Let x represent the number of question with correct answer and y represent the number of questions with wrong answers.
By the given data, x + y 100 ............... (1)
x - \(\frac { 1 }{ 4 } \)y = 80
Multiplying by 4 we get we get
4x - y = 320...............(2)
From (1) and (2)
Δ = \(\\ \left| \begin{matrix} 1 & 1 \\ 4 & -1 \end{matrix} \right| \)= -1 - 4 = -5
Δ₁ = \(\left| \begin{matrix} 100 & 1 \\ 320 & -1 \end{matrix} \right| \) = -100 - 320 = -420
Δ₂ = \(\left| \begin{matrix} 1 & 100 \\ 4 & 320 \end{matrix} \right| \) = 320 - 400 = -80
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -720 }{ -5 } \) = +84
and y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -80 }{ -5 } \) = 16
Hence, the number of questions with correct answer is 84 and wrong question is 16.

5x − 2y + 16 = 0, x + 3y − 7 = 0
Given Δ = \(\left| \begin{matrix} 5 & -2 \\ 1 & 3 \end{matrix} \right| \) = 15+2 = 17
Δ₁ = \(\left| \begin{matrix} -16 & -2 \\ 7 & 3 \end{matrix} \right| \) = -48+14 = -34
Δ₂ = \(\left| \begin{matrix} 5 & -16 \\ 1 & 7 \end{matrix} \right| \) = 35+16 = 51
∴ x = \(\frac { \triangle _{ 1 } }{ \triangle } =\frac { -34 }{ 7 } \) = -2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 51 }{ 17 } \)
∴ Solution set is {-2, 3}

2x+5y = -2, x+2y = -3
The matrix form of the system is
\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
⇒ AX = B where
A =\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) ,B=\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
X =\(\left( \begin{matrix} x \\ y \end{matrix} \right) \)
⇒ = A^-1B
|A| = \(\left| \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right| \)= 4 - 5 = -1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ -1 } \left[ \begin{matrix} 2 & -5 \\ -1 & 2 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \)
∴ X = A^-1B =\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \left[ \begin{matrix} -2 \\ -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 4-15 \\ -2+6 \end{matrix} \right] =\left[ \begin{matrix} -11 \\ 4 \end{matrix} \right] \)
∴ Solution set is x = -11, y = 4

\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
A =\(\left[ \begin{matrix} 1 \\ 2 \\ 5 \end{matrix}\begin{matrix} 1 \\ -1 \\ -1 \end{matrix}\begin{matrix} 1 \\ 3 \\ 7 \end{matrix}\begin{matrix} 3 \\ 4 \\ 11 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { { R } }_{ 2 }-2{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 5 \end{matrix}\begin{matrix} 1 \\ -3 \\ -1 \end{matrix}\begin{matrix} 1 \\ 1 \\ 7 \end{matrix}\begin{matrix} 3 \\ -2 \\ 11 \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-5{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -6 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}\begin{matrix} 3 \\ -2 \\ -4 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { { R } }_{ 3 }-2{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix}\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} 3 \\ -2 \\ 0 \end{matrix} \right] \)
The last equivalent matrix is in row echelon form it ha two non-zero row \(\rho \)(A) = 2

Given A\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \)
Let B =\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \) and 
C = \(\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \)
∴ AB = C
Post multiply by B^-1 we get
A(BB^-1) = CB^-1
⇒ A = CB^-1     [∵ BB^-1 = 1]
|B| = \(\left| \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right| \)
= -10 + 3 = -7 ≠ 0
∴ B^-1 exists
B^-1 = \(\frac { 1 }{ |B| } adjB=\frac { -1 }{ 7 } \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
A = CB^-1
=\(\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \left( \frac { -1 }{ 7 } \right) \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(7\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} \right] \left( \frac { -1 }{ 7 } \right) \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(-\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(-\left[ \begin{matrix} -4+1 & -6+5 \\ -2+1 & -3+5 \end{matrix} \right] =-\left[ \begin{matrix} -3 & -1 \\ -1 & 2 \end{matrix} \right] \)
A = \(\left[ \begin{matrix} 3 & 1 \\ 1 & -2 \end{matrix} \right] \).

Given A = \(\left[ \begin{matrix} 1 & \tan { x } \\ -\tan { x } & 1 \end{matrix} \right] \)
|A| = 1 + tan² x
∴ A^-1 = \(\frac { 1 }{ |A| } \)adjA
= \(\frac { 1 }{ 1+tan^{ 2 } } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of elements in the off diagonal]
A^T= \(\\ \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
∴ A^TA^-1
=\(\left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1-tan^{ 2 }x & -tanx-tanx \\ tanx+tanx & -tan^{ 2 }x+1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} \frac { 1-{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } & \frac { -2tanx }{ 1+{ tan }^{ 2 }x } \\ \frac { 2tanx }{ 1+{ tan }^{ 2 }x } & \frac { 1-{ tan }^{ 2 }x }{ 1+tan^{ 2 }x } \end{matrix} \right] \)
A^TA^-1 =\(\left[ \begin{matrix} cos2x & -sin2x \\ sin2x & cos2x \end{matrix} \right] \)
Hence proved.

Given adj A =\(\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
Now adj(adj A) =\(\left[ \begin{matrix} +\left| \begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix} \right| & -\left| \begin{matrix} 0 & 0 \\ -1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 0 & 2 \\ -1 & 0 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ -1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 1 & 0 \\ -1 & 0 \end{matrix} \right| \\ +\left| \begin{matrix} 0 & 1 \\ 2 & 0 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right| & +\left| \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(2-0) & -(0) & +(0+2) \\ -(0) & +(1+1) & -(0) \\ +(0+2) & -(0) & +(2-0) \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 2 & 0 & 2 \\ 0 & 2 & 0 \\ -2 & 0 & 2 \end{matrix} \right] ^{ T }\)
adj(adj A) =\(\left[ \begin{matrix} 2 & 0 & -2 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{matrix} \right] \)

Given adj A =\(\left[ \begin{matrix} 2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2 \end{matrix} \right] \)
We know that A = \(\pm \frac { 1 }{ \sqrt { |adjA| } } \) adj (adj A)..(1)
|adj A| = \(2\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| +4\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| +2\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \)
[Expanded along R₁]
= 2(24-0)+4(-6-14)+2(0+24)
= 2(24)+4(-20)+2(24) = 48-80+48
= 96-80 = 16
Now, adj (adj A)
=\(\left[ \begin{matrix} +\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| & -\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 2 \\ 0 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 2 \\ -2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -2 & 0 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 2 \\ 12 & -7 \end{matrix} \right| & -\left| \begin{matrix} 2 & 2 \\ -3 & -7 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ -3 & 12 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(24-0)-(6-14)+(0+24) \\ -(-8-0)+(4+4)-(0-8) \\ +(28-24)-(-14+6)+(24-12) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & 20 & 24 \\ 8 & 8 & 8 \\ 4 & 8 & 12 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & 8 & 4 \\ 20 & 8 & 8 \\ 24 & 8 & 12 \end{matrix} \right] \)
= \(4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
Substituting (2) and (3) in (1) we get,
A = \(\frac { 1 }{ \sqrt { 16 } } .4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
A = \(\pm \frac { 4 }{ 4 } \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] =\pm \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)

Given A =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] \)
AB =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] =\left[ \begin{matrix} -3+10 & -9+4 \\ -7+25 & -21+10 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 7 & -5 \\ 18 & -11 \end{matrix} \right] \)
|AB| = -77+90 = 13 ≠ 0 ⇒ (AB)^-1 exists
|A| = 15-14 = 1 ≠ 0 ⇒ A^-1 exists
|B| = -2+15 = 13 ≠ 0 ⇒ B^-1 exists
(AB)^-1 = \(\frac { 1 }{ |AB| } adj(AB)=\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ...............(1)
B^-1 = \(\frac { 1 }{ |B| } adj(B)=\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \)
A^-1 = \(\frac { 1 }{ |A| } \)(adj A)
= \(\frac { 1 }{ 1 } \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) =\left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
∴ B^-1A^-1 = \(\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} 10-21 & -4+9 \\ -25+7 & 10-3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ..............(2)
 From (1) or (2) it is proved that
(AB)^-1 = B^-1 A^-1

Given A = \(\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \)
A^T = \(\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] \)...............(1)
We know that (\(\lambda\)A) -1 = \(\frac { 1 }{ \lambda } \)A-1
A-1 =\(\left\{ \frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \right\} ^{ -1 }=\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] ^{ T }\)
where \(\lambda\) = \(\frac{1}{9}\)
A^-1= 9B^-1 where B =\(\left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \) ............(2)
Now, |B|=\(-8\left| \begin{matrix} 4 & 7 \\ -8 & 4 \end{matrix} \right| -1\left| \begin{matrix} 4 & 7 \\ 1 & 4 \end{matrix} \right| +4\left| \begin{matrix} 4 & 4 \\ 1 & -8 \end{matrix} \right| \)
= -8(16+56)-1(16-7)+4(-32-4)
= -8(72)-1(9)+4(-36) = -576-9-144
= -729
adj B =\(\left[ \begin{matrix} +\left| \begin{matrix} 4 & 7 \\ -8 & 4 \end{matrix} \right| & -\left| \begin{matrix} 4 & 7 \\ 1 & 4 \end{matrix} \right| & +\left| \begin{matrix} 4 & 4 \\ 1 & -8 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 4 \\ -8 & 4 \end{matrix} \right| & +\left| \begin{matrix} -8 & 4 \\ 1 & 4 \end{matrix} \right| & -\left| \begin{matrix} -8 & 1 \\ 1 & -8 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 4 \\ 4 & 7 \end{matrix} \right| & -\left| \begin{matrix} -8 & 4 \\ 4 & 7 \end{matrix} \right| & +\left| \begin{matrix} -8 & 1 \\ 4 & 1 \end{matrix} \right| \end{matrix} \right] \)
=\(\left[ \begin{matrix} +(16+56)-(16-7)+(-32-3) \\ -(4+32)+(-32-4)+(64-1) \\ +(7-16)-(-56-16)+(-3-4) \end{matrix} \right] \)
=\(\left[ \begin{matrix} 72 & -9 & -36 \\ -36 & -36 & -63 \\ -9 & 72 & -36 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{matrix} \right] \)
= \(9\left[ \begin{matrix} 8 & -4 & -1 \\ -1 & -4 & +8 \\ -4 & -7 & -4 \end{matrix} \right] \)
∴ B^-1=\(\frac { 1 }{ |B| } adjB=\frac { -9 }{ 729 } \left[ \begin{matrix} 8 & -4 & -1 \\ -1 & -4 & +8 \\ -4 & -7 & -4 \end{matrix} \right] \)
= \(\frac { 1 }{ 81 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] \)
Substituting this in (2) we get,
A^-1 = \(9.\frac { 1 }{ 81 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] =\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] \) ...............(3)
From (1) and (3)we get,
A^T = A^-1

Given A =\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \)
A² = \(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 25-3 & 15-6 \\ -5+2 & -3+4 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22 & 9 \\ -2 & 1 \end{matrix} \right] \)
∴ A²- 3A - 7I₂
=\(\left[ \begin{matrix} 22 & 9 \\ -3 & 1 \end{matrix} \right] -3\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -7\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22-15-7 & 9-9+0 \\ -3+3+0 & 1+6-7 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)= O₂
Hence proved.
∴ A²-3A-7I₂ = O₂
Postmultiplying by A^-1 we get,
A²-A^-1-3AA^-1-7I₂A^-1 = 0.A^-1
⇒ A(AA^-1)-3(AA^-1)-7(A^-1) = 0
[∵ I₂A^-1 = A^-1 and | (0)A^-1= 0]
⇒ AI-3I-7A-1 = 0  [∵ AA^-1= 1]
⇒ AI-3I = 7A^-1
⇒ A^-1 = \(\frac { 1 }{ 7 } \)[A - 3I] [∴ AI = A]
⇒ A^-1 = \(\frac { 1 }{ 7 } =\left[ \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \right] \)
⇒ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 5-3 & 3-0 \\ -1-0 & -2-3 \end{matrix} \right] =\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \).

Since A² = \(\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] =\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] \).
A² + xA + yI₂ = O₂ ⇒ \(\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] +x\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] +y\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 22+4x+y & 27+3x \\ 18+2x & 31+5x+y \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \).
So, we get 22 + 4x + y = 0, 31 + 5x + y = 0, 27 + 3x = 0 and 18 + 2x = 0.
Hence x = −9 and y = 14. Then, we get  A² - 9A + 14I₂ = O₂.
Postmultiplying this equation by A^-1, we get A - 9I₂ + 14A^-1 = O₂. Hence, we get
A^-1 = \(\frac { 1 }{ 14 } \) (9I₂ - A) = \(\frac { 1 }{ 14 } \left( 9\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \right) =\frac { 1 }{ 14 } \left[ \begin{matrix} 5 & -3 \\ -2 & 4 \end{matrix} \right] \).

Given C₂H₆ + O₂ ⟶ H₂O + CO₂
We have to find positive integers x₁ x₂, x₃ and x₄ such that
x₁C₂H₆ + x2O₆ ⟶ x₃H₂O + x₄CO₂ .....(1)
The number of carbon atoms on the LHS of (1) should be equal to the number of carbon atoms on the RHS of (1)
∴ 2x₁ = 1x₄
⇒ 2x₁-x₄ = 0 ...(2)
Considering hydrogen atoms we get,
6x₁ = 2x₃
⇒ 6x₁- 2x₃ = 0
⇒ 3x₁- x₃ = 0....(3)
Also, considering oxygen atoms we get,
2x₂ = 1x₃ + 2x₄
⇒ 2x₂ - x₃ - 2a₄ = 0 ....(4)
Equations (2), (3) and (4) forni a homogeneous system of linear equations in 4 unknowns
∴ The augmented matrix [A|0] is
\(\left[ \begin{matrix} 2 \\ 3 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ -1 \\ -1 \end{matrix}\begin{matrix} -1 \\ 0 \\ 2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 3 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ -1 \\ -1 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ 0 \\ -2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ 1 \\ -1 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ \frac { 3 }{ 2 } \\ -2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ -1 \\ 0 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ \frac { 3 }{ 2 } \\ \frac { -7 }{ 2 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ -1 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ \frac { -7 }{ 2 } \\ \frac { 3 }{ 2 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow 2{ R }_{ 2 }\\ R_{ 2 }\rightarrow 2{ R }_{ 2 } }{ \underset { { R }_{ 2 }\rightarrow 2{ R }_{ } }{ \longrightarrow } } \left[ \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 4 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ -2 \end{matrix}\begin{matrix} -1 \\ -7 \\ 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 1, the number of unknowns
∴ The system is consistent with one parameter family of solutions, so let x₄ = t
Writing the equations, from the row-echelon form we get
2x₁-x₄ = 0
⇒ 2x₁ = x₄
⇒ 2x = t
⇒ x₁ = \(\frac { t }{ 2 } \)
4x₂ - 7x₄ = 0
⇒ 4x₂ = 7t
⇒ x₂ = \(\frac { 7t }{ 4 } \)
-2x₃ + 3x₄ = 0
⇒ 2x₃ = 3x₄
⇒ x₃ = \(\frac { 3t }{ 4 } \)
Since x₁, x₂, x₃ and x₄ are positive integers, let us choose t = 4
∴ x₁ = \(\frac{4}{2}\) = 2

So, the balanced equation is
2C₂H₆ + 7O₂ ⟶ 6H₂O + 4CO₂

We are searching for positive integers x₁, x₂, x₃ and x₄ such that
x₁C₅H₈ + x₂O₂ ⟶ x₃CO₂ + x₄H₂O. ....(1)
The number of carbon atoms on the left-hand side of (1) should be equal to the number of carbon atoms on the right-hand side of (1) So we get a linear homogenous equation
5x₁= x₃ ⇒ 5x₁ - x₃ = 0 .........(2)
Similarly, considering hydrogen and oxygen atoms, we get respectively,
8x₁ = 2x₄ ⇒ 4x₁ - x₄ = 0 ............(3)
2x₂ = 2x₃ + x₄ ⇒ 2x₂ - 2x₄ = 0 .........(4)
Equations (2), (3), and (4) constitute a homogeneous system of linear equations in four unknowns.
The augmented matrix is [A | B] = \(\left[ \begin{matrix} 5 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 0 \\ 2 \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 0 \\ -2 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \right] \).
By Gaussian elimination method, we get
[A | B] \(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ \begin{matrix} 5 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 0 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} -1 \\ -2 \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 0 \\ -1 \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ \begin{matrix} 0 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -1 \\ 0 \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 4{ R }_{ 3 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} -2 \\ -4 \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -1 \\ 5 \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \right] \).
Therefore, ρ(A) = ρ([A | B]) = 3 < 4 = Number of unknowns.
The system is consistent and has infinite number of solutions.
Writing the equations using the echelon form, we get 4x₁ - x₄ = 0, 2x₂ - 2x₃ - x₄ = 0, -4x₃ + 5x₄ = 0.
So, one of the unknowns should be chosen arbitrarily as a non-zero real number.
Let us choose x₄ = t, t ≠ 0. Then, by back substitution, we get x₃ = \(\frac { 5t }{ 4 } \), x₂ = \(\frac { 7t }{ 4 } \), x₁ = \(\frac { t }{ 4 } \).
Since x₁, x₂, x₃ and x₄ are positive integers, let us choose t = 4.
Then, we get x₁ = 1, x₂ = 7, x₃ = 5 and x₄ = 4.
So, the balanced equation is C₅H₈ + 7O₂ ⟶ 5CO₂ + 4H₂O.

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 = \(\pm \frac { 1 }{ \sqrt { 36 } } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
= \(\pm \frac { 1 }{ 6 } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \).

Let A = \(\left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] \). Then, A^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }^{ T }=\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
So, we get
AA^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
= \(\left[ \begin{matrix} \cos ^{ 2 }{ \theta +\sin ^{ 2 }{ \theta } } & \cos { \theta \sin { \theta } } -\sin { \theta \cos { \theta } } \\ \sin { \theta \cos { \theta -\cos { \theta \sin { \theta } } } } & \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \) = I₂
Similarly, we get A^TA = I₂. Hence AA^T = A^TA = I₂ ⇒ A is orthogonal.

x + y + 3z = 0, 4x + 3y + λz = 0, 2x + y + 2z = 0
Reducing the augmented matrix to row - echelon form we get,
[A|0] = \(\left[ \begin{matrix} 1 & 1 & 3 \\ 4 & 3 & \lambda \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -4 \\ 0 & -1 & \lambda -2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \underset { { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } }{ \longrightarrow } } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -4 \\ 0 & 0 & \lambda -8 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -4 \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Case (i) when λ ≠ 8
[A|0] =\(\left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -4 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3, \(\rho \)([A|0] = 3
∴ \(\rho \)(A) = \(\rho \)([A|0]) = 3 = the number of unknowns
∴ The given system is consistent and has unique solution.
Case (ii) when λ = 8
Here \(\rho \)(A) = 2, \(\rho \)([A|0] = 2
∴ \(\rho \)(A) =\(\rho \)([A|0]) = 2 < 3 the number of unknowns,
∴ The system is consistent and has non-trivial solutions.

Assume that the system px + by + cz = 0, ax + qy + cz = 0, ax + by + rz = 0 has a non-trivial solution.
So, we have \(\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} \right| \) = 0, Applying R₂ ➝ R₂ - R₁ and R₃ ➝ R₃ - R₁ in the above equation,
we get \(\left| \begin{matrix} p & b & c \\ a-p & q-b & c \\ a-p & b & r-c \end{matrix} \right| \) = 0. That is,  \(\left| \begin{matrix} p & b & c \\ -\left( p-a \right) & q-b & c \\ -\left( p-a \right) & b & r-c \end{matrix} \right| \) = 0.
Since p ≠ a, q ≠ b, r ≠ c, we get (p - a)(q - b)(r - c) \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
So, we have \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
Expanding the determinant, we get \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 0.
That is, \(\frac { p }{ p-a } +\frac { q-\left( q-b \right) }{ q-b } +\frac { r-\left( r-c \right) }{ r-c } \) = 0
⇒ \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 2.

Here the number of unknowns is 3. So, if the system is consistent and has a non-trivial solution, then the rank of the coefficient matrix is equal to the rank of the augmented matrix and is less than 3.
So the determinant of the coefficient matrix should be 0.
Hence we get
\(\left| \begin{matrix} 3\lambda -8 & 3 & 3 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 or \(\left| \begin{matrix} 3\lambda -2 & 3\lambda -2 & 3\lambda -2 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by applying R₁ ➝ R₁ + R₂ + R₃)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by taking out (3λ − 2) from R₁)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -11 & 3 \\ 3 & 3 & 3\lambda -11 \end{matrix} \right| \) = 0 (by applying R₂ ➝ R₂ - 3R₁, R₃ ➝ R₃ - 3R₁)
or (3λ - 2)(3λ - 11)² 0. So λ = \(\frac { 2 }{ 3 } \) and λ = \(\frac { 11 }{ 3 } \).
We now give an application of system of linear homogeneous equations to chemistry. You are already aware of balancing chemical reaction equations by inspecting the number of atoms present on both sides.

2x+3y = 9, 7x+3y-5z = 8, 2x+3y+⋋z = μ
The matrix form of the system is AX = B where
A =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \)
Applying elementary row operations augmented matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 2 & 3 & 5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 8 \\ 9 \\ \mu \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-\frac { 2 }{ 7 } { R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & \frac { 15 }{ 7 } & \frac { 45 }{ 7 } \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ \frac { 45 }{ 7 } \\ 4-9 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 7 }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ 47 \\ \mu -9 \end{matrix} \right] \)
Case (i): when λ = 5
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ -4 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B]
Hence the system is inconsistent and has no solution
Case (ii) : When λ ≠ 5, μ ≠ 9
[A|B] =\(\\ \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} -8 \\ 47 \\ not\quad zero \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
∴ \(\rho \)(A) = \(\rho \)[A|B] = 3 = number of unknowns
Hence, the system is consistent with solution
Case (iii) : When λ = 5, μ = 9
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2
∴ The system is consistent and has infinite number of solutions.

kx-2y+z = 1, -2ky+z = -2, x-2y+k = 1 
The matrix form of the system is AX = B where
\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
Applying elementary row operation on the augment matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k & 1 \\ k & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-k{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k+2 & k \\ 0 & -2+2k & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ - \\ 1-k \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & k \\ 0 & 0 & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ -3 \\ 1-k \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & { k }^{ 2 }-k+2 \end{matrix}\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & (k+2)(1-k) \end{matrix}|\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \).........(1)
Case (i): when k = 1
\([A|B]\rightarrow \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -3 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B] ⇒ The system has no solution
Case (ii): when k ≠ 2, k ≠ -2 
\(\left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} 1 \\ -3 \\ not\quad zero \end{matrix} \right] \)
⇒ \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
so, \(\rho \)(A) =\(\rho \)[A|B] = 3 = the number of unknowns Hence, the system has unique solution.
Case (iii): when k = -2
\(\rho [A|B]\rightarrow \left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} -2 \\ 6 \\ 0 \end{matrix}\begin{matrix} -2 \\ 3 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \) (A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2<3 the number of unknowns so the system is consistent with infinitely many solutions.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where A = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix}|\begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 3 & -5 \\ 1 & 1 & \lambda \end{matrix}|\begin{matrix} 7 \\ 5 \\ \mu \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & -1 & \lambda -1 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -7 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & 0 & \lambda -7 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -9 \end{matrix} \right] \).
(i) If λ =7 and μ \(\neq\) 9, then ρ(A) = 2 and ρ([A | B]) = 3. So ρ(A) ≠ ρ([A | B]) Hence the given system is inconsistent and has no solution.
(ii) If λ ≠ 7 and μ is any real number, then ρ(A) = 3 and ρ([A | B]) = 3.
So, ρ(A) = ρ([A | B]) = 3 = Number of unknown. Hence the given system is consistent and has a unique solution.
(iii) If λ = 7 and μ = 9, then ρ(A) = 2 and ρ([A | B]) = 2.
So, ρ(A) = ρ([A | B]) = 2 < Number of unknown. Hence the given system is consistent and has infinite number of solutions.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where A = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 5 & 7 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} a \\ b \\ c \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 5 & 7 \end{matrix}|\begin{matrix} a \\ b \\ c \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & 4 \end{matrix}|\begin{matrix} a \\ b-a \\ c-3a \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} a \\ b-a \\ \left( c-3a \right) -2\left( b-a \right) \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} a \\ b-a \\ \left( c-2b-a \right) \end{matrix} \right] \).
In order that the system should have one parameter family of solutions, we must have ρ(A) = ρ([A, B]) = 2. So, the third row in the echelon form should be a zero row.
So, c − 2b − a = 0 ⇒ c = a + 2b.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix[A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix}|\begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -9 \\ 0 \\ 0 \end{matrix} \right] \).
So, ρ(A) = ρ ([A | B]) = 1 < 3.
From the echelon form, we get the equivalent equations x - y + z = -9, 0 = 0, 0 = 0.
The equivalent system has one non-trivial equation and three unknowns.
Taking y = s, z = t arbitrarily, we get x - s + t = -9; x = -9 + s - t.
So, the solution is (x = -9 + s - t, y = s, z = t), where s and t are parameters.
The above solution set is a two-parameter family of solutions.
Here, the given system of equations is consistent and has infinitely many solutions which form a two parameter family of solutions.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 3 \\ \begin{matrix} 1 \\ \begin{matrix} 3 \\ -1 \end{matrix} \end{matrix} \end{matrix} \right] \).
The augmented matrix is [A | B] = \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 3 & -1 & 2 & 1 \\ 1 & -2 & 3 & 3 \\ 1 & -1 & 1 & -1 \end{array}\right]\).
Applying Gaussian elimination method on [A | B], we get
[A | B] \(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 }, \\ { R }_{ 4 }\longrightarrow { R }_{ 4 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -7 \\ \begin{matrix} -4 \\ -3 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 5 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} -8 \\ \begin{matrix} 0 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow \left( -1 \right) { R }_{ 2 }, \\ { R }_{ 3 }\longrightarrow \left( -1 \right) { R }_{ 3 } \\ { R }_{ 4 }\longrightarrow \left( -1 \right) { R }_{ 4 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 4 \\ 3 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} -4 \\ -2 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} 0 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 3 }\longrightarrow 7{ R }_{ 3 }-4{ R }_{ 2 } \\ { R }_{ 4 }\longrightarrow 7{ R }_{ 4 }-3{ R }_{ 2 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} -8 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} -32 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -8 \right) }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} 4 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 4 }\longrightarrow { R }_{ 4 }-{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \)
There are three non-zero rows in the row-echelon form of [A | B]. So, ρ([A | B]) = 3.
So, the row-echelon form of A is \(\left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \). There are three non-zero rows in it. So ρ(A) = 3.
Hence, ρ(A) = ρ([A | B]) = 3.
From the echelon form, we write the equivalent system of equations
x + 2y − z = 3,7y − 5z = 8, z = 4, 0 = 0.
The last equation 0=0 is meaningful. By the method of back substitution, we get
z = 4
7y − 20 = 8 ⇒ y = 4 ,
x = 3 − 8 + 4 ⇒ x = −1.
So, the solution is (x = −1, y = 4, z = 4).(Note that A is not a square matrix.)
Here the given system is consistent and the solution is unique.

Giveny = ax² + bx + c .............(1)
(-6, 8) lies on (1)
⇒ 8 = a(-6)²+b(-6)+c
⇒ 8 = 36z-6b+c ..........(2)
(-2,12) lies on (1)
⇒ -12 = a(-2)²+b(-2)+c
⇒ -12 = 4a-2b+c ...........(3)
Also (3, 8) lies on (1)
⇒ 8 = a(3)²+b(3)+c
⇒ 8 = 9a+3b+c ............(4)
Reducing the augment matrix to an equivalent row-echelon form by using elementary. row operations, we get,
\(\left[ \begin{matrix} 36 & -6 & 1 \\ 4 & -2 & 1 \\ 0 & 3 & 1 \end{matrix}|\begin{matrix} 8 \\ -12 \\ 8 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { 9R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow 4{ R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -12 & 8 \\ 0 & 18 & 3 \end{matrix}|\begin{matrix} 8 \\ -116 \\ 24 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div 4\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 3 }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -8 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -50 \end{matrix} \right] \)
Writing the equivalent equation from the row echelon matrix, we get
36a - 6b + c = 8 ........(1)
-3b+2c = -29 ....(2)
5c = -50
⇒ c = \(\frac{-50}{5}\) = -10
Substituting c = -10 in (2) we get,
-3b+2(-10)= -29
⇒ -3b+2(-10) = -29
⇒ -3b-20 = -29
⇒ -3b = -9
⇒ b = \(\frac{-9}{-3}\) = 3
Substituting b = 3 and c = -10 in (1) we get,
36a-6(3)-10 = 8
⇒ 36a-18-10 = 8
⇒ 36a-28 = 8
⇒ 6a = 8+28 = 36
⇒ a = \(\frac{36}{36}\) = 1
∴ a = 1, b = 3, c = -10
Hence the path of the boy is
y = 1(x²)+3(x)-10
⇒ y = x²+3x-10
Since his friend is at P(7, 60),
60 = (7)²+3(7)-10
⇒ 60 = 49+21-10
⇒ 60 = 70-10 = 60
⇒ 60 = 60
Since (7, 60) satisfies his path, he can meet his friend who is at P(7, 60)

Let the price of bond invested in 6%, 8% and 9% rates be let Rs. x, Rs. y and Rs. z respectively
∴ By the given data, x + y + z = 65000 ..........(1)
\(\frac { 6\times x\times 1 }{ 100 } +\frac { 8\times y\times 1 }{ 100 } +\frac { 9\times z\times 1 }{ 100 } \) = 4800
[∵ Intrest = \(\frac { PNR }{ 100 } \)]
⇒ \(\frac { 6x }{ 100 } +\frac { 8y }{ 100 } +\frac { 9z }{ 100 } \)= 4800
⇒ 6x+8y+9z = 480000 ............(2)
Also, \(\frac { 9z }{ 100 } =600+\frac { 8y }{ 100 } \)
⇒ \(\frac { -8y }{ 100 } +\frac { 9y }{ 100 } \) = 600
⇒ -8y+9z = 60000 ............(3)
Reducing the augmented matrix to an equivalent row-echelon form by using elementary row operation, we get
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & 8 & 9 \\ 0 & -8 & 9 \end{matrix}|\begin{matrix} 65000 \\ 480000 \\ 60000 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-6{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & -8 & 9 \end{matrix}|\begin{matrix} 65000 \\ 90000 \\ 60000 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 21 \end{matrix}|\begin{matrix} 65000 \\ 90000 \\ 420000 \end{matrix} \right] \)
Writing the equivalent from the row echelon matrix we get,
x+y+z = 65000 ...........(1)
2y+z = 90000 ...........(2)
21z = 42000
⇒ z = \(\frac { 420000 }{ 21 } \) = 20000
Substituting z = 20,000 in (2),
2y + 3(20,000) = -90000
⇒ 2y+60,000 = 90,000
⇒ 2y = 90,000 - 60,000
= 30,000
⇒ y = \(\frac { 30,000 }{ 2 } \) = 15,000
Substitutingy = 15,000 and z = 20,000 in (1) we get,
x + 15,000 + 20,000 = 65000
⇒ x + 35,000 = 65000
⇒ x = 65,000 - 35,000
⇒ 30,000
Thus the price of 6% bond is f 30,000 the price of 8% bond is f 15,000 and the price of 9% bond is f 20,000 is Rs. 20,000.

Let P(x) = ax²+ bx + c
Given P(-3) = 21
[∵ P(x) ÷ x + 3, the remainder is 21]
⇒ a(-3)2 + b(-3) + c = 21
⇒ 9a - 3b + c = 21
Also, P(5) = 61
⇒ a(5)2 + b(5) + c = 61
[using remainder theorem]
⇒ 25a +5b + c = 61..........(2)
and P(1) = 9
⇒ a(1)2 + b(1) + c = 9
⇒ a + b + c = 9 ............(3)
Reducing the augment matrix to an equivalent row-echelon form using elementary row operations, we get
\(\left[ \begin{matrix} 9 & - & 1 \\ 25 & 5 & 1 \\ -1 & 1 & 1 \end{matrix}|\begin{matrix} 21 \\ 61 \\ 9 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 25 & 5 & 1 \\ 9 & -3 & 1 \end{matrix}|\begin{matrix} 9 \\ 61 \\ 21 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-9{ R }_{ 1 }\\ { R }_{ 2 }\rightarrow { R }_{ 2 }-25{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -20 & -24 \\ 0 & -12 & -8 \end{matrix}|\begin{matrix} 9 \\ -164 \\ -60 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div \\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 4 }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & -3 & -2 \end{matrix}|\begin{matrix} 9 \\ -41 \\ -15 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 3 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & \frac { 8 }{ 5 } \end{matrix}|\begin{matrix} 9 \\ -41 \\ \frac { 48 }{ 5 } \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow 5{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & 8 \end{matrix}|\begin{matrix} 9 \\ -41 \\ 48 \end{matrix} \right] \)
Writing the equivalent equations from the row-echelon matrix we get,
a + b + c = 9 ............(1)
5b + 6c = 41 ................(2)
-8c = -48
⇒ c = 6
Substituting c = 6 
⇒  5b + 36 = 41
⇒ 5b = 5  
b = 1
Substituting b = 1, c = 6
a + 1 + 6 = 9
⇒ a + 7 = 9
⇒ a = 9 - 7
⇒ a = 2
∴ a = 2, b = 1, and c = 6

Since v(3) = 64, v(6) = 133,and v(9) = 208 , we get the following system of linear equations
9a + 3b + c = 64 ,
36a + 6b + c = 133 ,
81a + 9b + c = 208 .
We solve the above system of linear equations by Gaussian elimination method.
Reducing the augmented matrix to an equivalent row-echelon form by using elementary row
operations, we get
[A | B] = \(\left[ \begin{matrix} 9 & 3 & 1 \\ 36 & 6 & 1 \\ 81 & 9 & 1 \end{matrix}|\begin{matrix} 64 \\ 133 \\ 208 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 },{ R }_{ 3 }\longrightarrow { R }_{ 3 }-9{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & -6 & -3 \\ 0 & -18 & -8 \end{matrix}|\begin{matrix} 64 \\ -123 \\ -368 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -3 \right) ,{ R }_{ 3 }\div \left( -2 \right) }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 9 & 4 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 184 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\longrightarrow 2{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 18 & 8 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 368 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-9{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 64 \\ 41 \\ -1 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow \left( -1 \right) { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 1 \end{matrix} \right] \).
Writing the equivalent equations from the row-echelon matrix, we get
9a + 3b + c = 64, 2b + c = 41, c = 1.
By back substitution, we get c = 1, b = \(\frac { \left( 41-c \right) }{ 2 } =\frac { \left( 41-1 \right) }{ 2 } \) = 20, a = \(\frac { 64-3b-c }{ 9 } =\frac { 64-60-1 }{ 9 } =\frac { 1 }{ 3 } \).
So, we get v(t) = \(\frac { 1 }{ 3 }\)t² + 20t + 1. Hence, v(15) = \(\frac { 1 }{ 3 }\) (225) + 20(15) + 1 = 75 + 300 + 1 = 376.

Let the cost of one dosa be Rs. x
The cost of one idli be Rs. y
and the cost of one vadai be Rs. z
By the given data,
2x+ 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
∴ Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 2 & 4 \\ 4 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 2 \end{matrix} \right| +2\left| \begin{matrix} 2 & 2 \\ 5 & 4 \end{matrix} \right| \)
= 2(4 - 16) - 3(4 - 20) + 2(8 - 10)
= 2(- 12) - 3(- 16) + 2(- 2)
= - 24 + 48 - 4 = 20
Δ₁ = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
Taking 50 common from C3 we get,
= 100\(\left| \begin{matrix} 3 & 3 & 1 \\ 4 & 2 & 2 \\ 5 & 4 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 2 & 2 \\ 4 & 1 \end{matrix} \right| -3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 4 & 2 \\ 5 & 4 \end{matrix} \right| \right] \)
= 100[3(2 - 8) - 3(4 - 10) + 1(16 - 10)]
= 100[3(-6) - 3(- 6) + 6]
= 100[- 18 + 18 + 6] = 600
Δ₂ = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| =100\left| \begin{matrix} 2 & 3 & 1 \\ 2 & 4 & 2 \\ 5 & 5 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| \right] \)
= 100[2(4 - 10) - 3(2 - 10) + 1(10 - 20)]
= 100[2(- 6) - 3(- 8) + 1(- 10)]
= 100[- 12 + 24 - 10] = 100 [2] = 200
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| =50\left| \begin{matrix} 2 & 3 & 3 \\ 2 & 2 & 4 \\ 5 & 4 & 5 \end{matrix} \right| \)
= \(50\left[ 2\left| \begin{matrix} 2 & 4 \\ 4 & 5 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| +3\left| \begin{matrix} 2 & 2 \\ 4 & 4 \end{matrix} \right| \right] \)
= 50 [2(10 - 16) - 3(10 - 20) + 3(8 - 10)]
= 50[2(- 6) - 3(- 10) +3(- 2)]
= 50 [- 12 + 30 - 6] = 50 [12] = 600
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 200 }{ 20 } \) = 10
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
Hence, the price of one dosa be Rs. 30, one idli be Rs. 10 and the price of 1 vadai be Rs. 30.
Also the cost on dosa, six idlies and six vadai is
= 3x + 6y + 6z = 3(30) + 6(10) + 6(30)
= 90 + 60 + 180 = Rs. 330
Since the family had Rs. 350 in hand, they will be able to manage to pay the bill.

Let the pump A can fill the tank in x minutes, and the pump B can fill. the tank in y minutes In 1 minute A can fill \(\frac { 1 }{ x } \) units and in 1 minute B can fill \(\frac { 1 }{ y } \) units.
∴ \(\frac { 1 }{ x } +\frac { 1 }{ y } \) = 10
and \(\frac { 1 }{ x } -\frac { 1 }{ y } \) = 30
Put \(\frac { 1 }{ x } \) = a and \(\frac { 1 }{ y } \) = b
⇒ a + b = \(\frac { 1 }{ 10 } \) ..............(1)
and a - b = \(\frac { 1 }{ 30 } \) .............(2)
Δ = \(\left| \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix} \right| \)= -1-1= -2
Δ₁ = \(\left| \begin{matrix} \frac { 1 }{ 10 } & 1 \\ \frac { 1 }{ 30 } & -1 \end{matrix} \right| =\frac { -1 }{ 10 } -\frac { 1 }{ 30 } \)
= \(\frac { -3-1 }{ 30 } =\frac { -4 }{ 30 } =\frac { -2 }{ 15 } \)
Δ₂ = \(\left| \begin{matrix} 1 & \frac { 1 }{ 10 } \\ 1 & \frac { 1 }{ 30 } \end{matrix} \right| =\frac { 1 }{ 30 } -\frac { 1 }{ 10 } =\frac { 1-3 }{ 30 } \)
= \(\frac { -2 }{ 30 } =\frac { -1 }{ 15 } \)
∴ a = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -2 }{ \frac { 15 }{ -2 } } =\frac { 1 }{ 15 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 15 } \Rightarrow \)= x = 15
b = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -1 }{ \frac { 15 }{ -2 } } =\frac { 1 }{ 30 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 30 } \Rightarrow \)y = 30
Hence the pump A can fill the tank in 15 minutes and the pump B can fill the tank in 30 minutes.

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.

The path y = ax² + bx + c passes through the points (10, 8), (20, 16), (40, 22). So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c = 16, 1600a + 40b + c = 22. To apply Cramer’s rule, we find
Δ = \(\left| \begin{matrix} 100 & 10 & 1 \\ 400 & 20 & 1 \\ 1600 & 40 & 1 \end{matrix} \right| =1000\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{matrix} \right| \) = 1000 [-2 + 12 - 6] = -6000,
Δ₁ = \(\left| \begin{matrix} 8 & 10 & 1 \\ 16 & 20 & 1 \\ 22 & 40 & 1 \end{matrix} \right| =20\left| \begin{matrix} 4 & 1 & 1 \\ 8 & 2 & 1 \\ 11 & 4 & 1 \end{matrix} \right| \) = 20[-8 + 3 + 10] = 100,
Δ₂ = \(\left| \begin{matrix} 100 & 8 & 1 \\ 400 & 16 & 1 \\ 1600 & 22 & 1 \end{matrix} \right| =200\left| \begin{matrix} 1 & 4 & 1 \\ 4 & 8 & 1 \\ 16 & 11 & 1 \end{matrix} \right| \) = 200[-3 + 48 - 84] = -7800,
Δ₃ = \(\left| \begin{matrix} 100 & 10 & 8 \\ 400 & 20 & 16 \\ 1600 & 40 & 22 \end{matrix} \right| =2000\left| \begin{matrix} 1 & 1 & 4 \\ 4 & 2 & 8 \\ 16 & 4 & 11 \end{matrix} \right| \) = 2000[-10 + 84 - 64] = 20000.
By Cramer’s rule, we get a = \(\frac { { \Delta }_{ 1 } }{ \Delta } =-\frac { 1 }{ 60 } \), b = \(\frac { { \Delta }_{ 2 } }{ \Delta } =\frac { 7800 }{ 6000 } =\frac { 78 }{ 60 } =\frac { 13 }{ 10 } \), c = \(\frac { { \Delta }_{ 3 } }{ \Delta } =\frac { 20000 }{ 6000 } =-\frac { 20 }{ 6 } =-\frac { 10 }{ 3 } \).
So, the equation of the path is y = \(\frac { 1 }{ 60 } { x }^{ 2 }+\frac { 13 }{ 10 } x-\frac { 10 }{ 3 } \).
When x = 70, we get y = 6. So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before the boundary line to jump and catch the ball. Hence the ball went for a super six and the team won the match.

Let the prices per unit for the commodities A, B and C be Rs. x, Rs. y and Rs. z.
By the given data,
2x - 4y + 5z = 15000
3x + y - 2z = 1000
-x + 3y + z = 4000
The matrix form of the system of equations is
\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 12 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
and B =\(\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| +4\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| +5\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \)
= 2 (1 + 6) + 4 (3 - 2) + 5 (9 + 1)
= 2 (7) + 4 (1) + 5(10) = 14 + 4 + 50 = 68.
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 5 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 5 \\ -1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -1 & 3 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 5 \\ 1 & -2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 5 \\ 3 & -2 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ 3 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
= \(\left[ \begin{matrix} +(1+6) & -(3-2) & +(9+1) \\ -(-4-15) & +(2+5) & -(6-4) \\ +(8-5) & -(4-15) & +(2+12) \end{matrix} \right] \)
= \(\left[ \begin{matrix} 7 & -1 & 10 \\ 19 & 7 & -2 \\ 3 & 19 & 14 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adj=\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 105000+19000+12000 \\ -15000+7000+76000 \\ 150000-2000+56000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 136000 \\ 68000 \\ 204000 \end{matrix} \right] =\left[ \begin{matrix} 2000 \\ 1000 \\ 3000 \end{matrix} \right] \)
∴ x = 2000, y = 1000, z = 3000
Hence the prices per unit of the commodities A, B and C are Rs. 2000, Rs. 1000 and Rs. 3000 respectively.

Let the time by one man alone be x days and one woman alone be y days
∴ By the given data,
\(\frac { 4 }{ x } +\frac { 4 }{ y } =\frac { 1 }{ 3 } \)
and \(\frac { 2 }{ x } +\frac { 5 }{ y } =\frac { 1 }{ 4 } \)
put \(\frac { 1 }{ x } \) = s and \(\frac { 1 }{ y } \) = t
∴ 4s + 4t = \(\frac { 1 }{ 3 } \)
and 2s + 5t = \(\frac { 1 }{ 4 } \)
The matrix form of the system of equation is
\(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} s \\ t \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \) ⇒ AX = B where
A = \(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \)
X = A^-1B
Now |A| = \(\left| \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right| \) = 20 - 8 =12 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \left[ \frac { \begin{matrix} 1 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 4 \end{matrix} } \right] \)
=\(\frac { 1 }{ 12 } \left[ \begin{matrix} \frac { 5 }{ 3 } & -1 \\ \frac { -2 }{ 3 } & +1 \end{matrix} \right] \)
= \(\frac { 1 }{ 12 } \left[ \frac { \begin{matrix} 2 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 3 \end{matrix} } \right] =\left[ \begin{matrix} \frac { 2 }{ 3 } \times \frac { 1 }{ 12 } \\ \frac { 1 }{ 3 } \times \frac { 1 }{ 12 } \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 18 } \\ \frac { 1 }{ 36 } \end{matrix} \right] \)
∴ \(\frac { 1 }{ 18 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 18 } \Rightarrow \)x = 18
t = \(\frac { 1 }{ 36 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 36 } \Rightarrow \)y = 36
one man can do 18 days
one woman can do 36 days.

Let the man's starting salary be Rs. x and his annual increment be Rs. y.
By the given data x + 3y = 19800 and x + 9y = 23,400.
The matrix form of the given system of equations is
\(\left[ \begin{matrix} 1 & 3 \\ 1 & 9 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 19800 \\ 23400 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 1 & 3 \\ 1 & 9 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 19800 \\ 23400 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 1 & 3 \\ 1 & 9 \end{matrix} \right| \)= 9-3 = 6 ≠ 0
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 6 } \left[ \begin{matrix} 9 & -3 \\ -1 & 1 \end{matrix} \right] \)
∴ X = A^-1B
= \(\frac { 1 }{ 6 } \left[ \begin{matrix} 9 & -3 \\ -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 19800 \\ 23400 \end{matrix} \right] \)
= \(\frac { 1 }{ 6 } \left[ \begin{matrix} 178200 & -70200 \\ -19800 & +23400 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} 108000 \\ 3600 \end{matrix} \right] \)
= \(\\ \\ \left[ \begin{matrix} 18000 \\ 600 \end{matrix} \right] \)
∴ x = 18000, y = 600
Hence the man's starting salary is Rs. 18000 and his annual increment is Rs. 600.

We find AB = \(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] =\left[ \begin{matrix} -4+4+8 & 4-8+4 & -4-8+12 \\ -7+1+6 & 7-2+3 & -7-2+9 \\ 5-3-2 & -5+6-1 & 5+6-3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] \) = 8I₃
and BA = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] =\left[ \begin{matrix} -4+7+5 & 4-1-3 & 4-3-1 \\ -4+14-10 & 4-2+6 & 4-6+2 \\ -8-7+15 & 8+1-9 & 8+3-3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] \) = 8I₃
So we get AB = BA = 8I₃. That is, (\(\frac { 1 }{ 8 } A\))B = B(\(\frac { 1 }{ 8 } A\)) = I₃. Hence, B^-1 = \(\frac { 1 }{ 8 } A\).
Writing the given system of equations in matrix form, we get
\(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \).
That is B \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \).
So, \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ={ B }^{ -1 }\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
= \(\left( \frac { 1 }{ 8 } A \right) \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] =\frac { 1 }{ 8 } \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] =\frac { 1 }{ 8 } \left[ \begin{matrix} -16+36+4 \\ -28+9+3 \\ 20-27-1 \end{matrix} \right] =\frac { 1 }{ 8 } \left[ \begin{matrix} 24 \\ -16 \\ -8 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ -1 \end{matrix} \right] \)
Hence, the solution is (x = 3, y = -2, z = -1).

\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
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=\(\left[ \begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)

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]

Given F (\(\alpha\)) = \(\left[ \begin{matrix} \cos { \alpha } & 0 & \sin { \alpha } \\ 0 & 1 & 0 \\ -\sin { \alpha } & 0 & \cos { \alpha } \end{matrix} \right] \)
Expanding along R₁ we get,
|F(\(\alpha\))| = cos \(\alpha\) \(\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| -0+sin\alpha \left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \)
= cos \(\alpha\) (cos - 0) + sin \(\alpha\) (0 + sin \(\alpha\))
= cos² + sin² \(\alpha\) = 1 ≠ 0
Since F (\(\alpha\)) is a non-singular matrix, [F(\(\alpha\))]^-1 exists
Now, adj (F(\(\alpha\))) = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} 0 & 0 \\ sin\alpha & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & sin\alpha \\ 0 & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & 0 \\ sin\alpha & 0 \end{matrix} \right| \\ +\left| \begin{matrix} 0 & sin\alpha \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & sin\alpha \\ 0 & 0 \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & 0 \\ 0 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(cos\alpha -0) & -(0) & +(0+sin\alpha ) \\ -(0) & +(cos^{ 2 }\alpha +sin^{ 2 }\alpha & -(0) \\ +(0-sin\alpha ) & -(0) & +(cos-0) \end{matrix} \right] ^{ T }\)
\(\left[ \begin{matrix} cos\alpha & 0 & +sin\alpha \\ 0 & 1 & 0 \\ -sin\alpha & 0 & cos\alpha \end{matrix} \right] =\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
∴ F(\(\alpha\))^-1 = \(\frac { 1 }{ |F(\alpha )| } \) adj (F(\(\alpha\)))
[F(\(\alpha\))]^-1 = \(\frac { 1 }{ 1 } \left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
= \(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(1)
Now, F(-\(\alpha\))=\(\left[ \begin{matrix} cos(-\alpha ) & 0 & sin(-\alpha ) \\ 0 & 1 & 0 \\ -s9n(-\alpha ) & 0 & cos(-\alpha ) \end{matrix} \right] \)
=\(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(2)
[∵ cos \(\alpha\) is an even function, cos (-\(\alpha\)) = cos \(\alpha\) and sin \(\alpha\) is an odd function, sin (-\(\alpha\)) = -sin\(\alpha\)]
From (1) and (2)
[F(\(\alpha\))]^-1 = F (-\(\alpha\))