Given rank of \(\left[ \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right] \) is 2
⇒ The value of the third order determinant is zero
⇒ \(\left[ \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right] \)=0
⇒ λ\(\left| \begin{matrix} \lambda & -1 \\ 0 & \lambda \end{matrix} \right| +1\left| \begin{matrix} 0 & -1 \\ -1 & \lambda \end{matrix} \right| \)+0 = 0
⇒ λ(λ² - 0) + 1(0 - 1) = 0
⇒ λ³ - 1 = 0 ⇒ λ³ = 1 ⇒ λ = 1
∴ λ = 1

Augmented matrix [A|B] =\(\left[ \begin{matrix} 1 & 1 & 3 \\ 2 & 2 & 6 \\ 2 & 1 & 1 \end{matrix}|\begin{matrix} 4 \\ 7 \\ 10 \end{matrix} \right] \)
[A|B] \(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -5 \end{matrix}|\begin{matrix} 4 \\ -1 \\ 2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -5 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 4 \\ 2 \\ -1 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 [only 2 two-non zero rows]
And \(\rho\) ([A|B]) = 3 [There are 3 non-zero rows]
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
∴ The system is inconsistent.

Δ = \(\left| \begin{matrix} 3 & a \\ 2 & a \end{matrix} \right| \) = 3a - 2a = a
Δ₁ =\(\left| \begin{matrix} 4 & a \\ 2 & a \end{matrix} \right| \) = 4a - 2a = 2a
Δ₂ = \(\left| \begin{matrix} 3 & 4 \\ 2 & 2 \end{matrix} \right| \)= 6 - 8 = -2

\(\therefore x=\frac{\Delta_{1}}{\Delta}=\frac{2 \not a}{\not a}=2=y y=\frac{\Delta_{2}}{\Delta}=\frac{-2}{a}\)
∴ Solution set is {2,\(\frac { -2 }{ a } \)}

\(\Delta = \left| \begin{matrix} t & 3 & -1 \\ 1 & 2 & 1 \\ -t & 1 & 2 \end{matrix} \right| =t\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ -t & 2 \end{matrix} \right| -\left| \begin{matrix} 1 & 2 \\ -t & 1 \end{matrix} \right| \)
= t(4 - 1) -3 (2 + t) -1(1 + 2t)
= 3t - 6i - 3t - 1 - 2t = - 7 - 2t
The system will fail to have unique solution if
Δ = 0 ⇒ -7-2t = 0 ⇒ -2t = 7 ⇒ t = \(\frac { -7 }{ 2 } \)
∴ t = \(\frac { -7 }{ 2 } \).

Δ = \(\left| \begin{matrix} 2 & 3 \\ 1 & 6 \end{matrix} \right| \) = 12 - 3 = 9 ≠ 0
Δ₁ = \(\left| \begin{matrix} 10 & 3 \\ 4 & 6 \end{matrix} \right| \) = 60 - 12 = 48
Δ₂ = \(\left| \begin{matrix} 2 & 10 \\ 1 & 4 \end{matrix} \right| \) = 8 - 10 = -2
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 48 }{ 9 } =\frac { 16 }{ 3 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -2 }{ 9 } \)
∴ Solution set is \(\left\{ \frac { 16 }{ 3 } ,\frac { -2 }{ 9 } \right\} \).

\(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \)
Let A = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \)
A = \(\left[ \begin{matrix} 1 \\ 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \right] \overset { { { R } }_{ 2 }\rightarrow { { R } }_{ 2 }3{ { R } }_{ 1 }\\ { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-{ { R } }_{ 1 } }{ \underset { { { R } }_{ 4 }\rightarrow { { R } }_{ 4 }-{ { R } }_{ 1 } }{ \longrightarrow } } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} -4 \\ -3 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }\div 4 }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} -1 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 1 \\ -1 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 4 }\rightarrow { { R } }_{ 4 }-3{ { R } }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 1 \\ -1 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow 7{ { R } }_{ 3 }-{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 2 \\ -1 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 4 }\rightarrow 2{ { R } }_{ 4 }-{ { R } }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix} \right] \)
The last equivalent matrix is in row echelon form It has three non-zero rows.
∴ \(\rho \)(A) = 3

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 =\(\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
adj A =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of the elements in the off diagonal]
|A| = 24 - 20 = 4
∴ A(adj A) =\(\\ \left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & 32-32 \\ -15+15 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) ....(1)
(adj A)(A) =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] =\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & -12+12 \\ 40-40 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \)...(2)
|A|I₂ = 4\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) .....(3)
From (1), (2) and (3), it is proved that
A (adj A) = (adj A) A = |A|I₂

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

Let the encryption matrix be A =\(\left[ \begin{matrix} -1 & -1 \\ 2 & 1 \end{matrix} \right] \)
|A| = -1 + 2 = 1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 1 } \left[ \begin{matrix} 1 & 1 \\ -2 & -1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 \\ -2 & -1 \end{matrix} \right] \)
Hence the decryption matrix is \(\left[ \begin{matrix} 1 & 1 \\ -2 & -1 \end{matrix} \right] \)

So, the sequence of decoded row matrices is [8 5], [12 16]
Now the 8^th English alphabet is H.
5^th English alphabet is E.
12^th English alphabet is L.
and the 16^th English alphabet is P.
Thus the receiver reads the message as "HELP".