விரிவுபடுத்தப்பட்ட அணியின் மீது காஸ்-ஜோர்டான் முறையைப் பயன்படுத்த, நமக்குக் கிடைப்பது,
[A|I₃] =  \(\left[ \begin{matrix} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\rightarrow \frac { 1 }{ 2 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & (1/2) & (1/2) \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} (1/2) & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2R_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & (1/2) & (1/2) \\ 0 & (1/2) & -(1/2) \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} (1/2) & 0 & 0 \\ -(3/2) & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & (1/2) & (1/2) \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} (1/2) & 0 & 0 \\ -3 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \\ \left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 2 & -1 & 0 \\ - & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 3 }\\ { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1 \end{matrix} \right] \)
எனவே, A^-1 =\(\left[ \begin{matrix} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1 \end{matrix} \right] \)

\(A=\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
|A| = 24 - 20 = 4 ≠ 0.
A^-1 ஐக் காண இயலும்.
\(\operatorname{adj} A=\left[\begin{array}{ll} 3 & 4 \\ 5 & 8 \end{array}\right]\)
\(\begin{aligned} \mathrm{A}(\operatorname{adj} \mathrm{~A}) & =\left[\begin{array}{cc} 8 & -4 \\ -5 & 3 \end{array}\right]\left[\begin{array}{ll} 3 & 4 \\ 5 & 8 \end{array}\right] \\ & =\left[\begin{array}{cc} 24-20 & 32-32 \\ -15+15 & -20+24 \end{array}\right] \\ & =\left[\begin{array}{cc} 4 & 0 \\ 0 & 4 \end{array}\right] ....(1) \end{aligned}\)
\((\operatorname{adj} A) A=\left[\begin{array}{ll} 3 & 4 \\ 5 & 8 \end{array}\right]\left[\begin{array}{cc} 8 & -4 \\ -5 & 3 \end{array}\right]\)
             \(\begin{aligned} & =\left[\begin{array}{cc} 24-20 & -12+12 \\ 40-40 & -20+24 \end{array}\right] \\ & =\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right] ....(2) \end{aligned}\)
\(|\mathrm{A}| \mathrm{I}_2=4\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right]\) ...(3)
(1), (2) மற்றும் (3) லிருந்து,
A(adj A) = (adj A) A = |A| I₂.

\(\left[ \begin{matrix} 0 \\ -1 \\ 4 \end{matrix}\begin{matrix} 3 \\ 0 \\ 2 \end{matrix}\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}\begin{matrix} 6 \\ 5 \\ 0 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 0 \\ 4 \end{matrix}\begin{matrix} 0 \\ 3 \\ 2 \end{matrix}\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}\begin{matrix} 5 \\ 6 \\ 0 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 3 \\ 2 \end{matrix}\begin{matrix} 2 \\ 1 \\ 8 \end{matrix}\begin{matrix} 5 \\ 6 \\ 20 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 2 }{ 3 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 3 \\ 0 \end{matrix}\begin{matrix} 2 \\ 1 \\ \frac { 22 }{ 3 } \end{matrix}\begin{matrix} 5 \\ 6 \\ 16 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { 3R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 3 \\ 0 \end{matrix}\begin{matrix} 2 \\ 1 \\ 22 \end{matrix}\begin{matrix} 5 \\ 6 \\ 48 \end{matrix} \right] \)

|adj A| = \(\left| \begin{matrix} -1 & 2 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right| \) = 9
எனவே A^-1 = \(\pm \frac { 1 }{ \sqrt { |adj(A)| } } adj(A)=\pm \frac { 1 }{ \sqrt { 9 } } \left[ \begin{matrix} -1 & 2 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] =\pm \frac { 1 }{ 3 } \left[ \begin{matrix} -1 & 2 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] \).

A = \(\left[ \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right] \) என்க.
|A| = \(\left| \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right| \) = 2(7) + (-12) + 3(-1) = -1 ≠ 0 
எனவே , A⁻¹ காண இயலும்.
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} 3 & 1 \\ 2 & 3 \end{matrix} \right| & -\left| \begin{matrix} -5 & 1 \\ -3 & 3 \end{matrix} \right| & +\left| \begin{matrix} -5 & 3 \\ -3 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} -1 & 3 \\ 2 & 3 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ -3 & 3 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ -3 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} -1 & 3 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ -5 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ -5 & 3 \end{matrix} \right| \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 7 & 12 & -1 \\ 9 & 15 & -1 \\ -10 & -17 & 1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] \)
எனவே,  A^-1 = \(\frac { 1 }{ |A| } (adj\quad A)=\frac { 1 }{ (-1) } \left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -7 \\ -1 & -1 & 1 \end{matrix} \right] =\left[ \begin{matrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{matrix} \right] \)

A, B மற்றும் C பொருட்களின் ஓரலகு விலை ரூ. x, ரூ.y மற்றும் ரூ.z என்க.
கொடுக்கப்பட்ட தரவின்படி,
2x-4y+5z  = 15000
3x+y-2z  = 1000
-x+3y+z  = 4000
சமன்பாடுகளின் தொகுப்பின் அணி வடிவம் 
\(\left[ \begin{matrix} 2 \\ 3 \\ -1 \end{matrix}\begin{matrix} -4 \\ 1 \\ 3 \end{matrix}\begin{matrix} 5 \\ -2 \\ 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
AX=B \(\Rightarrow X={ A }^{ -1 }B\) இங்கு \(A=\left[ \begin{matrix} 2 \\ 3 \\ -1 \end{matrix}\begin{matrix} -4 \\ 1 \\ 3 \end{matrix}\begin{matrix} 5 \\ -2 \\ 1 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
\(B=\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
மற்றும் \(|A|=\left| \begin{matrix} 2 \\ 3 \\ -1 \end{matrix}\begin{matrix} -4 \\ 1 \\ 3 \end{matrix}\begin{matrix} 5 \\ -2 \\ 1 \end{matrix} \right| \)
\(=2\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| +4\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| +5\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \)
=2(1+6)+4(3-2)+5(9+1)
=2(7)+4(1)+5(10)=14+4+50=68
A^-1 ஐக் காண இயலும் 
\(={ \left[ \begin{matrix} +(1+6) \\ -(-4-15) \\ +(8-5) \end{matrix}\begin{matrix} -(3-2) \\ +(2+5) \\ -(-4-15) \end{matrix}\begin{matrix} +(9+1) \\ -(6-4) \\ +(2+12) \end{matrix} \right] }^{ T }\)
\(={ \left[ \begin{matrix} 7 \\ 19 \\ 3 \end{matrix}\begin{matrix} -1 \\ 7 \\ 19 \end{matrix}\begin{matrix} 10 \\ -2 \\ 14 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 7 \\ -1 \\ 10 \end{matrix}\begin{matrix} 19 \\ 7 \\ -2 \end{matrix}\begin{matrix} 3 \\ 19 \\ 14 \end{matrix} \right] \)
\(\therefore { A }^{ -1 }=\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ 68 } \left[ \begin{matrix} 7 \\ -1 \\ 10 \end{matrix}\begin{matrix} 19 \\ 7 \\ -2 \end{matrix}\begin{matrix} 3 \\ 19 \\ 14 \end{matrix} \right] \)
\(\therefore X={ A }^{ -1 }B=\frac { 1 }{ 68 } \left[ \begin{matrix} 7 \\ -1 \\ 10 \end{matrix}\begin{matrix} 19 \\ 7 \\ -2 \end{matrix}\begin{matrix} 3 \\ 19 \\ 14 \end{matrix} \right] \left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
\(=\frac { 1 }{ 68 } \left[ \begin{matrix} 105000 \\ -15000 \\ 150000 \end{matrix}\begin{matrix} +19000 \\ +7000 \\ -2000 \end{matrix}\begin{matrix} +12000 \\ +76000 \\ +56000 \end{matrix} \right] \)
\(=\frac { 1 }{ 68 } \left[ \begin{matrix} 136000 \\ 68000 \\ 204000 \end{matrix} \right] =\left[ \begin{matrix} 2000 \\ 1000 \\ 3000 \end{matrix} \right] \)
∴ ரூ.2000, ரூ.1000, ரூ.3000 
A, B மற்றும் C பொருட்களின் ஒரலகு விலைகள் முறையே ரூ.2000, ரூ.1000, மற்றும் ரூ.3000 ஆகும்.

(i) \(A=\left[ \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right] \)= \(\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]\)என்க.
[ ∵ a₂₂, a₁₁ ஆகியவற்றை பரிமாற்றம் செய்க மற்றும் a₁₂, a₂₁ ஆகியவற்றை குறி மாற்றம் செய்க.]
adj A = \(\left[ \begin{matrix} 2 & -4 \\ -6 & -3 \end{matrix} \right] \)
(ii) A = \(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
adj A = \(\left( \begin{matrix} +\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix} \right| \end{matrix}\begin{matrix} -\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| \end{matrix}\begin{matrix} +\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \\ -\left| \begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix} \right| \\ +\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| \end{matrix} \right) ^{ T }\)
= \({ \left[ \begin{matrix} +(8-7) \\ -(6-7) \\ +(3-4) \end{matrix}\begin{matrix} -(6-3) \\ +(4-3) \\ -(2-3) \end{matrix}\begin{matrix} +(21-12) \\ -(14-9) \\ +(8-9) \end{matrix} \right] }^{ T }\)
= \({ \left[ \begin{matrix} 1 \\ 1 \\ -1 \end{matrix}\begin{matrix} -3 \\ 1 \\ 1 \end{matrix}\begin{matrix} 9 \\ -5 \\ -1 \end{matrix} \right] }^{ T }\)
\(adj\ A=\left[ \begin{matrix} 1 \\ -3 \\ 9 \end{matrix}\begin{matrix} 1 \\ 1 \\ -5 \end{matrix}\begin{matrix} -1 \\ 1 \\ -1 \end{matrix} \right] \)
(iii) \(A=\frac{1}{3}\left[\begin{array}{ccc} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{array}\right]_{3 \times 3}\)
adj (kA) = k^n-1 adj A
\(=\operatorname{adj} \mathrm{A}=\left(\frac{1}{3}\right)^2{ \left[ \begin{matrix} +\left| \begin{matrix} 1 & 2 \\ -2 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 2 & 1 \\ -2 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| \end{matrix}\begin{matrix} -\left| \begin{matrix} -2 & 2 \\ 1 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 2 & 1 \\ -2 & 2 \end{matrix} \right| \end{matrix}\begin{matrix} +\left| \begin{matrix} -2 & 1 \\ 1 & -2 \end{matrix} \right| \\ -\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| \\ +\left| \begin{matrix} 2 & 2 \\ -2 & 1 \end{matrix} \right| \end{matrix} \right] }^{ T }\)
\(=\frac { 1 }{ 9 } { \left[ \begin{matrix} +(2+4) \\ -(4+2) \\ +(4-1) \end{matrix}\begin{matrix} -(-4-2) \\ +(4-1) \\ -(4+2) \end{matrix}\begin{matrix} +(4-1) \\ +(-4-2) \\ +(2+4) \end{matrix} \right] }^{ T }\)
\(=\frac { 1 }{ 9 } { \left[ \begin{matrix} 6 \\ -6 \\ 3 \end{matrix}\begin{matrix} 6 \\ 3 \\ -6 \end{matrix}\begin{matrix} 3 \\ 6 \\ 6 \end{matrix} \right] }^{ T }=\frac { 1 }{ 9 } { \left[ \begin{matrix} 6 \\ 6 \\ 3 \end{matrix}\begin{matrix} -6 \\ 3 \\ 6 \end{matrix}\begin{matrix} 3 \\ -6 \\ 6 \end{matrix} \right] }\)
\(=\frac { 1 }{ 3 } \left[ \begin{matrix} 2 & -2 & 1 \\ 2 & 1 & -2 \\ 1 & +2 & 2 \end{matrix} \right] \)

adj (adj A) = |A|^{n - 2} A
|A|^3-2 = A = |A| A 
IA| = 3(-3 + 4) + 3(2) + 4(-2)
= 3 + 6 - 8 = 1
adj (adj A) = 1 A = A

\(\text { adj } A =\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \)
\(A(\operatorname{adj} A) =\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \)
\({\left[\begin{array}{ll} k & 0 \\ 0 & k \end{array}\right] } =\left[\begin{array}{cc} \cos ^2 \theta+\sin ^2 \theta & 0 \\ 0 & \sin ^2 \theta+\cos ^2 \theta \end{array}\right] \)
\(k =\cos ^2 \theta+\sin ^2 \theta=1\)

\(A B=I \Rightarrow B=A^{-1}=\frac{1}{1+\tan ^2 \frac{\theta}{2}}\left[\begin{array}{cr} 1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1 \end{array}\right] \)
\(=\frac{1}{\sec ^2 \frac{\theta}{2}}\left[\begin{array}{cc} 1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1 \end{array}\right]=\cos ^2 \frac{\theta}{2} A^{\top}\)

|A| = 14 - 12 = 2 ≠ 0
\( \text { adj } A=\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right] \)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A\)
\(A^{-1}=\frac{1}{2}\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]=2 A^{-1} = \left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]\)
\(\begin{aligned} & 9I-A=9\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 7 & 3 \\ -4 & -3 \end{array}\right]=\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right] \end{aligned}\)
\(=\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right] = 2A^{-1}\)

|adj(adj A)| = |A|⁹
\(|A|^{(n-1)^2}=|\mathrm{A}|^9\)
(n - 1)² = 9 = 3²
n - 1 = 3 \(\Rightarrow\) n = 4