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

Since \(R_2 \sim R_3\)

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

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

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

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

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

Given equations are
x - y + 2z = 3, 2x + z = 1 and 3x + 2y +z = 4.
The given equations can be written in matrix form as 
\(\begin{bmatrix} 1&-1&2\\2&0&1\\3&2&1 \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 3\\1\\4 \end{bmatrix}\)
\(A X=B \Rightarrow X=A^{-1} B\)
\(\text {Where } A=\left(\begin{array}{ccc} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{array}\right), X=\left(\begin{array}{l} x \\ y \\ z \end{array}\right)\)
\(B=\left(\begin{array}{l} 3 \\ 1 \\ 4 \end{array}\right)\)
\(|A|=1(0-2)+1(2-3)+2(4-0)\)
\(=-2-1+8=5 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 1 & 4 \\ 5 & -5 & -5 \\ -1 & 3 & 2 \end{array}\right)\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} \mathrm{A}=\frac{1}{5}\left(\begin{array}{ccc} -2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2 \end{array}\right)\)
\(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\)
\(={{1}\over{5}}\begin{bmatrix} -2&5&-1\\1&-5&3\\4&-5&2 \end{bmatrix}\begin{bmatrix} 3\\1\\4 \end{bmatrix}={{1}\over{5}}\begin{bmatrix} -6+5-4\\3-5+12\\12-5+8 \end{bmatrix}={{1}\over{5}}\begin{bmatrix} -5\\10\\15 \end{bmatrix}=\begin{bmatrix} -1\\2\\3 \end{bmatrix}\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right)\)
\(x=-1, y=2, z=3\)

To show that A is inverse of A itself it is enough if, we prove that
\(A . A=I \left(\because \mathrm{AA}^{-1}=\mathrm{I}\right. and \left.\mathrm{A}^{-1}=\mathrm{A}\right)\)
\(A \cdot A=\left(\begin{array}{ccc} -1 & 2 & -2 \\ 4 & -3 & 4 \\ 4 & -4 & 5 \end{array}\right)\left(\begin{array}{ccc} -1 & 2 & -2 \\ 4 & -3 & 4 \\ 4 & -4 & 5 \end{array}\right)\)
\(=\left(\begin{array}{ccc} 1+8-8 & -2-6+8 & 2+8-10 \\ -4-12+16 & 8+9-16 & -8-12+20 \\ -4-16+20 & 8+12-20 & -8-16+25 \end{array}\right)\)
\(=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=I\)
\(\therefore \mathrm{A} \text { is inverse of } \mathrm{A}\)

\(A=\left(\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right)\)
\(|A|=3(12-10)+1(-30+25)+1(30-30)\)
\(=6-5=1 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right)\)

Given A \(=\begin{bmatrix} 1&1&1\\3&4&7\\1&-1&1 \end{bmatrix}\)
\(=(4+7)-1(3-7)+1(-3-4)\)
\(=11+4-7=8\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 11 & 4 & -7 \\ -2 & 0 & +2 \\ 3 & -4 & 1 \end{array}\right)\)
\(\operatorname{adj} A=\left(\begin{array}{ccc} 11 & -2 & 3 \\ 4 & 0 & -4 \\ -7 & 2 & 1 \end{array}\right)\)
\(\mathrm{A}(\operatorname{adj} A)=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 3 & 4 & 7 \\ 1 & -1 & 1 \end{array}\right)\left(\begin{array}{ccc} 11 & -2 & 3 \\ 4 & 0 & -4 \\ -7 & 2 & 1 \end{array}\right)\)
\(=\left(\begin{array}{ccc} 11+4-7 & -2+0+2 & 3-4+1 \\ 33+16-49 & -6+0+14 & 9-16+7 \\ 11-4-7 & -2+0+2 & 3+4+1 \end{array}\right)\)
\(=\left(\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}\right)=8\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=|A| I_3\)
\((adj\ A)\ A=\begin{bmatrix}11&-2&3\\4&0&-4\\-7&2&1\end{bmatrix}\begin{bmatrix} 1&1&1\\3&4&7\\1&-1&1 \end{bmatrix}\)
\(=\begin{bmatrix} 11-6+3&11-8-3&11-14+3\\4+0-4&4+0+4&4+0-4\\-7+6+1&-7+8-1&-7+14+1 \end{bmatrix}=\begin{bmatrix} 8&0&0\\0&8&0\\0&0&8 \end{bmatrix}\) ...(2)
\(|A|.{I}_{3}=8\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}=\begin{bmatrix} 8&0&0\\0&8&0\\0&0&8\end{bmatrix}\)  ...(3)
From (1), (2) and (3)
A (adj A) = (adj A) A = |A|I₃.

a₁₁ = 4, a₂ = 3, x₁ = 20
a₂₁ = 5, a₂₂ = 4, x₂ = 12
\({b}_{11}={{{a}_{11}}\over{x_1}}={{4}\over{20}}={{1}\over{4}}={{1}\over{5}};{b}_{12}={{{a}_{12}}\over{{x}_{2}}}={{3}\over{}12}={{1}\over{}4}\)
\({b}_{21}={{{a}_{21}}\over{x_1}}=\frac{5}{20}=\frac{1}{4};{b}_{22}=\frac{{a}_{22}}{{x}_{2}}=\frac{4}{12}=\frac{1}{3}\)
i) The technology matrix is B \(=\begin{bmatrix} \frac{1}{5}&\frac{1}{4} \\ \frac{1}{4}&\frac{1}{3} \end{bmatrix}\)
ii) \(I-B=\begin{bmatrix} 1&0\\0&1\end{bmatrix}-\begin{bmatrix} \frac{1}{5} &\frac{1}{4}\\ \frac{1}{4}&\frac{1}{3} \end{bmatrix}\)\(= \begin{bmatrix} \frac { 4 }{ 5 } & \frac { -1 }{ 4 } \\ \frac { -1 }{ 4 } & \frac { 2 }{ 3 } \end{bmatrix}\)
\(|I-B|=\frac{4}{5} \times \frac{2}{3}-\frac{1}{16}=\frac{8}{15}-\frac{1}{16}=\frac{128-15}{240}=\frac{113}{240}\)
Since the diagonals elements of I - B are positive and |I - B| are positive, the system is viable
\(\therefore\ {(I-B)}^{-1}={{1}\over{|I-B|}}adj\ (I-B)={{240}\over{}113}\begin{bmatrix} {{2}\over{3}}&{{1}\over{4}}\\ {{1}\over{4}}&{{4}\over{5}} \end{bmatrix}\)
X = (I - B)^-1 D where D \(=\begin{bmatrix} 23\\3\end{bmatrix}={{240}\over{113}}\begin{bmatrix} {{2}\over{3}} &{{1}\over{4}}\\ {{1}\over{4}}&{{4}\over{5}} \end{bmatrix}\begin{bmatrix} 23\\3 \end{bmatrix}\)
\(=\frac{240}{113}\left(\begin{array}{c} 15.33+0.75 \\ 5.75+2.4 \end{array}\right)=\frac{240}{113}\left(\begin{array}{c} 16.08 \\ 8.15 \end{array}\right)\)
\(X=\left(\begin{array}{l} 34.15 \\ 17.31 \end{array}\right)\)
The output for sector 1 is 34.15 units and 2 is 17.31 units

\(A B=\left(\begin{array}{cc} 2 & 3 \\ 1 & -6 \end{array}\right)\left(\begin{array}{cc} -1 & 4 \\ 1 & -2 \end{array}\right)\)
\(=\left(\begin{array}{cc} -2+3 & 8-6 \\ -1-6 & 4+12 \end{array}\right)=\left(\begin{array}{cc} 1 & 2 \\ -7 & 16 \end{array}\right)\)
\(\mathrm{LHS}=\operatorname{adj}(\mathrm{AB})=\left(\begin{array}{cc} 16 & -2 \\ 7 & 1 \end{array}\right)\)
\(\mathrm{RHS}=(\operatorname{adj} B)(\operatorname{adj} \mathrm{A})\)
\(=\left(\begin{array}{cc} -2 & -1 \\ -1 & -1 \end{array}\right)\left(\begin{array}{cc} -6 & -3 \\ -1 & 2 \end{array}\right)\)
\(=\left(\begin{array}{cc} 12+4 & 6-5 \\ 6+1 & 3-2 \end{array}\right)=\left(\begin{array}{cc} 16 & -2 \\ 7 & 1 \end{array}\right)\)
\(a d j(A B)=(a d j\ B)(a d j\ A)\)

2x + 3y = 5
x - 2y = -1
The given system can be written as
\(\begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix}\)\(\begin{bmatrix} X\\Y \end{bmatrix}\)=\(\begin{bmatrix} 5\\-1 \end{bmatrix}\)
AX = B \(\Rightarrow \) \(X=A^{-1} B\)
Where A =\(\begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix}\), \(X= \left[ \begin{matrix} x \\ y \end{matrix} \right] \),B=\(\begin{bmatrix} 5\\-1 \end{bmatrix}\)
\(|A|=-4-3=-7 \neq 0\)
\(\therefore\) A^-1 exists.
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{-1}{7}\left(\begin{array}{cc} -2 & -3 \\ -1 & 2 \end{array}\right)\)
\(X=A^{-1} B\)
\(=\frac{-1}{7}\left(\begin{array}{cc} -2 & -3 \\ -1 & 2 \end{array}\right)\left(\begin{array}{c} 5 \\ -1 \end{array}\right)\)
\(=\frac{-1}{7}\left(\begin{array}{cc} -10 & +3 \\ -5 & -2 \end{array}\right)=\frac{-1}{7}\left(\begin{array}{l} -7 \\ -7 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 1 \\ 1 \end{array}\right)\)
\(\text {Solution is } x=1, y=1\)

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

\(AB=\begin{bmatrix} 3&-1\\2&1 \end{bmatrix}\begin{bmatrix} 3&0\\1&-2 \end{bmatrix}=\begin{bmatrix} 9-1&0+2\\6+1&0-2 \end{bmatrix}=\begin{bmatrix} 8&2\\7&-2 \end{bmatrix}\)
= -16 - 14 = -30
\(\therefore\) |AB| = -30

\(\left|\begin{array}{ccc} 3 & -2 & 4 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right|=3[0-2]+2[6-1]+4[4-0]\)
\(=-6+10+16=20\)

Let A = \(\left[ \begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix} \right| \)
= 4 – 4 = 0
\(\therefore\) A is a singular matrix

\(\left| \begin{matrix} 1 & 3 & 4 \\ 102 & 18 & 36 \\ 17 & 3 & 6 \end{matrix} \right| =6\left| \begin{matrix} 1 & 3 & 4 \\ 17 & 3 & 6 \\ 17 & 3 & 6 \end{matrix} \right| \)
= 0 (since R₂ ≡ R₃)

\(\left| \begin{matrix} x-1 & x & x-2 \\ 0 & x-2 & x-3 \\ 0 & 0 & x-3 \end{matrix} \right| =0\)
⇒(x – 1)(x – 2)(x – 3) = 0
x = 1, x = 2, x = 3

Let \(A=\begin{bmatrix} 1&-1\\2&3 \end{bmatrix}\)
\(|A|=\begin{vmatrix} 1&-1\\2&3 \end{vmatrix}=3+2=5\) ≠ 0
\(\therefore A^{-1} \text { exists }\)
Now, \({A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{5}}\begin{bmatrix} 3 & 1 \\ -2 & 1\end{bmatrix}\)

Given \(A=\begin{bmatrix}2&3\\1&4 \end{bmatrix}\)    
\(Adj\ A=​​\begin{bmatrix}4&-3\\-1&2 \end{bmatrix}\)