Let first, second and third numbers be denoted by x, y and z, respectively.
Then, according to the question, we get
\(x+y+z=6, y+3 z=11 \)
\(x+z=2 y \Rightarrow x-2 y+z=0 \) 
This system can be written as A X = B, where
where,\(A=\left[\begin{array}{rrr} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 6 \\ 11 \\ 0 \end{array}\right]\) 
Here \(|A|=1(1+6)-(0-3)+(0-1)=9 \neq 0\). Now we find adj A 
\(\begin{array}{lll} A_{11}=1(1+6)=7, & A_{12}=-(0-3)=3, & A_{13}=-1 \\ A_{21}=-(1+2)=-3, & A_{22}=0, & A_{23}=-(-2-1)=3 \\ A_{31}=(3-1)=2, & A_{32}=-(3-0)=-3, & A_{33}=(1-0)=1 \end{array}\)
Hence \(\operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{array}\right]\)
Thus \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj}(\mathrm{A})=\frac{1}{9}\left[\begin{array}{ccc} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{array}\right]\)
Since \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\)
\(X=\frac{1}{9}\left[\begin{array}{ccc} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{array}\right]\left[\begin{array}{c} 6 \\ 11 \\ 0 \end{array}\right]\)
or \(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{9}\left[\begin{array}{c} 42-33+0 \\ 18+0+0 \\ -6+33+0 \end{array}\right]=\frac{1}{9}\left[\begin{array}{c} 9 \\ 18 \\ 27 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]\)
x = 1, y = 2 and z = 3

\( \left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right| \)
\( =x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta) \)
\( =-x^3-x+x \sin ^2 \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^2 \theta \)
\( =-x^3-x+x\left(\sin ^2 \theta+\cos ^2 \theta\right) \)
\(=-x^3-x+x \)
\(=-x^3 \text { (Which is Independent of } \theta \text { ) }\)

\( \Delta=\left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right| \)
Expanding along \(\mathrm{C}_3\),
we have:
\( \Delta =-\sin \alpha\left(-\sin \alpha \sin ^2 \beta-\cos ^2 \beta \sin \alpha\right)+\cos \alpha\left(\cos \alpha \cos ^2 \beta+\cos \alpha \sin ^2 \beta\right) \)
\(=\sin ^2 \alpha\left(\sin ^2 \beta+\cos ^2 \beta\right)+\cos ^2 \alpha\left(\cos ^2 \beta+\sin ^2 \beta\right) \)
\(=\sin ^2 \alpha(1)+\cos ^2 \alpha(1)  =1 \)

\(A=\left[\begin{array}{rrr} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]\)
|A|=1(15-1)+2(-10-1)+1(-2-3)=14-22-5=-13
Now,
\(A_{11}=14, A_{12}=11, A_{13}=-5 \)
\(A_{21}=11, A_{22}=4, A_{23}=-3 \)
\(A_{31}=-5, A_{32}=-3, A_{13}=-1\)
\(\operatorname{adj} A  =\left[\begin{array}{lll} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]\)
\(A^{-1}  =\frac{1}{|A|}(\operatorname{adj} A) \)
\(=-\frac{1}{13}\left[\begin{array}{lll} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]=\frac{1}{13}\left[\begin{array}{lll} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{array}\right]\)
\(\mid \text { adj } A \mid =14(-4-9)-11(-11-15)-5(-33+20) \)
=14(-13)-11(-26)-5(-13)
=-182+286+65=169
We have,
\( \operatorname{adj}(\operatorname{adj} A)=\left[\begin{array}{ccc} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right]\)
\( \therefore[\operatorname{adj} A]^{-1}=\frac{1}{|\operatorname{adj} A|}(\operatorname{adj}(\operatorname{adj} A))\)
\(=\frac{1}{169}\left[\begin{array}{lll} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right]\)
\(=\frac{1}{13}\left[\begin{array}{lll} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right]\)
Now, \(A^{-1}=\frac{1}{13}\left[\begin{array}{lll}-14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\ -\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13}\end{array}\right]\)
\(\therefore \operatorname{adj}\left(A^{-1}\right) =\left[\begin{array}{lll} -\frac{4}{169}-\frac{9}{169} & -\left(-\frac{11}{169}-\frac{15}{169}\right) & -\frac{33}{169}+\frac{20}{169} \\ -\left(-\frac{11}{169}-\frac{15}{169}\right) & -\frac{14}{169}-\frac{25}{169} & -\left(-\frac{42}{169}+\frac{55}{169}\right) \\ -\frac{33}{169}+\frac{20}{169} & -\left(-\frac{42}{169}+\frac{55}{169}\right) & \frac{56}{169}-\frac{121}{169} \end{array}\right]\)
\( =\frac{1}{169}\left[\begin{array}{lll} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right]=\frac{1}{13}\left[\begin{array}{lll} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right]\)
Hence, \([\operatorname{adj} A]^{-1}=\operatorname{adj}\left(A^{-1}\right)\).
We have shown that:
\(A^{-1}=\frac{1}{13}\left[\begin{array}{lll} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{array}\right]\)
And, \(\operatorname{adj} A^{-1}=\frac{1}{13}\left[\begin{array}{lll}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]\)
Now,
\( \left|A^{-1}\right|=\left(\frac{1}{13}\right)^3[-14 \times(-13)+11 \times(-26)+5 \times(-13)]=\left(\frac{1}{13}\right)^3 \times(-169)=-\frac{1}{13} \)
\(\therefore\left(A^{-1}\right)^{-1}=\frac{\operatorname{adj} A^{-1}}{\left|A^{-1}\right|}=\frac{1}{\left(-\frac{1}{13}\right)} \times \frac{1}{13}\left[\begin{array}{lll} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right]=\left[\begin{array}{lll} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]=A\)
\(\left(A^{-1}\right)^{-1}=A\)

\(\mathrm{A}=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right] \text {, then } a d j \mathrm{~A}=\left[\begin{array}{lll} \mathrm{A}_{11} & \mathrm{~A}_{21} & \mathrm{~A}_{31} \\ \mathrm{~A}_{12} & \mathrm{~A}_{22} & \mathrm{~A}_{32} \\ \mathrm{~A}_{13} & \mathrm{~A}_{23} & \mathrm{~A}_{33} \end{array}\right]\)
Since sum of product of elements of a row (or a column) with corresponding cofactors is equal to |A| and otherwise zero, we have
\(\mathrm{A}(\operatorname{adj} \mathrm{A})=\left[\begin{array}{ccc} |\mathrm{A}| & 0 & 0 \\ 0 & |\mathrm{~A}| & 0 \\ 0 & 0 & |\mathrm{~A}| \end{array}\right]=|\mathrm{A}|\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=|\mathrm{A}| \mathrm{I}\)
Similarly, we can show (adj A) A = A I
Hence A (adj A) = (adj A) A = A I

The given system of equations is:
2x + 3y + 3z = 5
x - 2y + z = -4
3x - y - 2z = 3
These can be written as AX = B
\(\Rightarrow X=A^{-1}B\) ..(1)
where \(A=\begin{bmatrix} 2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix},X=\begin{bmatrix} x\\y\\z\end{bmatrix}and \ B=\begin{bmatrix}5\\-4\\3 \end{bmatrix}\)
\(\therefore|A|=\begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2 \end{bmatrix}\)
\(=2(4+1)-3(-2-3)+3(-1+6)\)
\(=10+15+15=40\neq0\Rightarrow A^{-1}\) exists.
Now \(adj\ A=\begin{bmatrix}5&5&\\3&-13&11\\9&1&-7 \end{bmatrix}=\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7 \end{bmatrix}\)
\(\therefore\ A^{-1}={1\over|A|}(Adj\ A)\)
\(={1\over40}\begin{bmatrix} 5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\)
From (1), \(X={1\over40}\begin{bmatrix} 5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix} 5\\-4\\3\end{bmatrix}\)
\(={1\over40}\begin{bmatrix} 25-12+27\\25+52+3\\25-44-21\end{bmatrix}={11\over40}\begin{bmatrix} 40\\80\\-40\end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} 1\\2\\-1\end{bmatrix}\)
Hence x = 1, y = 2, z = -1

The given equation are:
5x-y+4z=5
2x+3y+5z=2
5x-2y+6z=-1
Here A= \(\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6 \end{bmatrix}\)
Now |A|=\(\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6 \end{bmatrix}\)
=5(18+10)+1.(12-25)+4(-4-15)
=140-13-76=\(51\neq0\)
Hence, the given system of equations is consistent.

Let \(A=\left[\begin{matrix}-1&5\\ -3&2\end{matrix}\right]\)
\(|A|=\left[\begin{matrix}-1&5\\ -3&2\end{matrix}\right]=-2+15=13\neq0\)
A is non-singular  = A^-1 exists.
Now \(A_{11}=(-1)^{1+1}M_{11}(+1)(1 )=2\)
 \(A_{12}=(-1)^{1+2}M_{12}\\ =(-1)(-3)=3;\)
\(A_{21}=(-1)^{2+1}M_{21}=(-1)(5)=-5\)
\(A_{22}=(-1)^{2+2}M_{22}\\ =(+1)(-1)=-1\)
\(\therefore\ adj\ A=\begin{bmatrix}2&3\\-5&-1 \end{bmatrix}=\begin{bmatrix}2&-5\\3&-1 \end{bmatrix}\)
\(\therefore\ A^{-1}={1\over|A|}(adj\ A)\)
\(={1\over13}\begin{bmatrix} 2&-5\\3&-1\end{bmatrix}\)
=\(\left[\begin{matrix}{2\over13}&{-5\over13}\\ {3\over13}&-{1\over13}\end{matrix}\right]\)

\(A=\begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix}\)
\(\therefore 2 A=2\left[\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right] \)
\(\therefore \text { L.H.S. }=|2 A|=\left|\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right|=2 \times 4-4 \times 8=8-32=-24 \)
\(\text { Now, }|A|=\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|=1 \times 2-2 \times 4=2-8=-6 \)
\(\therefore \text { R.H.S. }=4|A|=4 \times(-6)=-24 \)
\(\therefore \text { L.H.S. }=\text { R.H.S. } \)

We have \(A^2=A \cdot A=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right]\)
Hence \(\mathrm{A}^2-4 \mathrm{~A}+\mathrm{I}=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right]-\left[\begin{array}{cc} 8 & 12 \\ 4 & 8 \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=\mathrm{O}\)
\(\begin{array}{l} \mathrm{A}^2-4 \mathrm{~A}+\mathrm{I}=\mathrm{O} \\ \mathrm{A} \mathrm{A}-4 \mathrm{~A}=-\mathrm{I} \end{array}\)
or \( \mathrm{A} \mathrm{A}\left(\mathrm{A}^{-1}\right)-4 \mathrm{AA}^{-1}=-\mathrm{IA}^{-1} \) (Post multiplying by A–1 because |A| ≠ 0)
or \( {A}\left(\mathrm{~A} \mathrm{~A}^{-1}\right)-4 \mathrm{I}=-\mathrm{A}^{-1} \)
or \(\mathrm{AI}-4 \mathrm{I}=-\mathrm{A}^{-1} \)
or \(\mathrm{A}^{-1}=4 \mathrm{I}-\mathrm{A}=\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right]-\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]\)
Hence \(A^{-1}=\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]\)

By definition of minors and cofactors, we have
\(\text {Minor of } a_{11}=\mathrm{M}_{11}=\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|=a_{22} a_{33}-a_{23} a_{32}\)
\(\text {Cofactor of } a_{11}=\mathrm{A}_{11}=(-1)^{1+1} \quad \mathrm{M}_{11}=a_{22} a_{33}-a_{23} a_{32}\)
\(\text {Minor of } a_{21}=\mathrm{M}_{21}=\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array}\right|=a_{12} a_{33}-a_{13} a_{32}\)
\(\text {Cofactor of } a_{21}=\mathrm{A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=(-1)\left(a_{12} a_{33}-a_{13} a_{32}\right)=-a_{12} a_{33}+a_{13} a_{32}\)

Minor of the element a_ij is M_ij
Here a₁₁ = 1.
So M₁₁ = Minor of a₁₁ = 3
M₁₂ = Minor of the element a₁₂ = 4
M₂₁ = Minor of the element a₂₁ = –2
M₂₂ = Minor of the element a₂₂ = 1
Now, cofactor of aij is Aij. So
A₁₁ = (–1)^{1 + 1} M₁₁ = (–1)² (3) = 3
A₁₂ = (–1)^{1 + 2} M₁₂ = (–1)³ (4) = – 4
A₂₁ = (–1)^{2 + 1} M₂₁ = (–1)³ (–2) = 2
A₂₂ = (–1)^{2 + 2} M₂₂ = (–1)⁴ (1) = 1

As if A = \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) then \(\left| A \right| =\begin{bmatrix} a & b \\ c & d \end{bmatrix}\)
\(KA=\begin{bmatrix} Ka & Kb \\ Kc & Kd \end{bmatrix}\) and \(\left| KA \right| =\begin{bmatrix} Ka & Kb \\ Kc & Kd \end{bmatrix}\)
\(={ K }^{ 2 }\begin{vmatrix} a & b \\ c & d \end{vmatrix}={ K }^{ 2 }|A|\)

\({ \triangle }_{ 1 }=\left| \begin{matrix} Ax & By & Cz \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \\ yz & zx & xy \end{matrix} \right| \)
C₁ ⟶ xC₁, C₂ ⟶ yC₂ and C₃ ⟶ zC₃ and dividing determinat by xyz
= \(\frac { 1 }{ xyz } \left| \begin{matrix} { Ax }^{ 2 } & { By }^{ 2 } & { Cz }^{ 2 } \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \\ xyz & xyz & xyz \end{matrix} \right| \)
Taking xyz common from R₃, we get
\(\frac { xyz }{ xyz } \left| \begin{matrix} { Ax }^{ 2 } & { By }^{ 2 } & { Cz }^{ 2 } \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \\ 1 & 1 & 1 \end{matrix} \right| \)
= \(\left| \begin{matrix} { Ax }^{ 2 } & { By }^{ 2 } & { Cz }^{ 2 } \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \\ 1 & 1 & 1 \end{matrix} \right| =\left| \begin{matrix} { Ax }^{ 2 } & x^{ 3 } & 1 \\ { By }^{ 2 } & { y }^{ 3 } & 1 \\ { Cz }^{ 2 } & { z }^{ 3 } & 1 \end{matrix} \right| \)=Δ
⇒ Δ₁ - Δ = 0

Δ = 6(-1)- 1(1) = -7

As \(\begin{vmatrix} 2x & -1 \\ 4 & 2 \end{vmatrix}=\begin{vmatrix} 3 & 0 \\ 2 & 1 \end{vmatrix}\)
⇒ 4x + 4 = 3 - 0
⇒ x = \(-\frac { 1 }{ 4 } \)