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.

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,
\(|A|=2\times4\begin{vmatrix} 3 & 5&3 \\ 1 & 4 & 1\\1 &1 &1\end{vmatrix}=8\times 0\ [\because C_1\equiv C_3]=0\)

Given = \(\begin{bmatrix} 1 & 2 \\ 4 & 2\end{bmatrix},\) then 2A = \(\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
\(\therefore\)   |2A| = \(\begin{vmatrix}2 & 4 \\8 & 4 \end{vmatrix}\) = 8 - 32 = -24   ...(1)
Also, |A| = \(\begin{vmatrix} 1&2 \\4 & 2 \end{vmatrix}\) = 2 - 8 =  -6
\(\therefore\)  4|A| = 4(-6) = -24   ....(2)
From (1) and (2), |2A| = 4.|A|

\(a_{ 11 }=15,\quad { a }_{ 12 }=10,\quad { x }_{ 1 }=35\)
\(a_{ 21 }=20,\quad { a }_{ 22 }=30,\quad { x }_{ 2 }=65\)
\(b_{ 11 }=\frac { { a }_{ 11 } }{ { x }_{ 1 } } =\frac { 15 }{ 35 } =\frac { 3 }{ 7 } ;\quad b_{ 12 }=\frac { { a }_{ 12 } }{ { x }_{ 2 } } =\frac { 10 }{ 65 } =\frac { 2 }{ 13 } \)
\(b_{ 21 }=\frac { { a }_{ 21 } }{ { x }_{ 1 } } =\frac { 20 }{ 35 } =\frac { 4 }{ 7 } ;\quad b_{ 22 }=\frac { { a }_{ 12 } }{ { x }_{ 2 } } =\frac { 30 }{ 65 } =\frac { 6 }{ 13 } \)
B = \(\begin{bmatrix} \frac { 3 }{ 7 } & \frac { 2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 6 }{ 13 } \end{bmatrix}\)
\(I-B=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} \frac { 3 }{ 7 } & \frac { 2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 6 }{ 13 } \end{bmatrix}=\begin{bmatrix} \frac { 4 }{ 7 } & \frac { -2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 7 }{ 13 } \end{bmatrix}\)
\(|I-B|=\frac{4}{7} \times \frac{7}{13}-\left(\frac{2}{13} \times \frac{4}{7}\right)=\frac{28-8}{91}=\frac{20}{91}\)
Since the diagonal elements of I - B are positive and |I - B| is positive the system is available
 \((I-B)^{-1}=\frac{1}{|I-B|} \operatorname{adj}(\mathrm{I}-\mathrm{B})=\frac{91}{20}\left(\begin{array}{cc} \frac{7}{13} & \frac{2}{13} \\ \frac{4}{7} & \frac{4}{7} \end{array}\right)\)
Now, X = (I - B)^-1 D where D \(=\left[ \begin{matrix} 12 \\ 18 \end{matrix} \right] =\frac { 91 }{ 20 } \begin{bmatrix} \frac { 7 }{ 13 } & \frac { 2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 4 }{ 7 } \end{bmatrix}\left[ \begin{matrix} 12 \\ 18 \end{matrix} \right]=\frac{91}{20}\left(\begin{array}{l} \frac{84+36}{13} \\ \frac{48+72}{7} \end{array}\right)\)
\(=\left(\begin{array}{l} 42 \\ 78 \end{array}\right)\)
The gross output for two sectors X and Y are 42 and 78 respectively

\(\left|\begin{array}{cc} 20 & 15 \\ 15 & 5 \end{array}\right|=5 \times 5\left|\begin{array}{ll} 4 & 3 \\ 3 & 1 \end{array}\right|\)
\(=5 \times 5 \times-5\)
\(=-125\)

5x - 16 = 0
5x = 16
\(x=\frac{16}{5}\)

(Since \(|\operatorname{adj} A|=|A|^{n-1} n=3\) ))
\(=|3|^2=9\)

\(|A|=4-2=2\)
\(A(\operatorname{adj} A)=|A| I=\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)\)

\(|A|=6-5=1\)
\(A^{-1}=\left(\begin{array}{cc} 2 & -1 \\ -5 & 3 \end{array}\right)\)

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

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

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 A = \(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| \)
Applying C₁➝C₁+9C₂ we get
A = \(\left| \begin{matrix} 2+63 & 7 & 65 \\ 3+72 & 7 & 65 \\ 5+81 & 9 & 86 \end{matrix} \right| =\left| \begin{matrix} 65 & 7 & 65 \\ 75 & 8 & 75 \\ 86 & 9 & 86 \end{matrix} \right| =0[ \because {C}_{1}\equiv{C}_{3}]\)

LHS = \(\begin{vmatrix}x+a &b&c \\a &x+b&c\\a&b&x+c \end{vmatrix}\)
Applying C₁ \(\rightarrow\) C₁ + C₂ + C₃ we get,
LHS = \(\begin{vmatrix}x+a+b+c &b&c \\x+a+b+c &x+b&c\\x+a+b+c&b&x+c \end{vmatrix}\)
Taking (x + a + b + c) common from C₁ we get,
= (x + a + b + c) \(\begin{vmatrix}1 &b&c \\1 &x+b&c\\1&b&x+c \end{vmatrix}\)
Applying R₂ \(\rightarrow\) R₂ - R₁ and R₃ \(\rightarrow\) R₁ we get,
= \((x+a+b+c)\begin{vmatrix}1 &b&c \\0&x&0\\0&0&x\end{vmatrix}\)
Expanding along C₁ we get,
\(=(x+a+b+c)\begin{bmatrix} 1\begin{bmatrix} x & 0 \\ 0 & x\end{bmatrix} +0+0\end{bmatrix}\)
= (x + a + b + C)(x²) = RHS
Hence proved.

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 = \({{1}\over{13}}\begin{bmatrix} 2& -5\\3 & -1 \end{bmatrix}\)

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
⇒ \(x={11\over 22}={1\over 2}\)

\(3 x-y+2 z=13 ; 2 x+y-z=3\)
\(x+3 y-5 z=-8\)
The given system can be written as
\(\left(\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(A X=B \Rightarrow X=A^{-1} B\)
\(\text {Where } A=\left(\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{array}\right), X=\left(\begin{array}{l} x \\ y \\ z \end{array}\right)\)
\(B=\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(|A|=3(-5+3)+1(-10+1)+2(6-1)\)
\(=-6-9+10=-5 \neq 0\)
\(\therefore A^{-1} exists\)
\(\mathrm{A}_{11}=\text {Co-factor of } 3=(-5+3)=-2 \)
\(\mathrm{A}_{12}=\text {Co-factor of }-1=-(-10+1)=9 \)
\(\mathrm{A}_{13}=\text {Co-factor of } 2=(6-1)=5\)
\(\mathrm{A}_{21}=\text { Co-factor of } 2=-(5-6)=1\)
\(\mathrm{A}_{22}=\text { Co-factor of } 1=-15-2=-17 \)
\(\mathrm{A}_{23}=\text { Co-factor of }-1=-(9+1)=-10\)
\(\mathrm{A}_{31}=\text {Co-factor of } 1=1-2=-1\)
\(\mathrm{A}_{32}=\text {Co-factor of } 3=-(-3-4)=7 \)
\(\mathrm{A}_{33}=\text {Co-factor of }-5=3+2=5\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 9 & 5 \\ 1 & -17 & -10 \\ -1 & 7 & 5 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{-5}\left(\begin{array}{ccc} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{array}\right)\)
\(X=A^{-1} B=\frac{-1}{5}\left(\begin{array}{ccc} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{array}\right)\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(=\frac{-1}{5}\left(\begin{array}{ccc} -26 & 3 & 8 \\ 117 & -51 & -56 \\ 65 & -30 & -40 \end{array}\right)\)
\(=\frac{-1}{5}\left(\begin{array}{c} -15 \\ 10 \\ -5 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 3 \\ -2 \\ 1 \end{array}\right)\)
\(x=3, \mathrm{y}=-2, \mathrm{z}=1\)

\(|A|=\left| \begin{matrix}1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right| =1+{ tan }^{ 2 }x={ sec }^{ 2 }x\neq 0\)
\(\Rightarrow\) 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)

Given A = \(\begin{bmatrix}1 &3&3 \\1 &4&3\\1&3&4 \end{bmatrix} \)
\(A_{11}=\text {Cofactor of } 1=16-9=7\)
\(A_{12}=\text {Cofactor of } 3=-(4-3)=-1\)
\(A_{13}=\text {Cofactor of } 3=3-4=-1\)
\(A_{21}=\text {Cofactor of } 1=-(12-9)=-3 \)
\(A_{22}=\text {Cofactor of } 4=4-3=1 \)
\(A_{23}=\text {Cofactor of } 3=-(3-3)=0 \)
\( A_{31}=\text {Cofactor of } 1=9-12=-3 \)
\(A_{32}=\text {Cofactor of } 3=3-3=0 \)
\(A_{33}=\text {Cofactor of } 4=4-3=1\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{array}\right)\)
A^-1 = \(\begin{bmatrix} 7&-3&-3\\-1&1&0\\-1&0&1 \end{bmatrix}\)
\(|A| =1(16-9)-3(4-3)+3(3-4) \)
\(=7-3-3=1 \neq 0\)
\(\therefore \mathrm{A}^{-1} \text { exists }\)
\(\mathrm{A}(\operatorname{adj} \mathrm{A})=\left(\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right)\left(\begin{array}{ccc} 7 & -3 & -1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right)\)
\(=\left(\begin{array}{lll} 7-3-3 & -3+3+0 & -3+0+3 \\ 7-4-3 & -3+4+0 & -3+0+3 \\ 7-3-4 & -3+3+0 & -3+0+4 \end{array}\right)\)
\(=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=|A| I\)
\(=\mathrm{A}(\operatorname{adj} \mathrm{A})=|A| I\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right)\)

The technology matrix is given under.

B =\(\begin{bmatrix} 0.4 & 0.1 \\ 0.7 & 0.7 \end{bmatrix}\)
I - B =\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.4 & 0.1 \\ 0.7 & 0.7 \end{bmatrix}=\begin{bmatrix} 0.6 & -0.1 \\ -0.7 & 0.3 \end{bmatrix}\)
|I - B| = \(\begin{bmatrix} 0.6 & -0.1 \\ -0.7 & 0.3 \end{bmatrix}\)
|I - B| = 0.18 - 0.07 = 0.11 > 0
Since the diagonal elements of (I - B) are positive and |I - B| is positive, the system is viable.
\(\therefore \) ( I - B)^-1 = \(\frac { 1 }{ |I-B| } adj(I-B)=\frac { 1 }{ 0.11 } \begin{bmatrix} 0.3 & 0.1 \\ 0.7 & 0.6 \end{bmatrix}\)
Now X = (I - B)^-1 D where D = \(\left[ \frac { 6.8 }{ 10.2 } \right] \)
X = \(\frac { 1 }{ 0.11 } \begin{bmatrix} 0.3 & 0.1 \\ 0.7 & 0.6 \end{bmatrix}\left[ \frac { 6.8 }{ 10.2 } \right] =\frac { 1 }{ 0.11 } \left[ \frac { 0.3\times6.8+0.1\times10.2 }{ 0.7\times6.8+0.6\times10.2 } \right] \)
=\(\frac { 1 }{ 0.11 } \left[ \frac { 2.04+1.02 }{ 4.76+6.12 } \right] =\frac { 1 }{ 0.11 } \left[ \frac { 3.06 }{ 10.88 } \right] =\left[ \frac { 27.81 }{ 98.90 } \right] \)
Gross production of commodity A and B are 27.81
Gross production of B is 98.91 tonnes