \end{matrix}\right]\\ &=\left[\begin{matrix} 1&2&:&674\\ 0&0&:&0 \end{matrix}\right] &&[\text{By}\ R_2\rightarrow R_2-3R_1] \end{align*}$$

The above is the row reduced echelon matrix. Therefore, we can get the following:

• $p(A)=1$
• $p(A|B)=1$
• $n=2$

Since $p(A)=p(A|B)<n$, therefore there exists an infinite number of solutions for the given set of equations.

(e) Define rank of a matrix. Find the rank of the following matrix:

$$\left(\begin{matrix} 1&0&0&0\\ 1&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{matrix}\right)$$

Rank of matrix: Rank of a matrix $A$, is defined to be the greatest positive integer $r$, such that $A$ has atleast one non-zero minor order of $r$. The rank of $A$ is also called the determinant rank of $A$.

Given matrix:

$$\begin{align*} &=\left[\begin{matrix} 1&0&0&0\\ 1&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{matrix}\right]\\ &=\left[\begin{matrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{matrix}\right]&&[\text{By}\ R_2\rightarrow R_2-R_1] \end{align*}$$

The above is the row-reduced echelon form. Therefore the rank of the matrix can be found by counting the number of non-zero rows in the row-reduced echelon form. Rank of the matrix is $4$.

(f) If $a, b, c, d$ be positive real numbers, then find the minimum value of the following. When does the minimum value occur?

$$\frac ab+\frac bc+\frac cd+\frac da$$

---

Group B

$6\times4=24$

2. Answer any four questions

(a) If $\tan\left(i\log\frac{x-iy}{x+iy}\right)=2$, then prove that $x^2-y^2=xy$.

(b) If $x,y,z$ are positive real numbers and $x+y+z=1$, prove that:

$$8xyz\le(1-x)(1-y)(1-z)\le\frac8{27}$$

(c) Obtain a row echelon matrix which is row equivalent to the following matrix and hence find its rank.

$$\left(\begin{matrix} 0&0&2&2&0\\ 1&3&2&4&1\\ 2&6&2&6&2\\ 3&9&1&10&6 \end{matrix}\right)$$

The matrix is:

$$\begin{align*} &=\left(\begin{matrix} 0&0&2&2&0\\ 1&3&2&4&1\\ 2&6&2&6&2\\ 3&9&1&10&6 \end{matrix}\right)\\ &=\left(\begin{matrix} 1&3&2&4&1\\ 0&0&2&2&0\\ 2&6&2&6&2\\ 3&9&1&10&6 \end{matrix}\right)&&[R_2\leftrightarrow R_1]\\ &=\left(\begin{matrix} 1&3&2&4&1\\ 0&0&2&2&0\\ 0&0&-2&-2&0\\ 0&0&-5&-2&3 \end{matrix}\right)&&[R_3\rightarrow R_3-2R_1,R_4\rightarrow R_4-3R_1]\\ &=\left(\begin{matrix} 1&3&2&4&1\\ 0&0&-5&-2&3\\ 0&0&-2&-2&0\\ 0&0&2&2&0 \end{matrix}\right)&&[R_2\leftrightarrow R_4]\\ &=\left(\begin{matrix} 1&3&0&2&1\\ 0&0&-3&0&3\\ 0&0&-2&-2&0\\ 0&0&0&0&0 \end{matrix}\right)&&[R_4\rightarrow R_4+R_3, R_2\rightarrow R_2-R_3,R_1\rightarrow R_1+R_3]\\ &=\left(\begin{matrix} 1&3&0&2&1\\ 0&0&1&0&-1\\ 0&0&1&1&0\\ 0&0&0&0&0 \end{matrix}\right)&&[R_3\rightarrow -\frac12 R_3, R_2\rightarrow -\frac13 R_2]\\ &=\left(\begin{matrix} 1&3&0&2&1\\ 0&0&1&0&-1\\ 0&0&0&1&1\\ 0&0&0&0&0 \end{matrix}\right)&&[R_3\rightarrow R_3-R_2]\\ &=\left(\begin{matrix} 1&3&0&0&-1\\ 0&0&1&0&-1\\ 0&0&0&1&1\\ 0&0&0&0&0 \end{matrix}\right)&&[R_1\rightarrow R_1-2R_3]\\ \end{align*}$$

The above is the row reduced echelon form of the matrix. The rank can be found out by counting the number of non-zero rows, which is $3$.

(d) State Cayley-Hamilton theorem. Use it to find the inverse of the following matrix:

$$\left( \begin{matrix} 1&-1&0\\ 1&2&-1\\ 3&2&-2 \end{matrix} \right)$$

Cayley-Hamilton theorem states that, any square matrix $A$ without exception satisfies its characteristic equation, $\psi_A(x)$, where $x_n$ can be replaced with $A$:

$$\psi_A(x)=c_0x^n+c_1x^{n-1}+\ldots+c_{n-1}x+c_n=0$$

Finding the characteristic equation of the above matrix:

$$\begin{align*} &|A-\lambda I_n|=0\\ \Rightarrow\ &\left|\left[ \begin{matrix} 1&-1&0\\ 1&2&-1\\ 3&2&-2 \end{matrix} \right]-\lambda\left[ \begin{matrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{matrix} \right]\right|=0\\ \Rightarrow\ &\left| \begin{matrix} 1-\lambda&-1&0\\ 1&2-\lambda&-1\\ 3&2&-2-\lambda \end{matrix} \right|=0\\ \Rightarrow\ &(1-\lambda)[(2-\lambda)(-2-\lambda)+2]+(-2-\lambda+3)=0\\ \Rightarrow\ &(1-\lambda)(\lambda^2-2)+(1-\lambda)=0\\ \Rightarrow\ &(1-\lambda)(\lambda^2-1)=0\\ \Rightarrow\ &\lambda^2-\lambda^3-1+\lambda=0\\ \Rightarrow\ &\lambda^3-\lambda^2-\lambda+1=0 \end{align*}$$

As per Cayley-Hamilton theorem: $\lambda=A$. Therefore,

$$\begin{align*} &A^3-A^2-A+I_n=0\\ \Rightarrow\ &A^{-1}(A^3-A^2-A+I_n)=0\\ \Rightarrow\ &A^2-A-I_n+A^{-1}=0\\ \Rightarrow\ &A^{-1}=A+I_n-A^2\\ \Rightarrow\ &A^{-1}=\left[ \begin{matrix} 1&-1&0\\ 1&2&-1\\ 3&2&-2 \end{matrix} \right]+\left[ \begin{matrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{matrix} \right]-\left[ \begin{matrix} 0&-3&1\\ 0&1&0\\ -1&-3&2 \end{matrix} \right]\\ \Rightarrow\ &A^{-1}=\left[ \begin{matrix} 2&2&-1\\ 1&2&-1\\ 4&5&-3 \end{matrix} \right] \end{align*}$$

Therefore, inverse of matrix A is:

$$A^{-1}=\left[ \begin{matrix} 2&2&-1\\ 1&2&-1\\ 4&5&-3 \end{matrix} \right]$$

(e) Let $\alpha,\beta,\gamma$ be the roots of the equation $2x^3+3x^2-x-1=0$, find the equation whose roots are

$$\frac{\beta+\gamma}{\alpha},\frac{\gamma+\alpha}{\beta},\frac{\alpha+\beta}{\gamma}$$

Given equation:

$$f(x)=2x^3+3x^2-x-1=0$$

Let $y=\frac{\beta+\gamma}{\alpha}$

$$\begin{align*} y&=\frac{\beta+\gamma}{\alpha}\\ \Rightarrow \alpha y&=\beta+\gamma\\ \Rightarrow \alpha y&=\sum\alpha-\alpha\\ \Rightarrow \alpha y&=-\frac32-\alpha\\ \Rightarrow \alpha(y+1)&=-\frac32\\ \therefore \alpha&=\frac{-3}{2(y+1)} \end{align*}$$

Now $\alpha$ is a root of the equation $f(x)$. Therefore:

$$\begin{align*} f(\alpha)&=2\alpha^3+3\alpha^2-\alpha-1=0\\ &\Rightarrow 2\left(\frac{-3}{2(y+1)}\right)^3+3\left(\frac{-3}{2(y+1)}\right)^2-\left(\frac{-3}{2(y+1)}\right)-1=0\\ &\Rightarrow\frac{-27}{4(y+1)^3}+\frac{27}{4(y+1)^2}+\frac{3}{2(y+1)}-1=0\\ &\Rightarrow-27+27(y+1)+6(y+1)^2-4(y+1)^3=0\\ &\Rightarrow-4y^3-6y^2+27y+2=0\\ &\Rightarrow4y^3+6y^2-27y-2=0 \end{align*}$$

Therefore the transformed equation $f(x)'$ is:

$$f(x)'=4x^3+6x^2-27x-2=0$$

(f) Solve by Cardan's method $x^3-3x-1=0$. Hence find $\cos{\frac\pi9}\cos{\frac{7\pi}9}\cos{\frac{13\pi}9}$.

Given equation:

$$f(x)=x^3-3x-1=0$$

Let $z=u^{1/3}+v^{1/3}$.

$$\begin{align*} &z=u^{1/3}+v^{1/3}\\ \Rightarrow\ &z^3=u+v+3u^{1/3}v^{1/3}(u^{1/3}+v^{1/3})\\ \Rightarrow\ &z^3-3u^{1/3}v^{1/3}z-(u+v)=0&&[\because u^{1/3}+v^{1/3}=z] \end{align*}$$

Comparing the above equation with $f(x)$, we get:

• $u^{1/3}v^{1/3}=1\Rightarrow uv=1$, and
• $u+v=1$

Creating a quadratic equation with the above roots:

$$\begin{align*} g(x)&=x^2-(u+v)x+uv=0\\ &\Rightarrow x^2-x+1=0\\ &\Rightarrow x=\frac{1\pm\sqrt{-3}}{2}\\ &\Rightarrow x=\frac12+i\frac{\sqrt3}2,\frac12-i\frac{\sqrt3}2\\ &\Rightarrow x=\cos{\frac\pi3}+i\sin{\frac\pi3}, \cos{\frac\pi3}-i\sin{\frac\pi3} \end{align*}$$

Considering:

$$\begin{align*} u&=\cos{\frac\pi3}+i\sin{\frac\pi3}\\ \Rightarrow u^{1/3}&=\cos\left(\frac{2k\pi+\pi/3}{3}\right)+i\sin\left(\frac{2k\pi+\pi/3}{3}\right)\\ &=\cos\left(\frac{6k\pi+\pi}{9}\right)+i\sin\left(\frac{6k\pi+\pi}{9}\right) \end{align*}$$

and,

$$\begin{align*} v&=\cos{\frac\pi3}-i\sin{\frac\pi3}\\ \Rightarrow v^{1/3}&=\cos\left(\frac{2k\pi+\pi/3}{3}\right)-i\sin\left(\frac{2k\pi+\pi/3}{3}\right)\\ &=\cos\left(\frac{6k\pi+\pi}{9}\right)-i\sin\left(\frac{6k\pi+\pi}{9}\right) \end{align*}$$

Now using these to find the roots of the cubic equation are unique for $k=0,1,2$:

For $k=0$:

$$\begin{align*} x_1&=u^{1/3}+v^{1/3}\\ &=2\cos\frac{\pi}{9} \end{align*}$$

For $k=1$:

$$\begin{align*} x_2&=u^{1/3}+v^{1/3}\\ &=2\cos\frac{7\pi}{9} \end{align*}$$

For $k=2$:

$$\begin{align*} x_3&=u^{1/3}+v^{1/3}\\ &=2\cos\frac{13\pi}{9} \end{align*}$$

Therefore, for $f(x)=x^3-3x-1=0$, the roots are:

$$x=2\cos\frac{\pi}{9},2\cos\frac{7\pi}{9},2\cos\frac{13\pi}{9}$$

We know for the above cubic equation:

$$\begin{align*} &x_1x_2x_3=-1\\ \Rightarrow\ &2^3\left(\cos\frac{\pi}{9}\cos\frac{7\pi}{9}\cos\frac{13\pi}{9}\right)1\\ \therefore\ &\cos\frac{\pi}{9}\cos\frac{7\pi}{9}\cos\frac{13\pi}{9}=\frac{1}{8} \end{align*}$$

---

Group C

$6\times2=12$

3. Answer any two questions

(a) (i) State De-Moivre's theorem for integer and rational indices. Use it to prove that

$$\cos 5\phi=16\cos^5\phi-20\cos^3\phi+5\cos\phi$$

De-Moivre's theorem: When $n$ is a fraction, positive or negative, and $\phi$ is a real number, then $\cos n\phi+i\sin n\phi$ is one of the values of $(\cos n\phi+i\sin n\phi)^n$. As:

$$(\cos\phi+i\sin\phi)^n=\cos n\phi+i\sin n\phi$$

Considering $(\cos\phi+i\sin\phi)^5$:

$$\begin{align*} (\cos\phi+i\sin\phi)^5&=\cos^5\phi+\ ^5C_1\cos^4\phi\cdot i\sin\phi+\ ^5C_2\cos^3\phi\cdot i^2\sin^2\phi+\ ^5C_3\cos^2\phi\cdot i^3\sin^3\phi+\ ^5C_4\cos\phi\cdot i^4\sin^4\phi+i^5\sin^5\phi\\ &=\cos^5\phi+5\cos^4\phi\cdot i\sin\phi-10\cos^3\phi\sin^2\phi-10\cos^2\phi\cdot i\sin^3\theta+5\cos\phi\sin^4\phi+i\sin^5\phi\\ &=(\cos^5\phi-10\cos^3\sin^2\phi+5\cos\phi\sin^4\phi)+i(5\cos^4\phi\sin\phi-10\cos^2\phi\sin^3\phi+\sin^5\phi) \end{align*}$$

But we know: $(\cos\phi+i\sin\phi)^5=\cos5\phi+i\sin5\phi$. Comparing this with the equation above, we get:

$$\begin{align*} \cos5\phi&=\cos^5\phi-10\cos^3\sin^2\phi+5\cos\phi\sin^4\phi\\ &=\cos^5\phi-10\cos^3\phi(1-\cos^2\phi)+5\cos\phi(1-\cos^2\phi)^2\\ &=\cos^5\phi-10\cos^3\phi+10\cos^5\phi+5\cos\phi(1+\cos^4\phi-2\cos^2\phi)\\ &=11\cos^5\phi-10\cos^3\phi+5\cos\phi+5\cos^5\phi-10\cos^3\phi\\ &=16\cos^5\phi-20\cos^3\phi+5\cos\phi \end{align*}$$

Hence proved that:

$$\cos 5\phi=16\cos^5\phi-20\cos^3\phi+5\cos\phi$$

(a) (ii) Find $\text{mod }z$ and $\arg z$ where $z=1+i\tan\frac{3\pi}{5}$.

(b) (i) Find the eigenvalues and the corresponding eigenvectors of the matrix

$$\left[ \begin{matrix} 2&2&1\\ 1&3&1\\ 1&2&2 \end{matrix} \right]$$

Let the matrix be $A$, as:

$$\begin{align*} A&=\left[ \begin{matrix} 2&2&1\\ 1&3&1\\ 1&2&2 \end{matrix} \right]\\ \Rightarrow\psi_A(x)&=|A-\lambda I_n|=0\\ &\Rightarrow\left| \begin{matrix} 2-\lambda&2&1\\ 1&3-\lambda&1\\ 1&2&2-\lambda \end{matrix} \right|=0\\ &\Rightarrow(2-\lambda)[(3-\lambda)(2-\lambda)-2]-2(2-\lambda-1)+(2-3+\lambda)=0\\ &\Rightarrow(2-\lambda)(6-5\lambda+\lambda^2-2)-2(1-\lambda)-(1-\lambda)=0\\ &\Rightarrow(2-\lambda)(\lambda^2-5\lambda+4)-3(1-\lambda)=0\\ &\Rightarrow2\lambda^2-10\lambda+8-\lambda^3+5\lambda^2-4\lambda-3+3\lambda=0\\ &\Rightarrow-\lambda^3+7\lambda^2-11\lambda+5=0\\ &\Rightarrow\lambda^3-7\lambda^2+11\lambda-5=0\\ &\Rightarrow(\lambda-1)(\lambda^2-6\lambda+5)=0\\ &\Rightarrow(\lambda-1)(\lambda-1)(\lambda-5)=0\\ &\therefore \lambda=1,1,5 \end{align*}$$

Therefore: $\lambda=1,1,5$ are the necessary eigenvalues. Now finding the eigenvectors:

For $\lambda=5$:

$$\psi_A(5)=|A-5I_n|=\left| \begin{matrix} -3&2&1\\ 1&-2&1\\ 1&2&-3 \end{matrix} \right|$$

Solving for $x_1,x_2,x_3$:

$$\begin{align*} &\frac{x_1}{\left| \begin{matrix} -2&1\\ 2&-3 \end{matrix} \right|}=\frac{-x_2}{\left| \begin{matrix} 1&1\\ 1&-3 \end{matrix} \right|}=\frac{x_3}{\left| \begin{matrix} 1&-2\\ 1&2 \end{matrix} \right|}\\ &=\frac{x_1}{4}=\frac{x_2}{4}=\frac{x_3}{4}\\ &=x_1=x_2=x_3 \end{align*}$$

Let $x_1=x_2=x_3=k$:

$$\therefore \left( \begin{matrix} x_1\\ x_2\\ x_3 \end{matrix} \right)_{\lambda=5}=k\left( \begin{matrix} 1\\ 1\\ 1 \end{matrix} \right)$$

For $\lambda=1$:

$$\begin{align*} &Ax=0\\ \Rightarrow\ &\left[ \begin{matrix} 1&2&1\\ 1&2&1\\ 1&2&1 \end{matrix} \right]\left( \begin{matrix} x_1\\ x_2\\ x_3 \end{matrix} \right)=\left( \begin{matrix} 0\\ 0\\ 0 \end{matrix} \right)\\ \Rightarrow\ &\left[ \begin{matrix} 1&2&1\\ 0&0&0\\ 0&0&0 \end{matrix} \right]\left( \begin{matrix} x_1\\ x_2\\ x_3 \end{matrix} \right)=\left( \begin{matrix} 0\\ 0\\ 0 \end{matrix} \right)&&[A:R_2\rightarrow R_2-R_1,R_3\rightarrow R_3-R_1] \end{align*}$$

Therefore, we get: $x_1+2x_2+x_3=0$.

Considering through:

• $x_2=0, x_3=k$, we get: $x_1=-k$. Therefore eigenvector:

  $$\therefore \left( \begin{matrix} x_1\\ x_2\\ x_3 \end{matrix} \right)_{\lambda=1}=k\left( \begin{matrix} -1\\ 0\\ 1 \end{matrix} \right)$$

• $x_2=k,x_3=0$, we get: $x_1=-2k$. Therefore eigenvector:

  $$\therefore \left( \begin{matrix} x_1\\ x_2\\ x_3 \end{matrix} \right)_{\lambda=1}=k\left( \begin{matrix} -2\\ 1\\ 0 \end{matrix} \right)$$

$$\therefore \left( \begin{matrix} x_1\\ x_2\\ x_3 \end{matrix} \right)_{\lambda=1}=k\left( \begin{matrix} -1\\ 0\\ 1 \end{matrix} \right),k\left( \begin{matrix} -2\\ 1\\ 0 \end{matrix} \right)$$

(b) (ii) Determine all values of $(1+\sqrt{3}i)^{\frac34}$ and show that their product is 8.

Given complex number is: $(1+\sqrt{3}i)^{\frac34}$. Converting it into trigonometric form:

$$\begin{align*} \Rightarrow(1+\sqrt{3}i)^{\frac34}&=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)^{\frac34}\\ &=8(\cos\pi+i\sin\pi)^{\frac14}\\ &=\sqrt[4]{8}\left[\cos\left(\frac{2k\pi+\pi}{4}\right)+i\sin\left(\frac{2k\pi+\pi}{4}\right)\right]\\ \end{align*}$$

The unique roots $z_i$ are found for $k=0,1,2,3$.

For $k=0$:

$$\begin{align*} z_1=\sqrt[4]{8}\left[\cos\frac\pi4+i\sin\frac\pi4\right]=\sqrt[4]{8}\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right) \end{align*}$$

For $k=1$:

$$\begin{align*} z_2=\sqrt[4]{8}\left[\cos\frac{3\pi}4+i\sin\frac{3\pi}4\right]=\sqrt[4]{8}\left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right) \end{align*}$$

For $k=2$:

$$\begin{align*} z_3=\sqrt[4]{8}\left[\cos\frac{5\pi}4+i\sin\frac{5\pi}4\right]=\sqrt[4]{8}\left(-\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right) \end{align*}$$

For $k=3$:

$$\begin{align*} z_1=\sqrt[4]{8}\left[\cos\frac{7\pi}4+i\sin\frac{7\pi}4\right]=\sqrt[4]{8}\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right) \end{align*}$$

Multiplying the roots, we get:

$$\begin{align*} z_1z_2z_3z_4&=(\sqrt[4]8)^4\left[\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)\left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)\left(-\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)\right]\\ &=8\left[(-1,0)\left(-\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)\right]\\ &=8[(-1,0)(-1,0)]\\ &=8 \end{align*}$$

Hence proved, $z_1z_2z_3z_4=8$.

(c) (i) Solve the system of linear equations given by:

$$2x+4y+6z+4w=4\\ 2x+5y+7z+6w=3\\ 2x+3y+5z+2w=5$$

Forming the augmented matrix:

$$\begin{align*} [A|B]&=\left[ \begin{matrix} 2&4&6&4&:&4\\ 2&5&7&6&:&3\\ 2&3&5&2&:&5 \end{matrix} \right]\\ &=\left[ \begin{matrix} 1&2&3&2&:&2\\ 2&5&7&6&:&3\\ 2&3&5&2&:&5 \end{matrix} \right]&&[R_1\rightarrow\frac12 R_1]\\ &=\left[ \begin{matrix} 1&2&3&2&:&2\\ 0&1&1&2&:&-1\\ 0&-1&-1&-2&:&-1 \end{matrix} \right]&&[R_2\rightarrow R_2-2R_1,R_3\rightarrow R_3-2R_1]\\ &=\left[ \begin{matrix} 1&0&1&-2&:&4\\ 0&1&1&2&:&-1\\ 0&0&0&0&:&0 \end{matrix} \right]&&[R_3\rightarrow R_3+R_2,R_1\rightarrow R_1-2R_2]\\ \end{align*}$$

The above is the row-reduced echelon form of the augmented matrix.

Since, $p[A]=p[A|B]<n$, where $n=3$, therefore there exists an infinite number of solutions for the above set of equations.

Deriving equations from it:

$$\begin{align*} &x+z-2w=4&&&&&&&&&&(1)\\ &y+z+2w=-1&&&&&&&&&&(2) \end{align*}$$

Taking, $z=t_1,w=t_2$, where: $\{t_1,t_2\}\in\R\in[-\infty,+\infty]$.

Substituting the values of $z,w$ in $(1)$, we get:

$$x=4-t_1+2t_2$$

Substituting the values of $z,w$ in $(2)$, we get:

$$y=4-t_1-2t_2$$

Since all the values $x,y,z,w$ dependes on $t_1,t_2$.

Thereofore: $\{x,y,z,w\}\in\R\in[-\infty,+\infty]$, i.e., an infinite number of solutions.

(c) (ii) If $A$ and $B$ be invertible matrices of the same order then show that $AB$ is invertible and $(AB)^{-1}=B^{-1}A^{-1}$. Hence show that $(A^{20})^{-1}=(A^{-1})^{20}$.

Considering two $n\times n$ matrices $A$ and $B$.

Since $A$ is invertible, it is non-singular, $A^{-1}$ exists, and $\det A\ne0$. Furthermore by the definition of an inverse, $A^{-1}A=AA^{-1}=I_n$.

Similarly since $B$ is invertible, it is non-singular, $B^{-1}$ exists, and $\det B\ne0$. Furthermore, by the definition of an inverse, $B^{-1}B=BB^{-1}=I_n$.

$$\det AB=\det A\cdot\det B\ne0$$

Therefore, $AB$ is a non-singular matrix, and its inverse exists.

Now:

$$\begin{align*} (AB)(B^{-1}A^{-1})&=A(BB^{-1})A^{-1}\\ &=(AI_n)A^{-1}\\ &=AA^{-1}\\ &=I_n \end{align*}$$

Similarly:

$$\begin{align*} (B^{-1}A^{-1})(AB)&=B^{-1}(A^{-1}A)B\\ &=(B^{-1}I_n)B\\ &=B^{-1}B\\ &=I_n \end{align*}$$

Therefore, we can say:

$$\begin{align*} (B^{-1}A^{-1})(AB)&=I_n\\ \Rightarrow (B^{-1}A^{-1})&=(AB)^{-1} \end{align*}$$

Hence proved: $(AB)^{-1}=B^{-1}A^{-1}$

Now:

$$\begin{align*} A^{20}(A^{-1})^{20}&=(AA^{-1})^20\\ &=I_n^{20}\\ &=I_n\\ \Rightarrow(A^{-1})^{20}&=(A^{20})^{-1} \end{align*}$$

Hence proved: $(A^{-1})^{20}=(A^{20})^{-1}$

(d) (i) If $\alpha,\beta,\gamma$ be the roots of the equation $x^3+qx+1=0$, find the equation whose roots are

$$\frac\alpha\beta+\frac\beta\alpha,\frac\beta\gamma+\frac\gamma\beta,\frac\gamma\alpha+\frac\alpha\gamma$$

Given equation:

$$f(x)=x^3+qx+1=0$$

Let $y$ be one of the new roots:

$$\begin{align*} y&=\frac\alpha\beta+\frac\beta\alpha\\ &=\frac{\alpha^2+\beta^2}{\alpha\beta}.\frac\gamma\gamma\\ &=\frac{\left(\sum\alpha^2-\gamma^2\right)\gamma}{-1}\\ &=\frac{\left((\sum\alpha)^2-2\sum\alpha\beta-\gamma^2\right)\gamma}{-1}\\ &=\frac{(-2q-\gamma^2)\gamma}{-1}\\ &=2q\gamma+\gamma^3\\ &\Rightarrow2q\gamma+\gamma^3-y=0=y_\gamma \end{align*}$$

Performing:

$$\begin{align*} y_\gamma-f(\gamma)&=2q\gamma+\gamma^3-y-\gamma^3-q\gamma-1=0\\ &\Rightarrow q\gamma-y-1=0\\ &\Rightarrow \gamma=\frac{y+1}{q} \end{align*}$$

Now, $\gamma$ is a root of the above equation:

$$\begin{align*} \therefore f(\gamma)&=\gamma^3+q\gamma+1=0\\ &\Rightarrow\left(\frac{y+1}{q}\right)^3+q\left(\frac{y+1}{q}\right)+1=0\\ &\Rightarrow(y+1)^3+q^3(y+1)+q^3=0\\ &\Rightarrow y^3+3y^2+(q^3+3)y+(2q^3+1)=0 \end{align*}$$

Therefore the transformed equation for the new roots is:

$$f(x)=x^3+3x^2+(q^3+3)x+(2q^3+1)=0$$

(d) (ii) If $\alpha,\beta,\gamma$ be the roots of the equation $x^3+qx+1=0$, then prove that

• (I) $\sum\alpha^3=-3$
• (II) $\sum\alpha^5=5q$

Given equation:

$$f(x)=x^3+qx+1=0$$

From here we can find out that:

• $\sum\alpha=-b/a=0$
• $\sum\alpha\beta=c/a=q$
• $\alpha\beta\gamma=-d/a=-1$
• $\sum\alpha^2=(\sum\alpha)^2-2\sum\alpha\beta=-2q$

(I) To prove: $\sum\alpha^3=-3$

We know: $f(\alpha)=f(\beta)=f(\gamma)=0$, since $\alpha,\beta,\gamma$ are the roots of $f(x)$.

$$\begin{align*} \therefore f(\alpha)+f(\beta)+f(\gamma)&=(\alpha^3+\beta^3+\gamma^3)+q(\alpha+\beta+\gamma)+3=0\\ &\Rightarrow\sum\alpha^3+q\sum\alpha+3=0\\ &\Rightarrow\sum\alpha^3=-3&&[\because \sum\alpha=0] \end{align*}$$

Hence proved: $\sum\alpha^3=-3$

(II) To prove: $\sum\alpha^5=5q$

Performing the following:

$$\alpha^2f(\alpha)=\alpha^5+q\alpha^3+\alpha^2=0\\ \beta^2f(\beta)=\beta^5+q\beta^3+\beta^2=0\\ \gamma^2f(\gamma)=\gamma^5+q\gamma^3+\gamma^2=0$$ $$\begin{align*} \therefore\ &\alpha^2f(\alpha)+\beta^2f(\beta)+\gamma^2f(\gamma)=0\\ \Rightarrow\ &(\alpha^5+\beta^5+\gamma^5)+q(\alpha^3+\beta^3+\gamma^3)+(\alpha^2+\beta^2+\gamma^2)=0\\ \Rightarrow\ &\sum\alpha^5+q\sum\alpha^3+\sum\alpha^2=0\\ \Rightarrow\ &\sum\alpha^5-3q-2q=0&&[\because\sum\alpha^3=-3,\sum\alpha^2=-2q]\\ \Rightarrow\ &\sum\alpha^5=5q \end{align*}$$

Hence proved: $\sum\alpha^5=5q$