\(|A|=\cos ^2 \theta+\sin ^2 \theta=1\)
\(|2 A|=2^2(1)=4\)

\(\text {Since }|A|=4-4=0\)

(Since \(|k A|=k^n|A|,\) n is the order of matrix A

\(\left|\begin{array}{lll} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{array}\right|=0 \Rightarrow-1\left[x^2-x\right]=0\)
\(\Rightarrow x(x-1)=0 \Rightarrow x=0,1\)

\(\left| \begin{matrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{matrix} \right| \) = 1 (Minor of 1) –2 (Minor of 2) + 4 (Minor of 4)
\(=1\left| \begin{matrix} 3 & 0 \\ 1 & 0 \end{matrix} \right| -2\left| \begin{matrix} -1 & 0 \\ 4 & 0 \end{matrix} \right| +4\left| \begin{matrix} -1 & 3 \\ 4 & 1 \end{matrix} \right| \)
= 0 – 0 – 52 = –52.

Let |A| = \(\begin{vmatrix} 6 &5 &12 \\ 2 & 4 &4 \\2 & 1 & 4 \end{vmatrix}\)
Taking 2 common from C₁ and 4 common from C₃, we get,
\(|A|=2\times4\begin{vmatrix} 3 & 5&3 \\ 1 & 4 & 1\\1 &1 &1\end{vmatrix}=8\times 0\ [\because C_1\equiv C_3]=0\)

B = \(\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}\) 
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}=\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
|I - B| =\(\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
= 0.08 - 0.18 = - 0.1 < 0
Since |I - B| is negative, Hawkins - Simon conditions are not satisfied.
Therefore the given system is not viable.

B \(=\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}\)
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}=\begin{bmatrix} 0.50 & -0.30 \\ -0.41 & 0.67 \end{bmatrix}\)
= (0.50) (0.67) - (0.30) (0.41)
\(|I-B|\) = 0.335 - 0.123 = 0.212 >  0
Since the main diagonal elements of I - B are positive and |I-B| is positive. Hawkins Simon conditions are satisfied. Therefore given 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 = \(\begin{vmatrix}5 & 3 \\-6 & 2\end{vmatrix}\)
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

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.

\(A^{-1}=\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)\)
\(A^2=\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)\)
\(=\left(\begin{array}{ll} 1-2 & -1-3 \\ 2+6 & -2+9 \end{array}\right)=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)\)
\(\mathrm{LHS}=A^2-4 A+5 l_2\)
\(=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)-4\left(\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right)+5\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)\)
\(=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)-\left(\begin{array}{cc} 4 & -4 \\ 8 & 12 \end{array}\right)+\left(\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right)\)
\(=\left(\begin{array}{cc} -1 & -4 \\ 8 & 7 \end{array}\right)+\left(\begin{array}{cc} 1 & 4 \\ -8 & -7 \end{array}\right)=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right)\)
\(=O=R H S\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A\)
\(|A|=3+2=5 \neq 0\)
\(A^{-1}=\frac{1}{5}\left(\begin{array}{cc} 3 & 1 \\ -2 & 1 \end{array}\right)\)

LHS = \(\begin{vmatrix}0 &ab^2 &ac^2 \\a^2b & 0 & bc^2\\a^2c&b^2c&0\end{vmatrix}\)
Taking a, b and c common from R₁, R₂, R₃
\(=\operatorname{abc}\left|\begin{array}{ccc} 0 & b^2 & c^2 \\ a^2 & 0 & c^2 \\ a^2 & b^2 & 0 \end{array}\right|\)
Taking a², b², c² common from C₁, C₂, C₃
= \(a^2b^2c^2\begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 0 \end{vmatrix}\)
= a³b³c³ [0 - 1 (0 - 1)+ 1(1 - 0)]
= a³b³c³ (1 + 1) = 2 a³b³c³ = RHS

Expanding the given determinant along R₁ we
\(\left|\begin{array}{ccc} 7 & 4 & 11 \\ -3 & 5 & x \\ -x & 3 & 1 \end{array}\right|=0\)
⇒ 7 (5 - 3x) - 4 (-3 + x²) + 11 (-9 + 5x) = 0
⇒ 35 - 21x + 12 - 4x² - 99 + 55x =  0
⇒ - 4x² + 34x - 52 = 0
\(\div\)  by - 2
\(2 x^2-17 x+26 =0\)
\(2 x^2-4 x-13 x+26 =0 \)
\(2 x(x-2)-13(x-2) =0 \)
\((2 x-13)(x-2) =0\)
\(x =2, \frac{13}{2}\)

Let the number of radios of type P be x and the radios of type Q be y.
Given 2x + 3y = 130 and 3x + 4y = 180
\(\Rightarrow \left( \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 130 \\ 180 \end{matrix} \right) \)
AX = B where A = \(\left( \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right) ,X=\left( \begin{matrix} x \\ y \end{matrix} \right) and\quad B=\left( \begin{matrix} 130 \\ 180 \end{matrix} \right) \)
|A|=\(\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| =8-9=-1\neq 0\Rightarrow { A }^{ -1 }\) existsadj A=\(\left( \begin{matrix} 4 & -3 \\ -3 & 2 \end{matrix} \right) \)    [\(\because\) A₁₁ = 4, A₁₂ = -3 A₂₁ = -3, A₂₂ = 2]
\(\therefore \ { A }^{ -1 }=\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ -1 } \left( \begin{matrix} 4 & -3 \\ -3 & 2 \end{matrix} \right) =\left( \begin{matrix} -4 & 3 \\ 3 & -2 \end{matrix} \right) \)
\(\therefore \ X={ A }^{ -1 }B=\left( \begin{matrix} -4 & 3 \\ 3 & -2 \end{matrix} \right) \left( \begin{matrix} 130 \\ 180 \end{matrix} \right) =\left( \begin{matrix} -520 & +540 \\ 390 & -360 \end{matrix} \right) =\left( \begin{matrix} 20 \\ 30 \end{matrix} \right) \)
\(\therefore\) Number of radios of Type P is 20.
Number of radios of Type Q is 30.

\(|A|=\left| \begin{matrix}1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right| =1+{ tan }^{ 2 }x={ sec }^{ 2 }x\neq 0\)
\(\Rightarrow\) A^-1 exists
Let C_ij be the cofactor of a_ij  in A
C₁₁ = (-1)¹⁺¹ M₁₁ = (-1)²(1) = 1
C₁₂ = (-1)¹⁺² (-tan x) = tan x
C₂₁ = (-1)²⁺²(1) = 1
\(\therefore \quad adj\quad A={ \left[ \begin{matrix} 1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ |A| } adjA=\frac { 1 }{ 1+{ tan }^{ 2 }x } \left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] \)
\(\therefore \quad { A }^{ T }{ A }^{ -1 }=\left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] \left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } +\frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] \)
\(=\left[ \begin{matrix} \frac { 1-tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -2tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { 2tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1-{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] =\left[ \begin{matrix} cos\quad 2x & -sin2x \\ sin\quad 2x & cos\quad 2x \end{matrix} \right] \)  (Using multiple angle formula)

\(x-y+z=2\)
\(2 x-y=0 \)
\(2 y-z=1\)
The given equations can be written in matrix form as
\(\begin{bmatrix} 1&-1&1\\2&-1&0\\0&2&-1 \end{bmatrix}\begin{bmatrix} x \\ y\\z \end{bmatrix}=\begin{bmatrix} 2\\0\\1 \end{bmatrix} \)
\(A X=B \Rightarrow X=A^{-1} B\)
Where \(A=\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & -1 \end{matrix} \right] ,\ X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,\ B=\left[ \begin{matrix} 2 \\ 0 \\ 1 \end{matrix} \right] \)
 \(|A|=\begin{vmatrix} 1&-1&1\\2&-1&0\\0&2&-1 \end{vmatrix}=-1\begin{vmatrix} -1&0\\2&-1\end{vmatrix}+1\begin{vmatrix} 2&0\\0&-1 \end{vmatrix}+1\begin{vmatrix} 2&-1\\0&2 \end{vmatrix}\)
\(|A|=1(1-0)+1(-2-0)+1(4-0)\)
\(=1-2+4=3\neq0\)
\(\therefore\) A^-1 exists.
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 1 & 2 & 4 \\ 1 & -1 & -2 \\ 1 & 2 & 1 \end{array}\right)\)
\(\therefore\ {A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{3}}\begin{bmatrix} 1&1&1\\2&-2&2\\4&-2&1 \end{bmatrix}\)
Now \(X={A}^{-1}B={{1}\over{3}}\begin{bmatrix} 1&1&1\\2&-1&2\\4&-2&1 \end{bmatrix}\begin{bmatrix} 2\\0\\1 \end{bmatrix}=\frac{1}{3}\left(\begin{array}{l} 2+0+1 \\ 4+0+2 \\ 8+0+1 \end{array}\right)=\frac{1}{3}\left(\begin{array}{l} 3 \\ 6 \\ 9 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right)\)
\(x=1, y=2, z=3\)

Let A \(=\begin{vmatrix} 1 & a&a^2&-bc \\1 &b&{b}^{2}&-ca\\1&c&c^2&-ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|+\left|\begin{array}{ccc} 1 & a & -b c \\ 1 & b & -c a \\ 1 & c & -a b \end{array}\right|\)
\(A=\begin{vmatrix} 1 & a&{a}^{2} \\ 1 &b&b^2\\1&c&c^2 \end{vmatrix}-\begin{vmatrix} 1 & a&bc \\1 &b&ca\\1&c&ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{1}{a b c}\left|\begin{array}{ccc} a & a^2 & a b c \\ b & b^2 & a b c \\ c & c^2 & a b c \end{array}\right|\)
(Multiplying R₁, R₂ and R₃ of II det by a, b, c respectively)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right|\)
\(\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|=0\)