Here the technology matrix is given under

The technology matrix is B = \(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \)
I - B = \(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)-  \(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \) = \(\left[ \begin{matrix} 0.6 & -0.1 \\ -0.7 & 0.4 \end{matrix} \right] \)
|I - B| = \(\left| \begin{matrix} 0.6 & -0.1 \\ -0.7 & 0.4 \end{matrix} \right|\)
= (0.6)(0.4) – (–0.7)(–0.1)
= 0.24 – 0.07 = 0.17
Since the diagonal elements of I – B are positive and value of |I-B| is positive, the system is viable.
adj(I - B) = \(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \)
(I - B)^-1 = \(\frac{1}{|I -B |} \)adj(I - B)
= \(\frac{1}{0.17} \) \(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \)
X = (I – B)^–1D, where D =\(\left[ \begin{matrix} 50 \\ 100 \end{matrix} \right] \)
= \(\frac{1}{0.17} \)\(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \)\(\left[ \begin{matrix} 50 \\ 100 \end{matrix} \right] \)
= \(\frac{1}{0.17} \)\(\left[ \begin{matrix} 30 \\ 95 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 176.5 \\ 558.8 \end{matrix} \right] \)
Steel output = 176.5 tonnes
Coal output = 558.8 tonnes
Total labour days required
= 5(steel output) + 2( coal output)
= 5(176.5) + 2(558.8)
= 882.5 + 1117.6 = 2000.1
\(\simeq\) 2000 labour days.

The given system can be written as
\(\left[ \begin{matrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 8 \\ 1 \\ 4 \end{matrix} \right] \)
i.e., AX = B
X = A^–1B
Here \(A=\left[ \begin{matrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)and \( B=\left[ \begin{matrix} 8 \\ 1 \\ 4 \end{matrix} \right] \)
\(|A|=\left[ \begin{matrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{matrix} \right] \)
= -17 ≠ 0
A^–1 exists
A₁₁ = –1 A₁₂ = –8 A₁₃ = –10
A₂₁ = –5 A₂₂ = –6 A₂₃ = 1
A₃₁ = –1 A₃₂ = 9 A₃₃ = 7
\(|A_{ij}|=\left[ \begin{matrix} -1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7 \end{matrix} \right] \)
adj A = [A_ij]^T
\(=\left[ \begin{matrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{matrix} \right] \)
A^–1 = \(\frac{1}{|A|}adjA\)
\(=-\frac{1}{17}\left[ \begin{matrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{matrix} \right] \)
X = A^-1B
\(=-\frac{1}{17}\left[ \begin{matrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{matrix} \right] \)\(\left[ \begin{matrix} 8 \\ 1 \\ 4 \end{matrix} \right] \)
\(=-\frac{1}{17}\left[ \begin{matrix} -17 \\ -34 \\ -51 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\3 \end{matrix} \right] \)
\(\left[ \begin{matrix}x\\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] \)
x = 1, y = 2 and z = 3.

\(\left| A \right| =\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| =1-2=-1\neq 0\)
\(\therefore\)` A ^-1 exists.
\(\left| B \right| =\left| \begin{matrix} 0 & -1 \\ 1 & 2 \end{matrix} \right| =0+1=1\neq 0\)
B ^-1also exists.
\(AB=\left[ \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} 0 & -1 \\ 1 & 2 \end{matrix} \right] =\left[ \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right] \)
\(AB=\left[ \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right] \) = -1 ≠ 0
(AB) ^-1 exists.
\(adj(AB)=\left[ \begin{matrix} 1 & -3 \\ -1 & 2 \end{matrix} \right] \)
\(({ AB) }^{ -1 }=\frac { 1 }{ \left| AB \right| } adj(AB)\)
\(=\frac { 1 }{ -1 } \left[ \begin{matrix} 1 & -3 \\ -1 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -1 & 3 \\ 1 & -2 \end{matrix} \right]\)  ...(1)
\(adj\ A=\left[ \begin{matrix} 1 & -2 \\ -1 & 1 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adjA\)
\(=\frac { 1 }{ -1 } \left[ \begin{matrix} 1 & -2 \\ -1 & 1 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -1 & 2 \\ 1 & -1 \end{matrix} \right] \)
\(adj\ B=\left[ \begin{matrix} 2 & 1 \\ -1 & 0 \end{matrix} \right] \)
\({ B }^{ -1 }=\frac { 1 }{ \left| B \right| } adjB\)
\(=\frac { 1 }{ 1 } \left[ \begin{matrix} 2 & 1 \\ -1 & 0 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 2 & 1 \\ -1 & 0 \end{matrix} \right] \)
\({ B }^{ -1 }{ A }^{ -1 }=\left[ \begin{matrix} 2 & 1 \\ -1 & 0 \end{matrix} \right] \left[ \begin{matrix} -1 & 2 \\ 1 & -1 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -1 & 3 \\ 1 & -2 \end{matrix} \right]\)  ...(2)
From (1) and (2),  (AB)^-1 = B^-1A^-1. Hence proved.

\(\left| A \right| =\left| \begin{matrix} 2 & 4 & 4 \\ 2 & 5 & 4 \\ 2 & 5 & 3 \end{matrix} \right| \)
\(=2\left| \begin{matrix} 5 & 4 \\ 5 & 3 \end{matrix} \right| -4\left| \begin{matrix} 2 & 4 \\ 2 & 3 \end{matrix} \right| +4\left| \begin{matrix} 2 & 5 \\ 2 & 5 \end{matrix} \right| \)
= 2(15 - 20) -4(6 - 8) + 4(10 - 10)
= 2(-5) -4 (-2) + 4(0)
= -10 + 8 + 0 = -2 ≠ 0
Since A is a nonsingular matrix, A ^-1 exists.
A₁₁ = -5; A₂₁ = 8; A₃₁ = -4
A₁₂ = 2; A₂₂ = -2; A₃₂ = 0
A₁₃ = 0; A₂₃ = -2; A₃₃ = 2
\(\left[ { A }_{ ij } \right] =\left[ \begin{matrix} -5 & 2 & 0 \\ 8 & -2 & -2 \\ -4 & 0 & 2 \end{matrix} \right] \)
\(adj\ A=\left[ \begin{matrix} -5 & 8 & -4 \\ 2 & -2 & 0 \\ 0 & -2 & 2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adj\ A=\frac { 1 }{ -2 } \left[ \begin{matrix} -5 & 8 & -4 \\ 2 & -2 & 0 \\ 0 & -2 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} \frac { 5 }{ 2 } & -4 & 2 \\ -1 & 1 & 0 \\ 0 & 1 & -1 \end{matrix} \right] \)  

\(X=\left(\begin{array}{ccc} 8 & -1 & -3 \\ -5 & 1 & 2 \\ 10 & -1 & -4 \end{array}\right)\)
\(|X|=8(-4+2)+1(20-20)-3(5-10)\)
\(=-16+15=-1 \neq 0\)
\(X^{-1} \text { exists }\)
\(A_{11}=\text {Co-factor of } 8=-4+2=-2 \)
\(A_{12}=\text {Co-factor of }-1=-(20-20)=0\)
\(A_{13}=\text {Co-factor of }-3=5-10=-5\)
\(\mathrm{A}_{21}=\text {Co-factor of }-5=-(4-3)=-1 \)
\(\mathrm{A}_{22}=\text {Co-factor of } 1=-32+30=-2 \)
\(\mathrm{A}_{23}=\text {Co-factor of } 2=-(-8+10)=-2\)
\(\mathrm{A}_{31}=\text {Co-factor of } 10=-2+3=1 \)
\(\mathrm{A}_{32}=\text {Co-factor of }-1=-(16-15)=-1 \)
\(\mathrm{A}_{33}=\text {Co-factor of }-4=8-5=+3\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 0 & -5 \\ -1 & -2 & -2 \\ 1 & -1 & +3 \end{array}\right)\)
\(\operatorname{Adj} X=\left(\begin{array}{ccc} -2 & -1 & 1 \\ 0 & -2 & -1 \\ -5 & -2 & +3 \end{array}\right)\)
\(X^{-1}=\frac{1}{|X|} \operatorname{adj} X\)
\(=\frac{-1}{1}\left(\begin{array}{ccc} -2 & -1 & 1 \\ 0 & -2 & -1 \\ -5 & -2 & +3 \end{array}\right)\)
\(\text {Given } Y=X^{-1}\)
\(\left(\begin{array}{ccc} 2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & p & q \end{array}\right)=\left(\begin{array}{ccc} 2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & 2 & -3 \end{array}\right)\)
\(\mathrm{p}=2 ; \mathrm{q}=-3\)