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

Let A =\(\left[ \begin{matrix} 4 \\ -2 \\ 1 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 0 \\ -1 \\ 8 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
A\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ -2 \\ 4 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 8 \\ -1 \\ 0 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 4 \\ 11 \\ -12 \end{matrix}\begin{matrix} 8 \\ 15 \\ -32 \end{matrix}\begin{matrix} 7 \\ 19 \\ -25 \end{matrix} \right] \)
A is in row - echelon form and it has 3 non-zero rows.
∴ \(\rho\) (A) = 3

Let A be the matrix. Performing elementary row operations, we get
A = \(\left[ \begin{matrix} 2 \\ \begin{matrix} -3 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -1 \\ 7 \end{matrix} \end{matrix} \right] \)\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} -6 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 8 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -4 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -2 \\ 7 \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -13 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -2 \end{matrix} \end{matrix} \right] \).
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -45 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -30 \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -15 \right) }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ 2 \end{matrix} \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has three non-zero rows. So, ρ(A) = 3.

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) = -576-9144
= -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
Post multiplying 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 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \).

We get AB = \(\left[ \begin{matrix} 0 & -3 \\ 1 & 4 \end{matrix} \right] \left[ \begin{matrix} -2 & -3 \\ 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0+0 & 0+3 \\ -2+0 & -3-4 \end{matrix} \right] =\left[ \begin{matrix} 0 & 3 \\ -2 & -7 \end{matrix} \right] \)
(AB)^-1 = \(\frac { 1 }{ \left( 0+6 \right) } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \)...(1)
A^-1 = \(\frac { 1 }{ \left( 0+3 \right) } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] \)
B^-1 = \(\frac { 1 }{ \left( 2-0 \right) } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] =\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \)
B^-1A^-1 = \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \) ...(2).
As the matrices in (1) and (2) are same, (AB)⁻¹ = B⁻¹A⁻¹ is verified.

The equations can be written in matrix form as
\(\left( \begin{matrix} 2 & -1 \\ 5 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 3 \\ 4 \end{matrix} \right) \) ⇒ AX = B where
A =\(\left( \begin{matrix} 2 & -1 \\ 5 & 1 \end{matrix} \right) ,X=\left( \begin{matrix} x \\ y \end{matrix} \right) ,B=\left( \begin{matrix} 3 \\ 4 \end{matrix} \right) \)
∴ X = A^-1B
|A| = \(\left| \begin{matrix} 2 & -1 \\ 5 & 1 \end{matrix} \right| \) = 2+5 =7 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ 7 } \left[ \begin{matrix} 1 & 1 \\ -5 & 2 \end{matrix} \right] \)
∴ X = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 1 & 1 \\ -5 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 4 \end{matrix} \right] =\frac { 1 }{ 7 } \left[ \begin{matrix} 3+4 \\ -15+8 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 7 \\ -7 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ -1 \end{matrix} \right] \)
∴ Solution set is {1, -1}

Let A =\(\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{matrix} \right] \)
Now |A| = \(\left| \begin{matrix} 6 & -5 \\ -2 & 2 \end{matrix} \right| +1\left| \begin{matrix} -15 & -5 \\ 5 & 2 \end{matrix} \right| +1\left| \begin{matrix} -15 & 6 \\ 5 & -2 \end{matrix} \right| \)
= 3 (12 - 10) + 1 (-30 + 25) + 1 (30 - 30)
= 3(2) + 1 (-5) + 0 = 6 - 5 = 1 ≠ 0
∴ Rank of A is 3.

\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 2
∴ \(\rho \)(A) ≤ min (3, 2) = 2
The highest order of minor of A is 2
It is \(\left| \begin{matrix} -1 & 3 \\ 4 & -7 \end{matrix} \right| \)= 7-12 = 5 ≠ 0
∴ \(\rho \)(A) = 2

\(\left[ \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right] \)
Let A = \(\left( \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right) \)
adj A = \(\left( \begin{matrix} 2 & -4 \\ -6 & -3 \end{matrix} \right) \)
[Interchange the elements in the leading diagonal and change the sign of the elements in off diagonal]

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.

The augmented matrix [A|B] is \(\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 1 & \lambda \\ 1 & -7 & 8 \end{matrix}|\begin{matrix} 12 \\ \mu \\ 17 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 3 & 1 & \lambda \\ 2 & -3 & 5 \end{matrix}|\begin{matrix} 17 \\ \mu \\ 12 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2R_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 22 & \lambda -51 \\ 0 & 11 & -11 \end{matrix}|\begin{matrix} 17 \\ \mu -51 \\ -22 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 3 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 11 }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 0 & \lambda -2 \\ 0 & 1 & -1 \end{matrix}|\begin{matrix} 17 \\ \mu -7 \\ -2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & \lambda -2 \end{matrix}|\begin{matrix} 17 \\ -2 \\ \mu -7 \end{matrix} \right] \)
Case (i) : when λ ≠ 2,
\(\rho\) ([A|B]) = 3 and \(\rho\)(A) = 3
∴ \(\rho\)([AIB])= \(\rho\)(A) = 3 = the number of unknowns
∴ The system has unique solution
Case (ii) : when λ = 2, μ =7
\(\left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 17 \\ -2 \\ 0 \end{matrix} \right] \)
Here \(\rho\)(A) = 2 and \(\rho\)([A|B]) = 2
∴ \(\rho\)(A) = \(\rho\)([A|B]) = 2 < number of unknowns
Thus the system is consistent with infinitely many solutions.
Case (iii) : When λ= 2 and μ ≠ 7
\(\rho\) (A) = 2 and \(\rho\) ([A|B]) = 3
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
Thus, the given system of equations is inconsistent.

Augmented matrix [A|B] is \(\left[ \begin{matrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} a \\ b \\ c \end{matrix} \right] \)
[A|B]\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ -2 & 1 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} b \\ a \\ c \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 3 & -3 \end{matrix}|\begin{matrix} b \\ a+2b \\ c-b \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} b \\ c+2b \\ a+b+c \end{matrix} \right] \)
Here \(\rho\) (A) = 2
The given system is consistent only when \(\rho\)([A|B]) = 2\(\rho\)([A|B]) = 2 only if a + b + c = 0 Hence proved.

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 (-\(\alpha\))