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.

Let A = \(\left| \begin{matrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x \end{matrix} \right| \)
Expanding along R₁ we get
|A| = x\(\left| \begin{matrix} -x & 1 \\ 1 & x \end{matrix} \right| -sin\theta \left| \begin{matrix} -sin\theta & 1 \\ cos\theta & x \end{matrix} \right| +cos​​\theta \begin{vmatrix} -sin\theta & -x \\ cos\theta & 1 \end{vmatrix}\)
= x(-x² - 1) - sin \(\theta\) (-x sin \(\theta\) - cos \(\theta\)) + cos \(\theta\) (-sin \(\theta\) + x cos \(\theta\))
\(=-x^{ 3 }-x+xsin^{ 2 }\theta +sin\theta cos\theta +xcos^{ 2 }\theta \)
= -x³ - x + x(sin²\(\theta\) + cos²\(\theta\))
= -x³ - x + x(1)    [\(\because\) sin²\(\theta\) + cos²\(\theta\) ] = -1
= -x³ which is independent of \(\theta\)