Let A = \(\left[ \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3.
The only third order minor is |A| = \(\left| \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right| \) = (-2)(5)(0) = 0. So ρ(A) ≤ 2.
There are several second order minors. We find that there is a second order minor, for example, \(\left| \begin{matrix} -2 & 2 \\ 0 & 5 \end{matrix} \right| \) = (-2)(5) = -10 ≠ 0. So, ρ(A) = 2.
Note that there are two non-zero rows. The third row is a zero row.

Given adj (A) =\(\left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
We know that A^-1 = ±\(\frac { 1 }{ \sqrt { |adjA| } } \) (adj A) ...............(1)
|adj A| = 0 + 2\(\left| \begin{matrix} 6 & -6 \\ -3 & 6 \end{matrix} \right| \) + 0
[Expanded along R₁]
= 2(36-18) = 2(18) = 36
∴ A^-1 = \(\pm \frac { 1 }{ \sqrt { 36 } } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
= \(\pm \frac { 1 }{ 6 } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \).

Here Δ = b²−4ac = 0 for equal roots. This implies 4(k + 2)² = 4(9)k. This implies k = 4 or 1.

We first find adj A. By definition, we get adj A = \({ \left[ \begin{matrix} +{ M }_{ 11 } & -{ M }_{ 12 } \\ -{ M }_{ 21 } & +{ M }_{ 22 } \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} d & -c \\ -b & a \end{matrix} \right] }^{ T }=\left[ \begin{matrix} d & -c \\ -c & a \end{matrix} \right] \).
Since A is non-singular, |A| = ad - bc ≠ 0.
As \({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } \) adj A, we get A^-1 = \(\frac { 1 }{ ad-bc } \left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right] \).

Since α, β, and γ are the roots of the equation x³+ px²+ qx + r = 0, we have
Σ₁ α + β + γ = -p and Σ₃ αβγ = -r
\(\Sigma \frac { 1 }{ \beta \gamma } =\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } +\frac { 1 }{ \alpha \beta } =\frac { \alpha +\beta +\gamma }{ \alpha \beta \gamma } =\frac { -p }{ -r } =\frac { p }{ r } \).

Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

Consider the polynomial equation
\(a_{n} x^{n}+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\cdots+a_{2} x^{2}+a_{1} x+a_{0}=0, \)  ......(1)
Replacing \(x \text { by } \frac{1}{x}\) in (1), we get
\(P\left(\frac{1}{x}\right)=\frac{a_{n}}{x^{n}}+\frac{a_{n-1}}{x^{n-1}}+\frac{a_{n-2}}{x^{n-2}}+\cdots+\frac{a_{2}}{x^{2}}+\frac{a_{1}}{x}+a_{0}=0\)    .....(2)
Multiplying both sides of (2) by xⁿ, we get
\(x^{n} P\left(\frac{1}{x}\right)=a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n-2} x^{2}+a_{n-1} x+a_{n}=0\)       .......(3)
Now, (1) is a reciprocal equation \(\Leftrightarrow P(x)=\pm x^{n} P\left(\frac{1}{x}\right) \Leftrightarrow\) (1) and (3) are same
This is possible \(\Leftrightarrow \frac{a_{n}}{a_{0}}=\frac{a_{n-1}}{a_{1}}=\frac{a_{n-2}}{a_{2}}=\cdots=\frac{a_{2}}{a_{n-2}}=\frac{a_{1}}{a_{n-1}}=\frac{a_{0}}{a_{n}} .\)
Let the proportion be equal to λ. Then, we get \(\frac{a_{n}}{a_{0}}=\lambda \ \text { and } \ \frac{a_{0}}{a_{n}}=\lambda\) Multiplying these equations, we get \(\lambda\)² = 1. So, we get two cases \(\lambda\) = 1and \(\lambda\) = −1.
Case (i) :
\(\lambda\) = 1 In this case, we have \(a_{n}=a_{0}, a_{n-1}=a_{1}, a_{n-2}=a_{2}, \cdots\)
That is, the coefficients of (1) from the beginning are equal to the coefficients from the end.
Case (ii) :
\(\lambda\) = −1 In this case, we have \(a_{n}=a_{0}, a_{n-1}=a_{1}, a_{n-2}=a_{2}, \cdots\)
That is, the coefficients of (1) from the beginning are equal in magnitude to the coefficients from the end, but opposite in sign.

Let A be non-singular. Then \(|A| \neq 0\) and A⁻¹ exists. By definition
\(A A^{-1}=A^{-1} A=I_{n}\)
(i) By (1), we get \(\left|A A^{-1}\right|=\left|A^{-1} A\right|=\left|I_{n}\right|\)
Using the product rule for determinants, we get \(|A|\left|A^{-1}\right|=\left|I_{n}\right|=1\)
Hence, \(\left|A^{-1}\right|=\frac{1}{|A|}\)
(ii) From (1), we get \(\left(A A^{-1}\right)^{T}=\left(A^{-1} A\right)^{T}=\left(I_{n}\right)^{T} .\)
Using the reversal law of transpose, we get \(\left(A^{-1}\right)^{T} A^{T}=A^{T}\left(A^{-1}\right)^{T}=I_{n}\). Hence \(\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}\)
(iii) Since λ is a non-zero scalar, from (1), we get \((\lambda A)\left(\frac{1}{\lambda} A^{-1}\right)=\left(\frac{1}{\lambda} A^{-1}\right)(\lambda A)=I_{n}\)
\(\text { So, }(\lambda A)^{-1}=\frac{1}{\lambda} A^{-1}\)

\(\left[ \begin{matrix} 3 & -1 & 2 \\ -6 & 2 & 4 \\ -3 & 1 & 2 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \)
Note
\(\left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }/8 }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \).
This is also a row-echelon form of the given matrix.
So, a row-echelon form of a matrix is not necessarily unique.

Given 2i + 3 is a root
∴ Its conjugate 3 - 2i is also a root of the polynomial equation.
∴ Sum of the roots 3 + 2i + 3 - 2i = 6
Product of the roots = (3 + 2i) (3 - 2i)
= 3² + 2² = 9+ 4 = 13
∴ The polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
⇒x² -x (6) + 13 = 0
⇒ x²- 6x + 13 = 0

The roots of x³+2x2+3x+4 = 0 are ∝, β, ૪
∴ ∝+β+૪ = -co-efficient of x² = -2 ...(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3 ....(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4 ...(3)
Form a cubic equation whose roots are 2∝, 2β, 2૪
2∝+2β+2૪ = 2(∝+β+૪) = 2(-2) = -4 [from (1)]
4∝β+4β૪+4૪∝ = 4(∝β+β૪+୪∝) = 4(3) = 12 [from (2)]
(2∝)(2β)(2૪) = 8(∝β૪) = 8(-4) = -32 [from (3)]
∴ The required cubic equation is
x³-(2∝+2β+2૪)x² + (2∝β+2β૪+2୪∝)x - (2∝)(2β)(2૪) = 0
⇒ x³+(-4)x²+12x+32 = 0
⇒ x³+4x²+12x+32 = 0

Let A = \(\left[ \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right] \). Then |A| = \(\left| \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right| \) = 2(7) + (-12) + 3(-1) = -1 ≠ 0.
Therefore, A⁻¹ exists. Now, we get
adj A = \({ \left[ \begin{matrix} +\left| \begin{matrix} 3 & 1 \\ 2 & 3 \end{matrix} \right| & -\left| \begin{matrix} -5 & 1 \\ -3 & 3 \end{matrix} \right| & +\left| \begin{matrix} -5 & 3 \\ -3 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} -1 & 3 \\ 2 & 3 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ -3 & 3 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ -3 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} -1 & 3 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ -5 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ -5 & 3 \end{matrix} \right| \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 7 & 12 & -1 \\ 9 & 15 & -1 \\ -10 & -17 & 1 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] \).
Hence, A^-1 = \(\frac { 1 }{ \left| A \right| } \)(adj A) = \(\frac { 1 }{ \left( -1 \right) } \left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] =\left[ \begin{matrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{matrix} \right] \).

x⁸- 3x + 1 = 0
Here a_n = 1, a_o = 1
If \(\frac{p}{q}\) is a root of the polynomial, then as 
(p, q) = 1p is a factor of a_o = 1 and q is a factor of a_n = 1
Since 1 has no factors, the given equation has no rational roots.

Transforming the augmented matrix to echelon form, we get
\(\left[ \begin{matrix} 4 & 3 & 6 \\ 1 & 5 & 7 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 25 \\ 13 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 4 & 3 & 6 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 13 \\ 25 \\ 1 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & -17 & -22 \\ 0 & -1 & -13 \end{matrix}|\begin{matrix} 13 \\ -27 \\ -25 \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -1 \right) \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -1 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 1 & 13 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 25 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 17{ R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 0 & 199 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 398 \end{matrix} \right] \).
The equivalent system is written by using the echelon form:
x + 5y + 7 = 13, … (1)
17y + 22z = 27, … (2)
199z = 398..... (3)
From (3), we get z = \(\frac { 398 }{ 199 } \) = 2.
Substituting z = 2 in (2), we get y = \(\frac { 27-22\times 2 }{ 17 } =\frac { -17 }{ 17 } \) = -1
Substituting z = 2, y = -1, in (1), we get x = 13 - 5 x (-1) - 7 \(\times\) 2 = 4.
So, the solution is (x = 4, y = -1, z = 2).

x = 1 
y = x² -5x + 6
y = 1 -5 + 6 = 2
x = 2
y = 4 -10 + 6 = 0
x = 0 
y = 6
x = 3
y = 9 -15 + 6 = 12
x = -1 
y = 1 + 5 + 6 = 12
x = 4
y = 16 - 20 + 6 = 2

(1, 2), (0, 6), (-1, 12)
P(x) = (x²-5x + 6) (x²-5x+16)
= x⁴- 5x³+16x²-5x+25x²- 80x + 6x²- 30x + 96 = 0
x⁴-10x³+ 47x² -110x + 96 = 0
It has two sign changes
\(\therefore\) it has two positive real roots
P(-x) = x⁴-10x³+ 47x² -110x + 96
It has no sign changes, no negative real roots
y = x²- 5x + 16

Let the amount of 50% acid be x and the amount of 25% acid be y litre
By the given data, x + y = 10 ..............(1)
and \(x\left( \frac { 50 }{ 100 } \right) +y\left( \frac { 25 }{ 100 } \right) =10\left( \frac { 40 }{ 100 } \right) \)
⇒ 50x + 25y = 400 ⇒ 2x + y = 16 ...............(2)
The matrix from of the equation is \(\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \end{matrix} \right] ,B=\left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
⇒ X = A^-1N |A| = \(\left| \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right| \) = 1 - 2 = -1
⇒ X = \(\frac { 1 }{ |A| } \)adj A.B
⇒ X= \(-1\left[ \begin{matrix} 1 & -1 \\ -2 & 1 \end{matrix} \right] \left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
= -\(\left[ \begin{matrix} 10-16 \\ -20+16 \end{matrix} \right] \)
⇒ X = -\(\left[ \begin{matrix} -6 \\ -4 \end{matrix} \right] =\left[ \begin{matrix} 6 \\ 4 \end{matrix} \right] \)
Thus, the amount of 50% acid is 6 litre and the amount of 25% acid is 4 litre = 10 litres of 40% acid solution.

Since the sum of the co-efficients = 2 - 1- 1 = 0
x = 1 is a root.

∴ x = 1 is a root and the remaining factor is
2x²+x+1
\(\Rightarrow x=\frac {-1\pm \sqrt { { 1 }^{ 2 }-4(2)(1) } }{ 2 } \)
\(\Rightarrow x=\frac { -1\pm \sqrt { -7 } }{ 2 } \) Which is a complex root.
\(\Rightarrow x=\frac { -1\pm \sqrt { -7 } }{ 2 } \)
∴ x = 1 is the rational root.

\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₂] =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -\frac { 5 }{ 2 } & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 0 & 1 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -5 & 2 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)
∴ We get A^-1=\(\left[ \begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)

Since the sum of the co-efficients is
2 - 9 + 10 - 3 = 12 - 12 = 0
⇒ x = 1 is a root of (x)
∴ (x - 1) is a factor of (x)
To find the other factor, let us divide f(x) by x-1 

[Using synthetic division]
f(x) = (x-1)(x - 3)(2x - 1) = 0
⇒x - 1 = 0, x - 3 = 0 or 2x - 1 = 0
⇒ x = 1, x = 3x, x = \(\frac{1}{2}\)
Hence the roots are 1, 3, \(\frac{1}{2}\).

Given A =\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
|A| = 0-1\(\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \)
= -1(0-1) + 1(1-0) = 1 + 1 = 2
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} (0-1) & -(0-1) & +(1-0) \\ -(0-1) & +(0-1) & -(0-1) \\ +(1+0) & -(0-1) & +(0-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) ................(1)
Now A² =\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{matrix} \right] =\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] \)
A²- 3I =\(\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 2-3 & 1-0 & 1-0 \\ 1-0 & 2-3 & 1-0 \\ 1-0 & 1-0 & 2-3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) .............(2)
From (1) and (2), it is proved that A^-1 = \(\frac{1}{2}\) [A² - 3I]

Let ∝, β, ૪ be the roots of the equation 
Given \(\frac { \alpha }{ \beta } =\frac { 3 }{ 2 } \Rightarrow 2\alpha =3\beta \Rightarrow \alpha =\frac { 3 }{ 2 } \beta \)
\(\therefore \frac { 3 }{ 2 } \beta ,\beta ,\gamma \) are the roots of the given equation
Then by Vieta's formula,
\(\frac { 3 }{ 2 } \beta +\beta +\gamma =\frac { -b }{ a } =\frac { -(-9) }{ 1 } =9\)
\(\frac { 5 }{ 2 } \beta +\gamma =9\Rightarrow \gamma =9-\frac { 5 }{ 2 } \beta \)
\(\Rightarrow \gamma =\frac { 18-5\beta }{ 2 } ...(2)\)
Also \(\frac { 3 }{ 2 } \beta (\beta )+\beta \gamma +\left( \frac { 3 }{ 2 } \beta \right) \gamma =\frac { c }{ a } =\frac { 14 }{ 1 } =14\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 5 }{ 2 } \beta \left( \frac { 18-5\beta }{ 2 } \right) =14\ [using\ (2)]\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 90\beta }{ 4 } -\frac { 25{ \beta }^{ 2 } }{ 4 } =14\)
Multiplying by \(4,6{ \beta }^{ 2 }+90{ \beta }-25{ \beta }^{ 2 }=56\)
\(19{ \beta }^{ 2 }-90{ \beta }+56=0\)
\(\Rightarrow ({ \beta }-4)(19{ \beta }-14)=0\)
\(\Rightarrow { \beta }=4\)
\({ \beta }=\frac { 14 }{ 19 } \)
When \(\beta\) = 4, the other roots are \(\frac { 3 }{ 2 } (4),4,\frac { 18-5 }{ 2 } (4)\)
\(\Rightarrow 6,4,-1\)

When \(\\ \beta =\frac { 14 }{ 19 } ,\) the other roots are \(\frac { 3 }{ 2 } \beta ,\beta \frac { 18-5\beta }{ 2 } [by(2)]\)
\(\Rightarrow \frac { 3 }{ 2 } \left( \frac { 14 }{ 19 } \right) ,\frac { 14 }{ 19 } ,\frac { 18-5\left( \frac { 14 }{ 19 } \right) }{ 2 } \Rightarrow \frac { 21 }{ 19 } ,\frac { 14 }{ 19 } ,\frac { 136 }{ 19 } \)

Given equation is 2x⁴- 8x + 6x²- 3 = 0
Here a = 2, b = -8, c = 6, d = 0, e = -3
Let ∝, β, ૪ and \(\delta \) be the roots of equation (1)
Then by Vieta's formula,
\(\sum { _{ 1 }= } \alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -(-8) }{ 2 } =4\)
\(\sum { _{ 2 } } =\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { 6 }{ 2 } =3\)
\(\sum { _{ 3 } } =\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =\frac { 0 }{ a } \)
\(\sum { _{ 4 } } =\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { -3 }{ 2 } \)
Now, (a+b+c+d)² = a²+b²+c²+d²+2(ab+ac+ad+bc+cd)
⇒ n∝²+β²+૪²+\(\delta\)² = (∝ + β + ૪ + \(\delta\))²-2(\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta \))
∝² + β² + ૪² = 4²-2(3) = 16 - 6 = 10