Given A =\(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\)
\(\therefore\)A² = A.A =\(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\)\(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 9+2 & 6+2 \\ 3+1 & 2+1 \end{bmatrix}=\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}\)
Given A² + aA + bI = 0
⇒ \(\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}+a\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}+b\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)= 0
⇒ \(\begin{bmatrix} 11+3a+b & 8+2a+0 \\ 4+a+0 & 3+a+b \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Equating the like terms we get,
4 + a = 0 ⇒ a = -4
3 + a + b = 0 ⇒ 3 - 4 + b = 0
-1 + b = 0 ⇒ b = 1

Let A =\(\left[ \begin{matrix} 2 & -1 & 3 \\ 0 & 5 & 1 \\ 3 & 6 & 8 \end{matrix} \right] \)
A₁₁ = 40 - 6 = 34
A₂₁ = -(-8 - 18) = 26
A₃₁ = -1 - 15 = -16
A₂₁ = -(0 - 3) = 3
A₂₂ = 16 - 9 = 7
A₃₂ = -(2 - 0) = -2
A₁₃ = 0 - 15 = -15
A₂₃ = -(12 + 3) = -15
A₃₃ = 10 - 0 = 10
\(\therefore\) adj A=\({ \left[ \begin{matrix} 34 & 3 & -15 \\ 26 & 7 & -15 \\ -16 & -2 & 10 \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 34 & 26 & -16 \\ 3 & 7 & -2 \\ -15 & -15 & 10 \end{matrix} \right] }\)

Let \(\triangle=\begin{vmatrix} 6&-1&5\\3&0&4\\-2&7&-3 \end{vmatrix}\)
Now M₁₂ = \(\begin{vmatrix} 3&4\\-2&-3 \end{vmatrix}=-9-(-8)=-1\)
M₂₂ = \(\begin{vmatrix} 6&5\\-2&-3 \end{vmatrix}=-18-(-10)=-8\)
M₃₂ = \(\begin{vmatrix} 6&5\\3&4 \end{vmatrix}=24-15=9\)
Now expansion of |A| using co-factors of elements of second column we get,
|A| = a₁₂ A₁₂ + a₂₁ A₂₁ + a₃₁A₃₁
A₁₂ = (-1)^{1 + 2} M₁₂ (-1) (-1) = 1
A₂₂ = (-1)^{2 + 2} M₂₂ = 1(-8) = -8
A₃₂ = (-1)^{3 + 2} M₃₂ = -(9) = -9
\(\therefore\) |A| = -1(1) + 0(-8) + 7(-9) = -1 - 63 = -64.

With the usual notation we have
a₁₁ = 16, a₁₂ = 12, x₁ = 40
a₂₁ = 12, a₂₂ = 8, x₂ = 24
Now, \({b}_{11} = {{{a}_{11}\over{x_1}}}={{16}\over{40}}={{2}\over{5}},{b}_{12}={{{a}_{12}}\over{{x}_{2}}}={{12}\over{24}}={{1}\over{2}}\)
\({b}_{21}={{{a}_{21}}\over{x_1}}={{12}\over{40}}={{3}\over{10}},{b}_{22}={{a}_{22}\over{x_2}}={{8}\over{24}}={{1}\over{3}}\)
The technology matrix is
\(B=\begin{bmatrix} {{2}\over{5}}&{{1}\over{2}}\\ {{3}\over{10}}&{{1}\over{3}} \end{bmatrix}\)
\(I-B=\begin{pmatrix} 1&0\\0&1 \end{pmatrix}-\begin{pmatrix} {{2}\over{5}}&{{1}\over{2}} \\ {{3}\over{10}}&{{1}\over{3}} \end{pmatrix}=\begin{pmatrix}{{3}\over{45}}&-{{1}\over{2}}\\-{{3}\over{10}}&{{2}\over{3}} \end{pmatrix}\)
The main diagonal elements in (I - B) namely \({{3}\over{5}}\) and \({{2}\over{3}}\) are positive.
Also, \(|I-B|=\begin{vmatrix} {{3}\over{5}} &{{-1}\over{2}}\\{{-3}\over{10}}&{{2}\over{3}} \end{vmatrix}=\left({{3}\over{5}} \right)\left( \frac{2}{3} \right)-\left(\frac{1}{2} \right)\left(\frac{3}{10} \right)\)
\(={{2}\over{5}}-{{3}\over{20}}={{8-3}\over{20}}={{5}\over{20}}={{1}\over{4}}\)
\(\therefore\)  |I - B| is positive.
\(\therefore\)  The two Hawkins' Simon conditions are satisfied. Hence the system is viable.

Given equations are 2x + 5y = 1; 3x + 2y = 7
This system of equations can be written in matrix form as \(\begin{pmatrix}2 & 5 \\3 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 1\\7 \end{pmatrix}\Rightarrow Ax=B\)
where A = \(\begin{pmatrix} 2 & 5 \\3 & 2 \end{pmatrix},X=\begin{pmatrix} x\\y \end{pmatrix}\) and B = \(\begin{pmatrix} 1 \\ 7 \end{pmatrix}\)
\(\therefore\)  X = A^-1 B.
\(|A|=\begin{vmatrix} 2 &5 \\3 & 2 \end{vmatrix}=4-15=-11\)
A₁₁ = 2, A₁₂ = -3, A₂₁ = -5, A₂₂ = 2
\(\therefore\)   adj A = \(\begin{bmatrix} 2 & -3 \\ -5 & 2 \end{bmatrix}^{T}=\begin{bmatrix} 2 & -5 \\-3 & 2 \end{bmatrix}\)
\(\therefore\)  \({A}^{-1}={{1}\over{|A|}}\) . adj A = \({{-1}\over{11}}\begin{bmatrix} 2 & -5\\ -3& 2 \end{bmatrix}\)
X = A^{- 1} B \(={{-1}\over{11}}\begin{bmatrix} 2 & -5\\ -3&2 \end{bmatrix}\begin{bmatrix} 1\\7 \end{bmatrix}\)
\(=\frac { 1 }{ 11 } \left[ \begin{matrix} +2-35 \\ -3+14 \end{matrix} \right] ={{-1}\over{11}}\begin{bmatrix} -33\\11 \end{bmatrix}=\begin{bmatrix} 3\\-1 \end{bmatrix}\)
\(\therefore\)  x = 3.and y = -1.

Let A = \(\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}\)
\(\therefore |A|=\begin{bmatrix} -1 &5 \\-3 & 2 \end{bmatrix}=-2+15=13\)
Now, A₁₁ = 2, A₁₂ = (-3) = 3, A₂₁ = -5, A₂₂ = -1
\(\therefore\) adj A = \({\begin{bmatrix}2 & 3 \\-5 &-1 \end{bmatrix}}^{T}=\begin{bmatrix}2 & -5 \\3 & -1 \end{bmatrix}\)
Now \({A}^{-1}={{1}\over{|A|}}\) adj A = \({{1}\over{13}}\begin{bmatrix} 2& -5\\3 & -1 \end{bmatrix}\)

Given A \(=\begin{bmatrix} 2&3\\-1&4 \end{bmatrix}\)
\(|A|=\begin{bmatrix} 2&3\\-1&4 \end{bmatrix}=8+3=11\)
Now, A₁₁ = 4, A₁₂ = - (-1) = 1, A₂₁  = 3, A₂₂ = 2
\(\therefore\ adj\ A={\begin{bmatrix} 4&1\\-3&2 \end{bmatrix}}^{T}=\begin{bmatrix} 4&-3\\1&2 \end{bmatrix} \)
\(\therefore\) A (adj A)\(=\begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix} { }\begin{bmatrix} 4 & -3 \\ 1 & 2 \end{bmatrix}\)
\(=\begin{bmatrix} 8+3&-6+6\\-1+4&3+8 \end{bmatrix}=\begin{bmatrix} 11&0\\0&11 \end{bmatrix}=11\begin{bmatrix} 1&0\\0&1 \end{bmatrix}=|A|I_2\)   ...(1)
Also ( adj A ) A = \(\begin{bmatrix} 4&-3\\1&2 \end{bmatrix}\begin{bmatrix} 2&3\\-1&4 \end{bmatrix}\)
\(=\begin{bmatrix} 8+3&12-12\\2-2&3+8 \end{bmatrix}=\begin{bmatrix} 11&0\\0&11 \end{bmatrix}=11\begin{bmatrix}1&0\\0&1 \end{bmatrix}=|A|I_2\)   ....(2)
From (1) and (2), A( adj A) = (adj A) A = |A|.I₂

Let A = \(\begin{vmatrix}5 & 3 \\-6 & 2\end{vmatrix}\)
Minor of 5 = M₁₁ = 2 and A₁₁ = (-1)¹⁺¹M₁₁ = 2
Minor of 3 = M₁₂ = -6 and A12 = (-1)¹⁺²M₁₂ = 6
Minor of -6 = M₂₁ = 3 and A₂₁ = (-1)²⁺¹ M₂₁ = -3
Minor of 2 = M₂₂ = 5 and A₂₂ = (-1)²⁺² M₂₂ = 5

Given that A^T = A^-1
\(\Rightarrow\) AA^T = AA^-1 
\(\Rightarrow\) AA^{T = I}
Now, AA^T = \(\left[ \begin{matrix} cos\ \alpha & sin\ \alpha \\ -sin\ \alpha & \ cos\ \alpha \ \end{matrix} \right] \)\(\left[ \begin{matrix} cos\ \alpha & -sin\ \alpha \\ sin\ \alpha & \ cos\ \alpha \ \end{matrix} \right] \)
\(=\left[ \begin{matrix} { cos }^{ 2 }\alpha +{ sin }^{ 2 }\alpha & -sin\alpha cos\alpha +sin\alpha cos\alpha \\ -sin\alpha cos\alpha +sin\alpha cos\alpha & +{ sin }^{ 2 }\alpha +{ cos }^{ 2 }\alpha \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)[\(\because\) sin²\(\alpha\) + cos²\(\alpha\) = 1]
Thus, AA^T = I is true for all \(\alpha\).
Hence \(\alpha\)  can take any real value.

The system of equations can be written in the form AX = B where,
\(A=\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 7 \\ 11 \\ 1 \end{matrix} \right] \)
Now, |A| = \(\left| \begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right| =1\left| \begin{matrix} 0 & 3 \\ -3 & 0 \end{matrix} \right| -2\left| \begin{matrix} 1 & 3 \\ 2 & 0 \end{matrix} \right| +1 \left| \begin{matrix} 1 & 0 \\ 2 & -3 \end{matrix} \right| \)
= 1(0 + 9) - 2(0 - 6) + 1(-3 - 0) = 9 + 12 - 3 = 18 \(\neq \) 0 
\(\Rightarrow\) A^-1 exists.
A₁₁ = 0 + 9 = 9, A₁₂ = -(0 - 6) = 6, A₁₃ = -3 - 0 = -3
A₂₁ = -(0 + 3) = -3, A₂₂ = 0 - 2 = -2, A₂₃ = -(-3 - 4) = 7
A₃₁ = 6 - 0 = 6, A₃₂ = -(3 - 1) = -2, A₃₃ = 0 - 2 = -2
\(\therefore adj\quad A={ \left[ \begin{matrix} 9 & 6 & -3 \\ -3 & -2 & 7 \\ 6 & -2 & -2 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ 18 } \left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \)
\(\therefore \ X={ A }^{ -1 }B=\frac { 1 }{ 18 } \left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \left[ \begin{matrix} 7 \\ 11 \\ 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 18 } \left[ \begin{matrix} 63 & -33 & +6 \\ 42 & -22 & -2 \\ -21 & +77 & -2 \end{matrix} \right] =\frac { 1 }{ 18 } \left[ \begin{matrix} 36 \\ 18 \\ 54 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ 3 \end{matrix} \right] \)
\(\therefore\) x = 2, y = 1, and z = 3.