The given system of equations can be written as
AX = B
where A = \(\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \)
X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] \)
\(\therefore\) X = A^-1B, if A^-1 exists
|A| = 40 => A^-1 exists
adj A = \(\left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \)
=> A^-1 = \(\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \)
\(\therefore\) X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
= \(\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right] \)
\(\therefore\) x = 1, y = 2 and z = -1

Let the award money for the values of honesty, regularity and hard work be Rs. x Rs. y and Rs. z respectively. Then the. system of equations is
x + y + z = 6,000 
x + 3z = 11,000
x - 2y + z = 0
Matrix equation is
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 6,000 \\ 11,000 \\ 0 \end{matrix} \right] \)
i.e., AX = B
\(|A|=\quad \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{matrix} \right| \)
= 1(0+6)-1(1-3)+1(-2-0)
= 6 \(\ne\)0
\(\therefore\) A^-1 exists.
adj A = \({ \left[ \begin{matrix} 6 & 2 & -2 \\ -3 & 0 & 3 \\ 3 & -2 & -1 \end{matrix} \right] }^{ T }\)
\(=\left[ \begin{matrix} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1 \end{matrix} \right] \)
\(X={ A }^{ -1 }B\)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1 \end{matrix} \right] \left[ \begin{matrix} 6,000 \\ 11,000 \\ 0 \end{matrix} \right] \quad \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 500 \\ 2,000 \\ 3,500 \end{matrix} \right] \)
x = 500, y = 2,000, z = 3,500 
Hence, award money given for the value of Honesty = Rs. 500, award money given for the value of Regularity = Rs. 2,000 and, award money given. for the value of Hard-work = Rs. 3,500.
Value: The school must include the value of Obedience for the awards.

Let x, y and z be the values of honesty, hard work and regularity respectively. Then the system of equations is
3x + 2y + z = 1.28 
4x + y + 3z= 1.54 
x + y + z = 0.57
Matrix equation is
\(\left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1.28 \\ 1.54 \\ 0.57 \end{matrix} \right] \)
i.e., AX=B
\(|A|=\left| \begin{matrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{matrix} \right| \)
= 3(1-3)-2(4-3)+1(4-1)
= -5 \(\ne\)0
A^-1 exists.
adj A=\({ \left[ \begin{matrix} -2 & -1 & 3 \\ -1 & 2 & -1 \\ 5 & -5 & -5 \end{matrix} \right] }^{ T }\)
\(=\left[ \begin{matrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{matrix} \right] \)
\(X={ A }^{ -1 }B\)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =-\frac { 1 }{ 5 } \left[ \begin{matrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{matrix} \right] \left[ \begin{matrix} 1.28 \\ 1.54 \\ 0.57 \end{matrix} \right] \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0.25 \\ 0.21 \\ 0.11 \end{matrix} \right] \)
\(\therefore\) x = 25,000, y = 21,000, z = 11,000
i.e., Rs. 25,000 for honesty, Rs. 21,000 for hard work and Rs. 11,000 for regularity.
Value: One more value like sincerity kindness etc.

The given system of equations in matrix form can be written as
AX = B,    ....(i)
where \(A=\left[ \begin{matrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{matrix} \right] ,\)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] and\quad B=\left[ \begin{matrix} 4 \\ -3 \\ 8 \end{matrix} \right] \)
Now,  \(|A|=\left| \begin{matrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{matrix} \right| \)
= 3(3-6)-4(-6-3)+7(4+1)
= -9 + 36 + 35 = 62 \(\ne\) 0
\(\therefore\) The given system of equations is consistent and have unique solution.
Let C_ij be the cofactor of the element a_ij in the determinant A, then
C₁₁ = -3, C₁₂ = 9, C₁₃ = 5
C₂₁ = 26, C₂₂ = -16, C₂₃ = -2
C₃₁ = 19, C₃₂ = 5, C₃₃ = -11
\(\therefore \ { A }^{ -1 }=\frac { 1 }{ \left| A \right| } (adj\quad A)=\frac { 1 }{ 62 } { [{ C }_{ ij }] }^{ T }\)
\(=\frac { 1 }{ 62 } { \left[ \begin{matrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 1- & 5 & -11 \end{matrix} \right] }^{ T }\)
\(=\frac { 1 }{ 62 } { \left[ \begin{matrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{matrix} \right] }\)
From (i), \(X={ A }^{ -1 }B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\frac { 1 }{ 62 } { \left[ \begin{matrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{matrix} \right] }\left[ \begin{matrix} 4 \\ -3 \\ 8 \end{matrix} \right] \)
\(=\frac { 1 }{ 62 } \left[ \begin{matrix} -12-78+152 \\ 36+48+40 \\ 20+6-88 \end{matrix} \right] \)
\(=\frac { 1 }{ 62 } \left[ \begin{matrix} 62 \\ 124 \\ -62 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right] \)
x = 1, y = 2 and z = -1

\(A=\left[ \begin{matrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{matrix} \right] \)
a₁₁ = (1 + 3) = 4
a₁₂ = -(-1 + 2) = -1
a₁₃ = (3 + 2) = 5
a₂₁ = -(-2 - 15) = 17
a₂₂ = (-1 - 10) = -11
a₂₃ = -(3 - 4) = 1
a₃₁ = (-2 + 5) = 3
a₃₂ = -(-1 - 5) = 6
a₃₃ = (-1 - 2) = -3
|A| = 1 x 4 + 2 x(-1) + 5 x (5)
= 4 - 2 + 25 = 27
\(\Rightarrow { A }^{ -1 }=\frac { 1 }{ 27 } \left[ \begin{matrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 4/27 & 17/27 & 1/9 \\ -1/27 & -11/27 & 2/9 \\ 5/27 & 1/27 & -1/9 \end{matrix} \right] \)
Given set of equations can be written as
x + 2y + 5z = 10
x - y - z = -2
and 2x + 3y - z = -11
\(\left[ \begin{matrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 10 \\ -2 \\ -11 \end{matrix} \right] \)
\(\Rightarrow AX=B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4/27 & 17/27 & 1/9 \\ -1/27 & -11/27 & 2/9 \\ 5/27 & 1/27 & -1/9 \end{matrix} \right] \left[ \begin{matrix} 10 \\ -2 \\ -11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} \frac { 40 }{ 27 } & -\frac { 34 }{ 27 } & -\frac { 33 }{ 27 } \\ -\frac { 10 }{ 27 } & \frac { 22 }{ 27 } & -\frac { 66 }{ 27 } \\ \frac { 50 }{ 27 } & -\frac { 2 }{ 27 } & \frac { 33 }{ 27 } \end{matrix} \right] =\left[ \begin{matrix} -\frac { 27 }{ 27 } \\ -\frac { 54 }{ 27 } \\ \frac { 81 }{ 27 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} -1 \\ -2 \\ 3 \end{matrix} \right] \)
\(\Rightarrow\) x = -1, y = -2 and z = 3

\(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -4+4+8 & 4-8+4 & -4-8+12 \\ -7+1+6 & 7-2+3 & -7-2+9 \\ 5-3-2 & -5+6-1 & 5+6-3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] =8I\)
where I is the identity matrix
Let \(A=\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \)
and \(B=\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(\Rightarrow AB=8I\)
Post Multiplyingbith sides by B ^-1, we get 
\(AB{ B }^{ -1 }=8I{ B }^{ -1 }\)
\(\Rightarrow A=8{ B }^{ -1 }\)
\(\Rightarrow { B }^{ -1 }=\frac { A }{ 8 } \)
Given Equations are:
x - y + z = 4
x - 2y - 2z = 9
and 2x + y + 3z = 1
\(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow BX=C\)
where \(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \ and\ C=\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow X={ B }^{ -1 }C\)
Using \({ B }^{ -1 }=\frac { A }{ 8 } \)
\(\Rightarrow X=\frac { A }{ 8 } C\)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -16+36+4 \\ -28+9+3 \\ 20-27-1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} 24 \\ -16 \\ -8 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ -1 \end{matrix} \right] \)
(x, y, z)  = (3, -2, -1)

\(A=\left[ \begin{matrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{matrix} \right] \)
Co factors are:
C₁₁=-6, C₂₁=17, C₃₁=13
C₁₂=14, C₂₂=5, C₃₂=-8
C₁₃=-15, C₂₃=9, C₃₃=-1
|A|=-6+28+45=67
\(\therefore \ { A }^{ -1 }=\frac { 1 }{ 67 } \left[ \begin{matrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{matrix} \right] \)
Now the equations are:
x+2y-3z=-4
2x+3y+2z=2
and 3x-3y-4z=11
They may be written as
\(\left[ \begin{matrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} -4 \\ 2 \\ 11 \end{matrix} \right] \)
where \(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \quad and\quad B=\left[ \begin{matrix} -4 \\ 2 \\ 11 \end{matrix} \right] \)
\(\Rightarrow AX=B\)
\(\Rightarrow X={ A }^{ -1 }B\)
\(\Rightarrow X=\frac { 1 }{ 67 } ​​\left[ \begin{matrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{matrix} \right] \left[ \begin{matrix} -4 \\ 2 \\ 11 \end{matrix} \right] \)
\(=\frac { 1 }{ 67 } \left[ \begin{matrix} 201 \\ -134 \\ 67 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ 1 \end{matrix} \right] \)
\(\Rightarrow x=3,\ y=-2\ and\ z=1\)

\(={ (x+y+z) }^{ 3 }\left| \begin{matrix} { (y+z) }^{ 2 } & x-y-z & x-y-z \\ { y }^{ 2 } & x+z-y & 0 \\ { z }^{ 2 } & 0 & x+y-z \end{matrix} \right| \)
LHS = \(\left| \begin{matrix} { (y+z) }^{ 2 } & xy & zx \\ xy & { (x+z) }^{ 2 } & yz \\ xz & yz & { (x+y) }^{ 2 } \end{matrix} \right| \)
Apply \(({ R }_{ 1 }\rightarrow x{ R }_{ 1 },\quad { R }_{ 2 }y\rightarrow { R }_{ 2 },{ R }_{ 3 }z\rightarrow { R }_{ 3 })\)
\(=\frac { 1 }{ xyz } \left| \begin{matrix} { x(y+z) }^{ 2 } & { x }^{ 2 }y & z{ x }^{ 2 } \\ x{ y }^{ 2 } & { y(x+z) }^{ 2 } & { y }^{ 2 }z \\ x{ z }^{ 2 } & y{ z }^{ 2 } & { z(x+y) }^{ 2 } \end{matrix} \right| \)
\(=\left| \begin{matrix} { (y+z) }^{ 2 } & { x }^{ 2 } & { x }^{ 2 } \\ { y }^{ 2 } & { (x+z) }^{ 2 } & { y }^{ 2 } \\ x{ z }^{ 2 } & { z }^{ 2 } & { (x+y) }^{ 2 } \end{matrix} \right| \)
Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 },{ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\)
\(=\left| \begin{matrix} { (y+z) }^{ 2 } & { x }^{ 2 }-{ (y+z) }^{ 2 } & { x }^{ 2 }-{ (y+z) }^{ 2 } \\ { y }^{ 2 } & { (x+z) }^{ 2 }-{ y }^{ 2 } & 0 \\ { z }^{ 2 } & 0 & { (x+y) }^{ 2 }-{ z }^{ 2 } \end{matrix} \right| \)
Taking (x + y + z) common from C₂ and C₃ . 
\(={ (x+y+z) }^{ 3 }\left| \begin{matrix} { (y+z) }^{ 2 } & x-y-z & x-y-z \\ { y }^{ 2 } & x+z-y & 0 \\ { z }^{ 2 } & 0 & x+y-z \end{matrix} \right| \)
Apply \({ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }-{ R }_{ 3 }\)
\(={ (x+y+z) }^{ 2 }\left| \begin{matrix} { (y+z) }^{ 2 } & x-y-z & x-y-z \\ { y }^{ 2 } & x+z-y & 0 \\ { z }^{ 2 } & 0 & x+y-z \end{matrix} \right| \)
Apply, \({ C }_{ 2 }\rightarrow y{ C }_{ 2 },{ C }_{ 3 }\rightarrow z{ C }_{ 3 }\)
\(=\frac { { (x+y+z) }^{ 2 } }{ yz } \left| \begin{matrix} 2yz & 0 & 0 \\ { y }^{ 2 } & x+z-y & { y }^{ 2 } \\ { z }^{ 2 } & { z }^{ 2 } & zx+yz \end{matrix} \right| \)
\(=\frac { { (x+y+z) }^{ 2 } }{ yz } \left| (2yz).({ x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 }-{ y }^{ 2 }{ z }^{ 2 } \right| \)
\(=2xyz{ (x+y+z) }^{ 3 }=RHS\)

Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 }\quad and\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA\cos ^{ 2 }{ B } +cosB & cosC-cosA\cos ^{ 2 }{ C } -cosC \\ \cos ^{ 2 }{ A } +cosA & \cos ^{ 2 }{ A } -cosA & -\cos ^{ 2 }{ A } -cosA \end{matrix} \right| =0\)
\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA & cosC-cosA \\ \cos ^{ 2 }{ A } -1 & \cos ^{ 2 }{ B } -\cos ^{ 2 }{ A } & \cos ^{ 2 }{ C } -cosA \end{matrix} \right| =0\)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }\)
(cos B - cos A ) x ( cos C -cos B)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & 1 & 0 \\ \cos ^{ 2 }{ A } -1 & cosB+cosA & cosC-cosA \end{matrix} \right| =0\)
\(\therefore \quad (cosB-cosA)\times (cosC-cosA)\times (cosC-cosB)\)
[1-0]=0  (Expanding along C₃)
\(\therefore \ -cosB=cosA\quad or\quad cosC=cosAorcosC=cosB\)
\(\Rightarrow cosB=cosAorcosC=cosAorcosC=cosB\)
\(\Rightarrow \angle B=\angle Aor\angle C=\angle Aor\angle C=\angle B\)
\(\Rightarrow \triangle ABC\)  is an isosceles triangle.