(Since \(\left|A^{-1}\right|=\frac{1}{|A|}\))

Since \(R_2 \sim R_3\)

\(A=\left(\begin{array}{cc} \frac{4}{5} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{1}{2} \end{array}\right)\)
\(|A|=\frac{2}{5}-\frac{1}{6}=\frac{12-5}{30}=\frac{7}{30}\)
\(A^{-1}=\frac{1}{|A|} \text { adjA }=\frac{30}{7}\left(\begin{array}{ll} \frac{1}{2} & \frac{5}{12} \\ \frac{2}{5} & \frac{4}{5} \end{array}\right)\)

\(\left|\begin{array}{ccc} 2 x+y & x & y \\ 2 y+z & y & z \\ 2 z+x & z & x \end{array}\right| \quad c_1 \rightarrow c_1-c_3\)
\(\left|\begin{array}{ccc} 2 x & x & y \\ 2 y & y & z \\ 2 z & z & x \end{array}\right|=0 \ \text {Since } c_1 \sim c_3\)

Let A = \(\begin{vmatrix} x &a &x+a \\ y & b &y+b \\z & c & z+c \end{vmatrix}\)
Applying the elementary transformation, \(C_1\rightarrow C_1+C_2\) we get,
\(A=\begin{vmatrix} x+a&a&a+x\\y+b&b&y+b\\z+c&c&z+c\end{vmatrix}=0[C_1\equiv C_3]\)
\(\therefore\) |A| = 0.

Given \(\begin{vmatrix}2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
\(\Rightarrow\)   2 - 20 = 2x² - 24
\(\Rightarrow\)  -18 = 2x² - 24
\(\Rightarrow\)  -18 + 24 = 2x²
\(\Rightarrow\)  6 = 2x²
\(\Rightarrow\) x² = 3 
\(\Rightarrow\) x = \(\pm\sqrt{3}\)

AB = \(\begin{bmatrix} 1 \\ -4 \\3 \end{bmatrix}\begin{bmatrix} -1 &2 & 1 \end{bmatrix}=\begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\-3 & 6 & 3 \end{bmatrix}\)
\(\therefore\)  \({(AB)}^{T}=\begin{bmatrix} -1 &4&-3 \\ 2 & -8&6\\1&-4&3 \end{bmatrix}\)  ....(1)
\({B}^{T}=\begin{bmatrix} -1 & 2 & 1 \end{bmatrix}^{T}=\begin{bmatrix} -1\\2\\1\end{bmatrix}\)and \({A}^{T}={\begin{bmatrix} 1\\-4\\3\end{bmatrix}}^{T}=\begin{bmatrix} 1&-4&3 \end{bmatrix}\)
\(\therefore\)  \({B}^{T}{A}^{T}=\begin{bmatrix} -1\\2\\1 \end{bmatrix}\begin{bmatrix} 1&-4&3 \end{bmatrix}=\begin{bmatrix} -4 & 4&-3 \\ 2&-8 &6\\1&-4&3 \end{bmatrix}\)  ....(2)
From (1) and (2), (AB)^T = B^T . A^T

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

B = \(\left[ \begin{matrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{matrix} \right] \)
I - B=\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)-\(\left[ \begin{matrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 0.2 & -0.2 \\ - 0.9 & 0.3 \end{matrix} \right] \)
|I - B|=  \(\left[ \begin{matrix} 0.2 & -0.2 \\ - 0.9 & 0.3 \end{matrix} \right] \)
= (0.2)(0.3) - (-0.2)(-0.9)
= 0.06 - 0.18
= 0.12 < 0
Since |I - B| is negative, Hawkins – Simon conditions are not satisfied.
Therefore, the given system is not viable.

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

LHS = \(\begin{vmatrix} a & a+b&a+b+c \\2a &3a+2b &4a+3b+2c\\3a&6a+3b&10a+6b+3c \end{vmatrix}\)
Applying R2 \(\rightarrow\) R2 - 2R1 and R3 \(\rightarrow\) 3R1 we get,
\(=\begin{vmatrix} a & a+b&a+b+c \\0 &0 &2a+b\\0&3a&7a+3b\end{vmatrix}\)

Expanding along C₁ we get
\(=a\begin{vmatrix}a&2a+b\\3a&7a+b\end{vmatrix}-0+0\)
= a(7a² + 3ab - 6a² - 3ab)
= a(7a² - 6a²) = a(a²) = a³ = RHS
Hence proved.

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

\(=\left|\begin{array}{lll} 10041 & 10042 & 10043 \\ 10045 & 10046 & 10047 \\ 10049 & 10050 & 10051 \end{array}\right|\)
Applying the elementary transformation R₂ \(\rightarrow\) R₂\(\rightarrow\)R₁, R₃ \(\rightarrow\) R₃\(\rightarrow\)R₂ we get
\(\left|\begin{array}{ccc} 10041 & 10042 & 10043 \\ 4 & 4 & 4 \\ 8 & 8 & 8 \end{array}\right|=0 \text { Since } R_2 \sim R_3\)

Expanding along R₁ we get,
\(\left|\begin{array}{lll} 2 & x & 3 \\ 4 & 1 & 6 \\ 1 & 2 & 7 \end{array}\right|=0\)
⇒ 2 (7 -12) -x (28- 6) + 3 (8 - 1) = 0
⇒ -10 - 22x + 21 = 0
⇒ 11 = 22x
⇒ \(x={11\over 22}={1\over 2}\)

Let x, y, z be the salary of each type of staff in A, B and C categories
4x + 2y + 3z = 4900
3x + 3y + 2z = 4500
4x + 3y + 4z = 5800
In matrix form\(\begin{bmatrix} 4&2&3\\3&3&2\\4&3&4 \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 4900\\4500\\5800 \end{bmatrix}\)
\(\Rightarrow\) \(A X=B \Rightarrow X=A^{-1} B\)
Where A  \(=\begin{bmatrix}4&2&3\\3&3&2\\4&3^4 \end{bmatrix},X=\begin{bmatrix} x\\y\\z \end{bmatrix},B=\begin{bmatrix} 4900 \\4500\\5800 \end{bmatrix}\)
= 4 (12 - 6) - 2(12 - 8) + 3 (9 - 12)
= 4 (6) - 2 (4) + 3 (-3) = 24 - 8 - 9 = 7 = 1
\(\therefore\) A^-1 exists.
\(Co-factor matrix=\begin{bmatrix} 6&-4&-3\\1&4&-4\\-5&1&6 \end{bmatrix}\)
\(\therefore\ {A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{7}}\begin{bmatrix} 6&1&-5\\-4&4&1\\-3&-4&6 \end{bmatrix}\)
\(\therefore\ X={A}^{-1}B={{1}\over{7}}\begin{bmatrix} 6&1&-5\\-4&4&1\\-3&-4&6 \end{bmatrix}\begin{bmatrix} 4900\\45000\\5800 \end{bmatrix}={{1}\over{7}}\begin{bmatrix} 29400&+45000&-29,000\\-19600&+18000&+5800\\-14700&-18,000&+34800 \end{bmatrix}\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)={{1}\over{7}}\begin{bmatrix} 49000\\4200\\2100\end{bmatrix}=\begin{bmatrix} 700\\600\\300 \end{bmatrix}\)
Salary for each type of staff under category A, B and C are respectively Rs 700,Rs 2600 and Rs 7300

Given A = \(\begin{bmatrix} 1&2&2\\2&1&2\\2&2&1 \end{bmatrix}\)
\({A}^{2}=A.A=\begin{bmatrix} 1&2&2 \\2&1&2\\2&2&1 \end{bmatrix}\begin{bmatrix} 1 &2&2 \\ 2 &1&2\\2&2&1 \end{bmatrix}=\begin{bmatrix} 9&8&8 \\ 8 &9&8\\8&8&9 \end{bmatrix}\)
\(4A=4\begin{bmatrix} 1&2&2 \\ 2 & 1 & 1\\2 & 2 & 1\end{bmatrix}=\begin{bmatrix} 4 & 8 & 8\\8 & 4 & 8\\ 8 & 8 & 9\end{bmatrix}\)and \({5I}_{3}=\begin{bmatrix} 5& 0& 0\\0 & 5 & 0\\0 & 0 & 5 \end{bmatrix}\)
\(\therefore\ A^2-4A-5I_3=\begin{bmatrix} 9&8&8\\8&9&8\\8&8&9 \end{bmatrix}-\begin{bmatrix} 4&8&8 \\ 8&4&8\\8&8&4 \end{bmatrix}-\begin{bmatrix} 5&0&0\\0&5&0\\0&0&5 \end{bmatrix}\)
\(=\begin{bmatrix} 9-4-5&8-8-0&8-8-0\\8-8-0&9-4-5&8-8-0\\8-8-0&8-8-0&9-4-5 \end{bmatrix}=\begin{bmatrix} 0&0&0 \\ 0&0&0\\0&0&0 \end{bmatrix}\)
\(\therefore\)  A² - 4A - 5I₃ = 0
\(\Rightarrow\)    A²-4A = 5I₃
Premultiplying throughout by A^-1 we get
A^-1 A² - 4A^-1 A = 5A^-1 I₃
\(\Rightarrow\) (A^-1 A)· A - 4 (A^-1 A) = 5 A^-1 I₃
\(\Rightarrow\) 5A^-1 = A - 4I
\(\Rightarrow\)   \({A}^{-1}=\frac{1}{5}[A-4I]\)
\(\Rightarrow\)  \({ A }^{ -1 }=\frac { 1 }{ 5 } \left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{matrix} \right] =\left[ \begin{matrix} \frac { -3 }{ 5 } & \frac { 2 }{ 5 } & \frac { 2 }{ 5 } \\ \frac { 2 }{ 5 } & \frac { -3 }{ 5 } & \frac { 2 }{ 5 } \\ \frac { 2 }{ 5 } & \frac { 2 }{ 5 } & \frac { -3 }{ 5 } \end{matrix} \right]\)

a₁₁ = 50, a₁₂ = 75, x₁ = 200
a₂₁ = 100, a₂₂ = 50, x₂ = 200
\({b}_{11}={{{a}_{11}}\over{{x}_{1}}}={{50}\over{200}}={{1}\over{4}};{b}_{12}=\frac{{a}_{12}}{{x}_{12}}=\frac{75}{200}=\frac{3}{8}\)
\({b}_{21}=\frac{{a}_{11}}{x_1}{{100}\over{200}}={{1}\over{2}};{b}_{22}={{{a}_{22}}\over{{x}_2{}}}={{50}\over{200}}={{1}\over{4}}\)
\(\therefore\)  The technology matrix is B \(=\begin{bmatrix} \frac{1}{4}&\frac{3}{8}\\\frac{1}{2}&\frac{1}{4} \end{bmatrix}\)
\(I-B=\begin{bmatrix}1&0\\0&1 \end{bmatrix}-\begin{bmatrix} \frac{1}{4} &\frac{3}{8}\\ \frac{1}{2}& \frac{1}{4} \end{bmatrix}=\begin{bmatrix} {{3}\over{4}}&{{-3}\over{8}}\\ -{{1}\over{2}} &{{3}\over{4}} \end{bmatrix}\)
\(|I-B| ={{9}\over{16}}-{{3}\over{16}}={{6}\over{16}}={{3}\over{8}}>0 \)
Now, X = (I - B)^-1 D Where D \(=\begin{bmatrix} 300 \\ 600 \end{bmatrix}\)
\(X=\frac{8}{3}\left(\begin{array}{ll} \frac{1}{4} & \frac{3}{8} \\ \frac{1}{2} & \frac{1}{4} \end{array}\right)\left(\begin{array}{l} 300 \\ 600 \end{array}\right)\)
\(=\frac{8}{3}\left(\begin{array}{c} 75+225 \\ 150+150 \end{array}\right)=\left(\begin{array}{l} 800 \\ 800 \end{array}\right)\)
The output of P and Q must be 800 each.

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)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right|\)
\(\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|=0\)