\(\Delta=\left|\begin{array}{ccc} 3 & -1 & 1 \\ 1 & 2 & -2 \\ 2 & 3 & 1 \end{array}\right|\)
= 3 (2+6) + 1 (1+4) -1 (3-4)
= 24 + 5 - 1
= 28\(\neq\) 0
\(\Delta_{x}=\left|\begin{array}{ccc} 5 & -1 & 1 \\ -3 & 2 & -2 \\ 4 & 3 & 1 \end{array}\right|\)
= 5 (2+6) +1 (-3+8) +1 (-9-8)
= 40 + 5 - 17 = 28
\(\Delta_{y}=\left|\begin{array}{ccc} 3 & 5 & 1 \\ 1 & -3 & -2 \\ 2 & 4 & 1 \end{array}\right|\)
= 3 (-3+5) -5 (1+4) +1 (4+6)
= 15 - 25+10 = 0
\(\Delta_{z}=\left|\begin{array}{ccc} 3 & -1 & 5 \\ 1 & 2 & -3 \\ 2 & 3 & 4 \end{array}\right|\)
= 3 (8+9) +1 (4+6) +5 (3-4)
= 51 + 10 -5 = 56
\( x=\frac{\Delta_{x}}{\Delta}=\frac{28}{28}=1 \)
\( y=\frac{\Delta_{y}}{\Delta}=\frac{0}{28}=0\)
\(z=\frac{\Delta_{2}}{\Delta}=\frac{56}{28}=2\)
The solution (x, y, z) = (1, 0, 2)

The matrix equation is
\(\left[\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 9 \\ 6 \\ 2 \end{array}\right]\)
AX = B
\(|A|=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{array}\right|\)
= 2 (1+1) + 1 (1-1) + 3 (-1-1)
= 4 - 6 = -2 \(\neq\) 0
A^-1 exists
\(\operatorname{adj} A=\left[\begin{array}{ccc} +(1+1) & -(1-1) & +(-1-1) \\ -(-1+3) & +(2-3) & -(-2+1) \\ +(-1-3) & -(2-3) & +(2+1) \end{array}\right]^{T}\)
\(=\left[\begin{array}{ccc} 2 & 0 & -2 \\ -2 & -1 & 1 \\ -4 & 1 & 3 \end{array}\right]^{T}\)
\(\operatorname{adj} A=\left[\begin{array}{ccc} 2 & -2 & -4 \\ 0 & -1 & 1 \\ -2 & 1 & 3 \end{array}\right]\)
\(\mathrm{A}^{-1}=\frac{1}{|A|}(\operatorname{adj} A)\)
\(A^{-1}=-\frac{1}{2}\left[\begin{array}{ccc} 2 & -2 & -4 \\ 0 & -1 & 1 \\ -2 & 1 & 3 \end{array}\right]\)
X = A^-1B
\(\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \quad=-\frac{1}{2}\left[\begin{array}{ccc} 2 & -2 & -4 \\ 0 & -1 & 1 \\ -2 & 1 & 3 \end{array}\right]\left[\begin{array}{l} 9 \\ 6 \\ 2 \end{array}\right]\)
\(=-\frac{1}{2}\left[\begin{array}{ccc} 18 & -12 & -8 \\ 0 & -6 & +2 \\ -18 & +6 & +6 \end{array}\right]\)
\(=-\frac{1}{2}\left[\begin{array}{l} -2 \\ -4 \\ -6 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]\)
x = 1, y = 2, z = 3

Using the Gauss-elimination method, we obtain the following sequence of equivalent augmented matrices:
\(\left[\begin{array}{ccc|c} 0 & 2 & 3 & 7 \\ 3 & 6 & -12 & -3 \\ 5 & -2 & 2 & -7 \end{array}\right] \stackrel{R_{1} \leftrightarrow R_{2}}{\longrightarrow}\left[\begin{array}{ccc|c} 3 & 6 & -12 & -3 \\ 0 & 2 & 3 & 7 \\ 5 & -2 & 2 & -7 \end{array}\right]\)
\(\stackrel{R_{1} \rightarrow \frac{1}{3} R_{1}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 2 & -4 & -1 \\ 0 & 2 & 3 & 7 \\ -5 & -2 & 2 & -7 \end{array}\right]\)
\(\underset{R_{3} \rightarrow R_{3}-5 R_{1}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 2 & -4 & -1 \\ 0 & 2 & 3 & 7 \\ 0 & -12 & 22 & -2 \end{array}\right]\)
\(\stackrel{\frac{1}{2} R_{2}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 2 & -4 & -1 \\ 0 & 1 & \frac{3}{2} & \frac{7}{2} \\ 0 & -12 & 22 & -2 \end{array}\right]\)
\(\underset{R_{3}+12 R_{2}}{\stackrel{R_{1} \rightarrow R_{1}-2 R_{2}}{\longrightarrow}}\left|\begin{array}{ccc|c} 1 & 0 & -7 & -8 \\ 0 & 1 & \frac{3}{2} & \frac{7}{2} \\ 0 & -12 & 22 & -2 \end{array}\right|\)
\(\stackrel{R_{3} \rightarrow \frac{1}{40} R_{3}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 0 & -7 & -8 \\ 0 & 1 & \frac{3}{2} & \frac{7}{2} \\ 0 & 0 & 1 & 1 \end{array}\right]\)
\(\underset{R_{3} \rightarrow R_{3}+12 R_{2}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 0 & -7 & -8 \\ 0 & 1 & \frac{3}{2} & \frac{7}{2} \\ 0 & 0 & 40 & 40 \end{array}\right]\)
\(\underset{R_{2} \rightarrow R_{2}-\frac{3}{2} R_{3}}{\stackrel{R_{1} \rightarrow R_{1}+7 R_{3}}{\longrightarrow}}\left[\begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 1 \end{array}\right]\)
Writing the equivalent equation from echelon form.
x = -1
y = 2
z = 1
Solution (x, y, z) = (-1, 2, 1)

Using the Gauss-elimination method, we obtain the following sequence of equivalent augmented matrices:
\(\left[\begin{array}{ccc|c} 3 & -2 & 8 & 9^{2} \\ -2 & 2 & 1 & 3 \\ 1 & 2 & -3 & 8 \end{array}\right]\)
\(\underset{R_{1} \rightarrow R_{1}+R_{2}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 0 & 9 & 12 \\ -2 & 2 & 1 & 3 \\ 1 & 2 & -3 & 8 \end{array}\right]\)
\(\underset{R_{3} \rightarrow R_{3}-R_{1}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 0 & 9 & 12 \\ 0 & 2 & 19 & 27 \\ 0 & 2 & -12 & -4 \end{array}\right]\)
\(\stackrel{R_{2} \leftrightarrow R_{3}}{\longrightarrow} \left[\begin{array}{ccc|c} 1 & 0 & 9 & 12 \\ 0 & 2 & -12 & -4 \\ 0 & 2 & 19 & 27 \end{array}\right] \)
\(\stackrel{\frac{1}{2} R_{2}}{\longrightarrow} \quad\left[\begin{array}{llc|c} 1 & 0 & 9 & 12 \\ 0 & 1 & -6 & -2 \\ 0 & 2 & 19 & 27 \end{array}\right]\)
\(\stackrel{R_{3}-2 R_{1}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 0 & 9 & 12 \\ 0 & 1 & -6 & -2 \\ 0 & 0 & 31 & 31 \end{array}\right]\)
\(\stackrel{\frac{1}{31} R_{3}}{\longrightarrow}\left[\begin{array}{ccc|c} 1 & 0 & 9 & 12 \\ 0 & 1 & -6 & -2 \\ 0 & 0 & 1 & 1 \end{array}\right]\)
\(\underset{R_{2} \rightarrow R_{2}+6 R_{3}}{\stackrel{R_{1} \rightarrow R_{1}-9 R_{3}}{\longrightarrow}}\left[\begin{array}{lll|l} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 1 \end{array}\right]\)
Writing the equivalent equation from echelon form
x = 3
y = 4
z = 1 
Solution (x, y, z) = (3, 4, 1)

\(I+A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{cc} 1 & 2 \\ -2 & -1 \end{array}\right]=\left[\begin{array}{cc} 2 & 2 \\ -2 & 0 \end{array}\right]\)
\(\mathrm{I}-\mathrm{A}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} 1 & 2 \\ -2 & -1 \end{array}\right]=\left[\begin{array}{cc} 0 & -2 \\ 2 & 2 \end{array}\right]\)
\(|I-A|=4 \neq 0\)
\(\therefore(I-A)^{-1}=\frac{1}{|I-A|} \operatorname{adj}(I-A)=\frac{1}{4}\left[\begin{array}{cc} 2 & 2 \\ -2 & 0 \end{array}\right]\)
\(\therefore \phi(A)=(\mathrm{I}+\mathrm{A})(\mathrm{I}-\mathrm{A})^{-1}\)
\(=\left[\begin{array}{cc} 2 & 2 \\ -2 & 0 \end{array}\right] \frac{1}{4}\left[\begin{array}{cc} 2 & 2 \\ -2 & 0 \end{array}\right]\)
\(=\frac{1}{4}\left[\begin{array}{cc} 0 & 4 \\ -4 & -4 \end{array}\right]\)
\(=\left[\begin{array}{cc} 0 & 1 \\ -1 & -1 \end{array}\right]\)

\( |D|= 1(1-3 \sin \theta \cos \theta)-3 \cos \theta \) \( (\sin \theta-3 \cos \theta)+1\left(\sin ^{2} \theta-1\right) \)
\(= 1-\cos ^{2} \theta-6 \sin \theta \cos \theta+9 \cos ^{2} \theta \)
\(= \sin ^{2} \theta-6 \sin \theta \cos \theta+9 \cos ^{2} \theta\)
\( =1-3 \sin \theta \cos \theta-3 \sin \theta \cos \theta+9 \cos ^{2} \theta-\cos ^{2} \theta \)
\( =(3 \cos \theta-\sin \theta)^{2}=(a \cos \alpha+b \sin \alpha)^{2} \)
\( \therefore-\sqrt{a^{2}+b^{2}} \leq \sqrt{D} \leq \sqrt{a^{2}+b^{2}} \)
\( \therefore-\sqrt{9+1} \leq(3 \cos \theta-\sin \theta) \leq \sqrt{9+1} \)
\(\therefore 0 \leq(3 \cos \theta-\sin \theta)^{2} \leq 10\)
Range = [0, 10]

On expanding
\( 1\left(c^{2}-a b\right)-a(c-a)+b(b-c)=0 \)
\(c^{2}-a b-a c+a^{2}+b^{2}-b c=0 \)
\(a^{2}+b^{2}+c^{2}-a b-b c-c a=0 \)
\(\therefore \frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0 \)
Provided a = b = c
It is equilateral triangle.
\(\Rightarrow \mathrm{A}=\mathrm{B}=\mathrm{C}=\frac{\pi}{3}\)
\(\therefore 64\left[\sin ^{2} \frac{\pi}{3}+\sin ^{2} \frac{\pi}{3}+\sin ^{2} \frac{\pi}{3}\right]=64\left[\frac{3}{4}+\frac{3}{4}+\frac{3}{4}\right]\)
\(=64\left(\frac{9}{4}\right)\)
= 144

Let \(\omega=\frac{-1+i \sqrt{3}}{2} \text { and } \omega^{2}=\frac{-1-i \sqrt{3}}{2}\)
\(\therefore \omega^{3}=1 \text { and } 1+\omega+\omega^{2}=0\)
\(\therefore A=\left[\begin{array}{cc} \frac{\omega}{i} & \frac{\omega^{2}}{i} \\ \frac{-\omega^{2}}{i} & \frac{-\omega}{i} \end{array}\right]=\frac{\omega}{i}\left[\begin{array}{cc} 1 & \omega \\ -\omega & -1 \end{array}\right]\)
\(A^{2}=A \times A\)
\(=\frac{\omega}{i}\left[\begin{array}{cc} 1 & \omega \\ -\omega & -1 \end{array}\right] \times \frac{\omega}{i}\left[\begin{array}{cc} 1 & \omega \\ -\omega & -1 \end{array}\right]\)
\(=\frac{\omega^{2}}{i^{2}}\left[\begin{array}{cc} 1-\omega^{2} & \omega-\omega \\ -\omega+\omega & -\omega^{2}+1 \end{array}\right]\)
\(=-\omega^{2}\left[\begin{array}{cc} 1-\omega^{2} & 0 \\ 0 & 1-\omega^{2} \end{array}\right]\)
\(=\left[\begin{array}{cc} -\omega^{2}+\omega^{4} & 0 \\ 0 & -\omega^{2}+\omega^{4} \end{array}\right]\)
\(=\left[\begin{array}{cc} -\omega^{2}+\omega & 0 \\ 0 & -\omega^{2}+\omega \end{array}\right]\)
\(=\left[\begin{array}{cc} 1+\omega+\omega & 0 \\ 0 & 1+\omega+\omega \end{array}\right]\)
\(=\left[\begin{array}{cc} 1+2 \omega & 0 \\ 0 & 1+2 \omega \end{array}\right]\)
\(f(A)=A^{2}+2 I=\left[\begin{array}{cc} 1+2 \omega & 0 \\ 0 & 1+2 \omega \end{array}\right]+\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\)
\(=\left[\begin{array}{cc} 3+2 \omega & 0 \\ 0 & 3+2 \omega \end{array}\right]\)
\(=(3+2 \omega)\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\)
\( =\left\{3+2\left(\frac{-1+i \sqrt{3}}{2}\right)\right\}\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \)
\( =(2+i \sqrt{3})\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] =\left[\begin{array}{cc} 2+i \sqrt{3} & 0 \\ 0 & 2+i \sqrt{3} \end{array}\right] \)

Let A = \(\left[\begin{array}{ccc} 4 & 8 & 1 \\ 2 & -3 & 2 \\ 1 & 7 & -3 \end{array}\right]\)
|A| = 4(9 -14)-8(-6 -2) + 1 (14 + 3)
= 4(-5) - 8 (- 8) + 1(17)
= - 20 + 64 + 17 = 61 \(\neq\) 0
A^-1 exists.
adj A = \(\left[\begin{array}{ccc} +(9-14) & -(-6-2) & +(14+3) \\ -(-24-7) & +(-12-1) & -(28-8) \\ +(16+3) & -(8-2) & +(-12-16) \end{array}\right]^{T}\)
\(=\left[\begin{array}{ccc} -5 & 8 & 17 \\ 31 & -13 & -20 \\ 19 & -6 & -28 \end{array}\right]^{T}=\left[\begin{array}{ccc} -5 & 31 & 19 \\ 8 & -13 & -6 \\ 17 & -20 & -22 \end{array}\right]\)
\( \mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} \mathrm{A} \\ \)
\( \mathrm{A}^{-1}=\frac{1}{61}\left[\begin{array}{ccc} -5 & 31 & 19 \\ 8 & -13 & -6 \\ 17 & -20 & -28 \end{array}\right] \)
X = A^-1B,
\(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{61}\left[\begin{array}{ccc} -5 & 31 & 19 \\ 8 & -13 & -6 \\ 17 & -20 & -28 \end{array}\right]\left[\begin{array}{c} -6 \\ 0 \\ -8 \end{array}\right]\)
\(=\frac{1}{61}\left[\begin{array}{c} 30-152 \\ -48+48 \\ -102+224 \end{array}\right]\)
\(\frac{1}{61}\left[\begin{array}{c} -122 \\ 0 \\ 122 \end{array}\right]=\left[\begin{array}{c} -2 \\ 0 \\ 2 \end{array}\right]\)
Solution (x, y, z) = (-2, 0 2)

The augmented matrix [A|B] is \(\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 1 & \lambda \\ 1 & -7 & 8 \end{matrix}|\begin{matrix} 12 \\ \mu \\ 17 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 3 & 1 & \lambda \\ 2 & -3 & 5 \end{matrix}|\begin{matrix} 17 \\ \mu \\ 12 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2R_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 22 & \lambda -51 \\ 0 & 11 & -11 \end{matrix}|\begin{matrix} 17 \\ \mu -51 \\ -22 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 3 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 11 }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 0 & \lambda -2 \\ 0 & 1 & -1 \end{matrix}|\begin{matrix} 17 \\ \mu -7 \\ -2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & \lambda -2 \end{matrix}|\begin{matrix} 17 \\ -2 \\ \mu -7 \end{matrix} \right] \)
Case (i) : when λ ≠ 2,
\(\rho\) ([A|B]) = 3 and \(\rho\)(A) = 3
∴ \(\rho\)([AIB])= \(\rho\)(A) = 3 = the number of unknowns
∴ The system has unique solution
Case (ii) : when λ = 2, μ =7
\(\left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 17 \\ -2 \\ 0 \end{matrix} \right] \)
Here \(\rho\)(A) = 2 and \(\rho\)([A|B]) = 2
∴ \(\rho\)(A) = \(\rho\)([A|B]) = 2 < number of unknowns
Thus the system is consistent with infinitely many solutions.
Case (iii) : When λ= 2 and μ ≠ 7
\(\rho\) (A) = 2 and \(\rho\) ([A|B]) = 3
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
Thus, the given system of equations is inconsistent.

Augmented matrix [A|B] is \(\left[ \begin{matrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} a \\ b \\ c \end{matrix} \right] \)
[A|B]\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ -2 & 1 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} b \\ a \\ c \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 3 & -3 \end{matrix}|\begin{matrix} b \\ a+2b \\ c-b \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} b \\ c+2b \\ a+b+c \end{matrix} \right] \)
Here \(\rho\) (A) = 2
The given system is consistent only when \(\rho\)([A|B]) = 2\(\rho\)([A|B]) = 2 only if a + b + c = 0 Hence proved.

Augmented matrix = [A|B] = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{matrix}|\begin{matrix} 1 \\ \lambda \\ { \lambda }^{ 2 } \end{matrix} \right] \)
[A|B] =  \(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 3 & 9 \end{matrix}|\begin{matrix} 1 \\ \lambda -1 \\ { \lambda }^{ 2 }-1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ \lambda -1 \\ { \lambda }^{ 2 }-1-3\lambda +3 \end{matrix} \right] \)
⟶ \(\left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ \lambda -1 \\ { \lambda }^{ 2 }-3\lambda +2 \end{matrix} \right] \)
Here \(\rho\)(A) = 2
The given system of equations is consistent only when \(\rho\)([AIB]) = 2
\(\rho\)([AIB]) = 2 only when λ²-3λ + 2 = 0
⇒ (λ-1) (λ-2) = 0 ⇒ λ = 1 or λ = 2
∴ The given system is consistent when the values of A are 1 and 2.

Let the required numbers be x, y and z
By the given data,
x + y + z = 20  ....(1)
2z + x = 23 ⇒ x + 2z = 23...(2)
y + z + 3x = 46 ⇒  3x + y + z = 46..(3)
Δ = \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{matrix} \right| =1\left| \begin{matrix} 0 & 2 \\ 0 & 1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -2 + 5 + 1 =4
Δ₁ = \(\left| \begin{matrix} 20 & 1 & 1 \\ 23 & 0 & 2 \\ 46 & 1 & 1 \end{matrix} \right| =20\left| \begin{matrix} 0 & 2 \\ 1 & 1 \end{matrix} \right| -1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| +1\left| \begin{matrix} 23 & 0 \\ 46 & 1 \end{matrix} \right| \)
= -40 + 69 + 23 = 52
Δ₂ = \(\left| \begin{matrix} 1 & 20 & 1 \\ 1 & 23 & 2 \\ 3 & 46 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| -20\left| \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| \)
= -69 + 100 - 23 = 8
Δ₃ = \(\left| \begin{matrix} 1 & 1 & 20 \\ 1 & 2 & 23 \\ 3 & 1 & 46 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 0 & 23 \\ 1 & 46 \end{matrix} \right| -1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| +20\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -23 + 23 + 20 = 20
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 52 }{ 4 } \) = 13
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 8 }{ 4 } \) = 2 and z =\(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 20 }{ 4 } \) = 5
Hence the required numbers are 13, 2 and 5.

Put \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
∴ 2a + 3b + 10c = 4  ....(1)
4a- 6b - 5c = 1 .....(2)
6a + 9b -20c = 2 ...(3)
Δ = \(\left| \begin{matrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{matrix} \right| \)
= \(2\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (120 - 45) -3 (-80 - 30) + 10 (36 + 36)
= 150 + 330 + 720 = 1200
Δ₁ = \(\left| \begin{matrix} 4 & 3 & 10 \\ 1 & -6 & 5 \\ 2 & 9 & -20 \end{matrix} \right| \)
= \(4\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| +10\left| \begin{matrix} 1 & -6 \\ 2 & 9 \end{matrix} \right| \)
= 4 (120 - 45) -3 (-20 - 10) + 10 (9 + 12)
= 300 + 90 + 120 = 600
Δ₂ = \(\left| \begin{matrix} 2 & 4 & 10 \\ 4 & 1 & 5 \\ 6 & 2 & -20 \end{matrix} \right| =2\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| -4\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| \)
= 2 (-2 - 10) - 4 (-80 - 30) + 10 (8 - 6)
= -60 + 440 + 20 = 400
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 4 \\ 4 & -6 & 1 \\ 6 & 9 & 2 \end{matrix} \right| 2\left| \begin{matrix} -6 & 1 \\ 9 & 2 \end{matrix} \right| -3\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (-12 - 9) -3 (8 - 6) + 4 (36 + 36)
= - 42 - 6 + 288 = 240
∴ a = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 1200 } =\frac { 1 }{ 2 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 2 } \) ⇒ x = 2
∴ b = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 400 }{ 1200 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 3 } \) ⇒ y = 1 
∴ c = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 240 }{ 1200 } =\frac { 1 }{ 5 } \Rightarrow \frac { 1 }{ z } =\frac { 1 }{ 5 } \) ⇒ z = 5
∴ Solution set is {2, 3, 5}

Given f(x) = ax² + bx + c
f(1) = 0
⇒ a(1)² + b(1) + c = 0
⇒ a + b + c = 0 ............(1)
f(2) = -2
⇒ a(2²) + b(2) + c =-2
⇒ 4a + 2b + c = -2 ..........(2)
f(3) = -6
⇒ a(3)² + b(3) + c = -6
⇒ 9a + 3b + c = -6 ...........(3)
Δ = \(\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{matrix} \right| =1\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| -1\left| \begin{matrix} 4 & 1 \\ 9 & 1 \end{matrix} \right| +1\left| \begin{matrix} 4 & 2 \\ 9 & 3 \end{matrix} \right| \)
= 1 (2 -3) -1 (4-9) +1 (12 - 18)
= - 1 + 5 - 6 = -2 ≠ 0
Δ₁ =\(\left| \begin{matrix} 0 & 1 & 1 \\ -2 & 2 & 1 \\ -6 & 3 & 1 \end{matrix} \right| =0-1\left| \begin{matrix} -2 & 1 \\ -6 & 1 \end{matrix} \right| +1\left| \begin{matrix} -2 & 2 \\ -6 & 1 \end{matrix} \right| \)
=-1 (-2 + 6) +1(-6 + 12) = -1 (4) +1 (6) = 2
Δ₂ =\(\left| \begin{matrix} 1 & 0 & 1 \\ 4 & -2 & 1 \\ 9 & -6 & 1 \end{matrix} \right| =1\left| \begin{matrix} -2 & 1 \\ -6 & 1 \end{matrix} \right| +0+1\left| \begin{matrix} 4 & -2 \\ 9 & -6 \end{matrix} \right| \)
-1 (-2 + 6) + 1(-24 + 18) = 4 - 6 = -2
Δ₃ =\(\left| \begin{matrix} 1 & 1 & 0 \\ 4 & 2 & -2 \\ 9 & 3 & -6 \end{matrix} \right| =1\left| \begin{matrix} 2 & -2 \\ 3 & -6 \end{matrix} \right| -1\left| \begin{matrix} 4 & -2 \\ 9 & -6 \end{matrix} \right| \)
= 1 (-12 + 6) -1 (-24 + 18) = -6 + 6 = 0
a = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 2 }{ -2 } \) = -1
b = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -2 }{ -2 } \) = 1
c = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 0 }{ -2 } \)
∴ f(x) = (-1)x² + 1x + 0
⇒ f(x) = x² + x.