Free BECC-104 Solved Assignment | July 2023-January 2024 | Mathematical Methods in Economics-II | IGNOU

BECC-104 Solved Assignment

July 2023-January 2024

Assignment A
Answer the following Long Category questions in about 500 words each. Each question carries 2 0 2 0 20\mathbf{2 0}20 marks. Word limit will not apply in the case of numerical questions.
2 × 20 = 40 2 × 20 = 40 2xx20=402 \times 20=402×20=40
  1. Consider the following two matrices
    A = ( 1 1 2 1 1 2 2 2 4 ) A = 1 1 2 1 1 2 2 2 4 A=([-1,1,2],[1,-1,-2],[-2,2,4])A=\left(\begin{array}{ccc}-1 & 1 & 2 \\ 1 & -1 & -2 \\ -2 & 2 & 4\end{array}\right)A=(112112224)
    B = ( 1 0 1 1 1 1 1 1 1 ) B = 1 0 1 1 1 1 1 1 1 B=([1,0,-1],[-1,1,1],[1,-1,-1])\mathrm{B}=\left(\begin{array}{ccc}1 & 0 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & -1\end{array}\right)B=(101111111)
    (i) Find the rank of ‘ A ‘ and ‘ B ‘
    (ii) Show that ( AB ) 1 = B 1 A 1 ( AB ) 1 = B 1 A 1 (AB)^(-1)=B^(-1)A^(-1)(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}(AB)1=B1 A1
    (iii) Show that ( A 1 ) 1 = A A 1 1 = A (A^(-1))^(-1)=A\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}(A1)1=A
    (iv) Show that ( B 1 ) 1 = B B 1 1 = B (B^(-1))^(-1)=B\left(\mathrm{B}^{-1}\right)^{-1}=\mathrm{B}(B1)1=B
  2. An individual consumer consumes two commodities X 1 & X 2 X 1 & X 2 X_(1)&X_(2)X_1 \& X_2X1&X2. The utility function is
U = X 1 0.4 X 2 0.6 U = X 1 0.4 X 2 0.6 U=X_(1)^(0.4)X_(2)^(0.6)U=X_1^{0.4} X_2^{0.6}U=X10.4X20.6
The price of commodity one is P 1 = P 1 = P_(1)=\mathrm{P}_1=P1= Rs. 3.00 , the price of commodity two is P 2 = P 2 = P_(2)=\mathrm{P}_2=P2= Rs.4.00, the individual’s income per period is Rs.108. Determine the utility maximizing level of X 1 & X 2 X 1 & X 2 X_(1)&X_(2)\mathrm{X}_1 \& \mathrm{X}_2X1&X2 and derive the demand curves for the two commodities.
Assignment B
Answer the following Middle Category questions in about 250 words each. Each question carries 1 0 1 0 10\mathbf{1 0}10 marks. Word limit will not apply in the case of numerical questions.
3 × 10 = 30 3 × 10 = 30 3xx10=303 \times 10=303×10=30
  1. Let Z = f ( x , y ) = 3 x 3 5 y 2 225 x + 70 y + 23 Z = f ( x , y ) = 3 x 3 5 y 2 225 x + 70 y + 23 Z=f(x,y)=3x^(3)-5y^(2)-225 x+70 y+23Z=f(x, y)=3 x^3-5 y^2-225 x+70 y+23Z=f(x,y)=3x35y2225x+70y+23.
    (i) Find the stationary points of z z zzz.
    (ii) Determine if at these points the function is at a relative maximum, relative minimum, infixion point, or saddle point.
  2. Solve the following differential equation
d 2 y d x 2 2 d y d x + 10 y = 0 d 2 y d x 2 2 d y d x + 10 y = 0 (d^(2)y)/(dx^(2))-2(dy)/(dx)+10 y=0\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+10 y=0d2ydx22dydx+10y=0
given Y ( 0 ) = 4 Y ( 0 ) = 4 Y(0)=4Y(0)=4Y(0)=4
d y d x ( 0 ) = 1 d y d x ( 0 ) = 1 (dy)/(dx)(0)=1\frac{d y}{d x}(0)=1dydx(0)=1
  1. If Z = f ( x , y ) = x y Z = f ( x , y ) = x y Z=f(x,y)=xyZ=f(x, y)=x yZ=f(x,y)=xy
Find the maximum value for f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) if x x xxx and y y yyy are constrained to sum to 1 (That is, x + y = 1 x + y = 1 x+y=1x+y=1x+y=1 ). Solve the problem in two ways: by substitution and by using the Lagrangian multiplier method.
Assignment C
Answer the following Short Category questions in about 100 words each
6. Define
a. Adjugate of a matrix
b. Decomposable matrix
c. Singular matrix
7. Evaluate ( 7 x 2 ) 3 x + 2 d x ( 7 x 2 ) 3 x + 2 d x int(7x-2)sqrt(3x+2)dx\int(7 x-2) \sqrt{3 x+2} d x(7x2)3x+2dx
8. Explain the concept of maximum value function.
  1. Let the production function by Q = A L a K b Q = A L a K b Q=AL^(a)K^(b)Q=A L^a K^bQ=ALaKb. Find the elasticity of production with respect to labour (L).
  2. Denote by a , b a , b a,b\mathbf{a}, \mathbf{b}a,b and c c c\mathbf{c}c the column vectors
a = ( 1 2 3 ) , b = ( 2 1 3 ) , c = ( 2 1 1 ) a = 1 2 3 , b = 2 1 3 , c = 2 1 1 a=([1],[2],[3]),b=([-2],[1],[-3]),c=([-2],[-1],[1])\mathbf{a}=\left(\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right), \mathbf{b}=\left(\begin{array}{l} -2 \\ 1 \\ -3 \end{array}\right), \mathbf{c}=\left(\begin{array}{l} -2 \\ -1 \\ 1 \end{array}\right)a=(123),b=(213),c=(211)
Calculate
2 a 5 b , 2 a 5 b + c , a b 2 a 5 b , 2 a 5 b + c , a b 2a-5b,2a-5b+c,a^(‘)*b2 \mathbf{a}-5 \mathbf{b}, \mathbf{2 a}-5 \mathbf{b}+\mathbf{c}, \mathbf{a}^{\prime} \cdot \mathbf{b}2a5b,2a5b+c,ab

Expert Answer:


Question:-1

Consider the following two matrices
A = ( 1 1 2 1 1 2 2 2 4 ) A = 1 1 2 1 1 2 2 2 4 A=([-1,1,2],[1,-1,-2],[-2,2,4])A=\left(\begin{array}{ccc}-1 & 1 & 2 \\ 1 & -1 & -2 \\ -2 & 2 & 4\end{array}\right)A=(112112224)
B = ( 1 0 1 1 1 1 1 1 1 ) B = 1 0 1 1 1 1 1 1 1 B=([1,0,-1],[-1,1,1],[1,-1,-1])B=\left(\begin{array}{ccc}1 & 0 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & -1\end{array}\right)B=(101111111)
(i) Find the rank of ‘ A ‘ and ‘ B ‘
(ii) Show that ( A B ) 1 = B 1 A 1 ( A B ) 1 = B 1 A 1 (AB)^(-1)=B^(-1)A^(-1)(AB)^{-1}=B^{-1}A^{-1}(AB)1=B1A1
(iii) Show that ( A 1 ) 1 = A ( A 1 ) 1 = A (A^(-1))^(-1)=A(A^{-1})^{-1}=A(A1)1=A
(iv) Show that ( B 1 ) 1 = B ( B 1 ) 1 = B (B^(-1))^(-1)=B(B^{-1})^{-1}=B(B1)1=B

Answer:

(i) Rank of ‘A’
Rank [ 1 1 2 1 1 2 2 2 4 ] Rank 1 1 2 1 1 2 2 2 4 Rank[[-1,1,2],[1,-1,-2],[-2,2,4]]\operatorname{Rank}\left[\begin{array}{ccc} -1 & 1 & 2 \\ 1 & -1 & -2 \\ -2 & 2 & 4 \end{array}\right]Rank[112112224]
Now, reduce this matrix interchanging rows R 1 R 3 R 1 R 3 R_(1)harrR_(3)R_1 \leftrightarrow R_3R1R3
= [ 2 2 4 1 1 2 1 1 2 ] R 1 R 1 ÷ 2 = [ 1 1 2 1 1 2 1 1 2 ] R 2 R 2 R 1 = 2 2 4 1 1 2 1 1 2 R 1 R 1 ÷ 2 = 1 1 2 1 1 2 1 1 2 R 2 R 2 R 1 {:[=[[-2,2,4],[1,-1,-2],[-1,1,2]]],[R_(1)larrR_(1)-:-2],[=[[1,-1,-2],[1,-1,-2],[-1,1,2]]],[R_(2)larrR_(2)-R_(1)]:}\begin{aligned} & =\left[\begin{array}{ccc} -2 & 2 & 4 \\ 1 & -1 & -2 \\ -1 & 1 & 2 \end{array}\right] \\ & R_1 \leftarrow R_1 \div-2 \\ & =\left[\begin{array}{ccc} 1 & -1 & -2 \\ 1 & -1 & -2 \\ -1 & 1 & 2 \end{array}\right] \\ & R_2 \leftarrow R_2-R_1 \end{aligned}=[224112112]R1R1÷2=[112112112]R2R2R1
= [ 1 1 2 0 0 0 1 1 2 ] = 1 1 2 0 0 0 1 1 2 =[[1,-1,-2],[0,0,0],[-1,1,2]]=\left[\begin{array}{ccc} 1 & -1 & -2 \\ 0 & 0 & 0 \\ -1 & 1 & 2 \end{array}\right]=[112000112]
R 3 R 3 + R 1 = [ 1 1 2 0 0 0 0 0 0 ] R 3 R 3 + R 1 = 1 1 2 0 0 0 0 0 0 {:[R_(3)larrR_(3)+R_(1)],[=[[1,-1,-2],[0,0,0],[0,0,0]]]:}\begin{aligned} & R_3 \leftarrow R_3+R_1 \\ & =\left[\begin{array}{ccc} 1 & -1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{aligned}R3R3+R1=[112000000]
The rank of a matrix is the number of non all-zeros rows
:.\therefore Rank = 1 = 1 =1=1=1
(ii) Rank of ‘B’
Rank [ 1 0 1 1 1 1 1 1 1 ] Now, reduce this matrix R 2 R 2 + R 1 = [ 1 0 1 0 1 0 1 1 1 ] R 3 R 3 R 1 = [ 1 0 1 0 1 0 0 1 0 ] R 3 R 3 + R 2 Rank 1 0 1 1 1 1 1 1 1  Now, reduce this matrix  R 2 R 2 + R 1 = 1 0 1 0 1 0 1 1 1 R 3 R 3 R 1 = 1 0 1 0 1 0 0 1 0 R 3 R 3 + R 2 {:[Rank[[1,0,-1],[-1,1,1],[1,-1,-1]]],[” Now, reduce this matrix “],[{:[R_(2)larrR_(2)+R_(1)],[=[[1,0,-1],[0,1,0],[1,-1,-1]]]:}],[{:[R_(3)larrR_(3)-R_(1)],[=[[1,0,-1],[0,1,0],[0,-1,0]]]:}],[R_(3)larrR_(3)+R_(2)]:}\begin{aligned} &\operatorname{Rank}\left[\begin{array}{ccc} 1 & 0 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & -1 \end{array}\right]\\ &\text { Now, reduce this matrix }\\ &\begin{aligned} & R_2 \leftarrow R_2+R_1 \\ & =\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & -1 & -1 \end{array}\right] \end{aligned}\\ &\begin{aligned} & R_3 \leftarrow R_3-R_1 \\ & =\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & -1 & 0 \end{array}\right] \end{aligned}\\ &R_3 \leftarrow R_3+R_2 \end{aligned}Rank[101111111] Now, reduce this matrix R2R2+R1=[101010111]R3R3R1=[101010010]R3R3+R2
= [ 1 0 1 0 1 0 0 0 0 ] = 1 0 1 0 1 0 0 0 0 =[[1,0,-1],[0,1,0],[0,0,0]]=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right]=[101010000]
The rank of a matrix is the number of non all-zeros rows
Rank = 2  Rank  = 2 :.” Rank “=2\therefore \text { Rank }=2 Rank =2
(ii) Show that ( A B ) 1 = B 1 A 1 ( A B ) 1 = B 1 A 1 (AB)^(-1)=B^(-1)A^(-1)(AB)^{-1}=B^{-1}A^{-1}(AB)1=B1A1
A × B = [ 1 1 2 1 1 2 2 2 4 ] × [ 1 0 1 1 1 1 1 1 1 ] = [ 1 × 1 + 1 × 1 + 2 × 1 1 × 0 + 1 × 1 + 2 × 1 1 × 1 + 1 × 1 + 2 × 1 1 × 1 1 × 1 2 × 1 1 × 0 1 × 1 2 × 1 1 × 1 1 × 1 2 × 1 2 × 1 + 2 × 1 + 4 × 1 2 × 0 + 2 × 1 + 4 × 1 2 × 1 + 2 × 1 + 4 × 1 ] = [ 1 1 + 2 0 + 1 2 1 + 1 2 1 + 1 2 0 1 + 2 1 1 + 2 2 2 + 4 0 + 2 4 2 + 2 4 ] = [ 0 1 0 0 1 0 0 2 0 ] | A × B | = | 0 1 0 0 1 0 0 2 0 | A × B = [ 1 1 2 1 1 2 2 2 4 ] × [ 1 0 1 1 1 1 1 1 1 ] = [ 1 × 1 + 1 × 1 + 2 × 1 1 × 0 + 1 × 1 + 2 × 1 1 × 1 + 1 × 1 + 2 × 1 1 × 1 1 × 1 2 × 1 1 × 0 1 × 1 2 × 1 1 × 1 1 × 1 2 × 1 2 × 1 + 2 × 1 + 4 × 1 2 × 0 + 2 × 1 + 4 × 1 2 × 1 + 2 × 1 + 4 × 1 ] = [ 1 1 + 2 0 + 1 2 1 + 1 2 1 + 1 2 0 1 + 2 1 1 + 2 2 2 + 4 0 + 2 4 2 + 2 4 ] = [ 0 1 0 0 1 0 0 2 0 ] | A × B | = | 0 1 0 0 1 0 0 2 0 | A × B = 1 1 2 1 1 2 2 2 4 × 1 0 1 1 1 1 1 1 1 = 1 × 1 + 1 × 1 + 2 × 1 1 × 0 + 1 × 1 + 2 × 1 1 × 1 + 1 × 1 + 2 × 1 1 × 1 1 × 1 2 × 1 1 × 0 1 × 1 2 × 1 1 × 1 1 × 1 2 × 1 2 × 1 + 2 × 1 + 4 × 1 2 × 0 + 2 × 1 + 4 × 1 2 × 1 + 2 × 1 + 4 × 1 = 1 1 + 2 0 + 1 2 1 + 1 2 1 + 1 2 0 1 + 2 1 1 + 2 2 2 + 4 0 + 2 4 2 + 2 4 = 0 1 0 0 1 0 0 2 0 | A × B | = 0 1 0 0 1 0 0 2 0 A × B = 1 1 2 1 1 2 2 2 4 × 1 0 1 1 1 1 1 1 1 = 1 × 1 + 1 × 1 + 2 × 1 1 × 0 + 1 × 1 + 2 × 1 1 × 1 + 1 × 1 + 2 × 1 1 × 1 1 × 1 2 × 1 1 × 0 1 × 1 2 × 1 1 × 1 1 × 1 2 × 1 2 × 1 + 2 × 1 + 4 × 1 2 × 0 + 2 × 1 + 4 × 1 2 × 1 + 2 × 1 + 4 × 1 = 1 1 + 2 0 + 1 2 1 + 1 2 1 + 1 2 0 1 + 2 1 1 + 2 2 2 + 4 0 + 2 4 2 + 2 4 = 0 1 0 0 1 0 0 2 0 | A × B | = 0 1 0 0 1 0 0 2 0 {:[{:[A xx B=[[-1,1,2],[1,-1,-2],[-2,2,4]]xx[[1,0,-1],[-1,1,1],[1,-1,-1]]],[=[[-1xx1+1xx-1+2xx1,-1xx0+1xx1+2xx-1,-1xx-1+1xx1+2xx-1],[1xx1-1xx-1-2xx1,1xx0-1xx1-2xx-1,1xx-1-1xx1-2xx-1],[-2xx1+2xx-1+4xx1,-2xx0+2xx1+4xx-1,-2xx-1+2xx1+4xx-1]]],[=[[-1-1+2,0+1-2,1+1-2],[1+1-2,0-1+2,-1-1+2],[-2-2+4,0+2-4,2+2-4]]],[=[[0,-1,0],[0,1,0],[0,-2,0]]],[|A xx B|=|[0,-1,0],[0,1,0],[0,-2,0]|]:}],[A xx B=[[-1,1,2],[1,-1,-2],[-2,2,4]]xx[[1,0,-1],[-1,1,1],[1,-1,-1]]],[=[[-1xx1+1xx-1+2xx1,-1xx0+1xx1+2xx-1,-1xx-1+1xx1+2xx-1],[1xx1-1xx-1-2xx1,1xx0-1xx1-2xx-1,1xx-1-1xx1-2xx-1],[-2xx1+2xx-1+4xx1,-2xx0+2xx1+4xx-1,-2xx-1+2xx1+4xx-1]]],[=[[-1-1+2,0+1-2,1+1-2],[1+1-2,0-1+2,-1-1+2],[-2-2+4,0+2-4,2+2-4]]],[=[[0,-1,0],[0,1,0],[0,-2,0]]],[|A xx B|=|[0,-1,0],[0,1,0],[0,-2,0]|]:}\begin{aligned} &\begin{aligned} & A \times B=\left[\begin{array}{ccc} -1 & 1 & 2 \\ 1 & -1 & -2 \\ -2 & 2 & 4 \end{array}\right] \times\left[\begin{array}{ccc} 1 & 0 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & -1 \end{array}\right] \\ & =\left[\begin{array}{ccc} -1 \times 1+1 \times-1+2 \times 1 & -1 \times 0+1 \times 1+2 \times-1 & -1 \times-1+1 \times 1+2 \times-1 \\ 1 \times 1-1 \times-1-2 \times 1 & 1 \times 0-1 \times 1-2 \times-1 & 1 \times-1-1 \times 1-2 \times-1 \\ -2 \times 1+2 \times-1+4 \times 1 & -2 \times 0+2 \times 1+4 \times-1 & -2 \times-1+2 \times 1+4 \times-1 \end{array}\right] \\ & =\left[\begin{array}{ccc} -1-1+2 & 0+1-2 & 1+1-2 \\ 1+1-2 & 0-1+2 & -1-1+2 \\ -2-2+4 & 0+2-4 & 2+2-4 \end{array}\right] \\ & =\left[\begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 0 \end{array}\right] \\ & |A \times B|=\left|\begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 0 \end{array}\right| \end{aligned}\\ &A \times B=\left[\begin{array}{ccc} -1 & 1 & 2 \\ 1 & -1 & -2 \\ -2 & 2 & 4 \end{array}\right] \times\left[\begin{array}{ccc} 1 & 0 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & -1 \end{array}\right]\\ &=\left[\begin{array}{ccc} -1 \times 1+1 \times-1+2 \times 1 & -1 \times 0+1 \times 1+2 \times-1 & -1 \times-1+1 \times 1+2 \times-1 \\ 1 \times 1-1 \times-1-2 \times 1 & 1 \times 0-1 \times 1-2 \times-1 & 1 \times-1-1 \times 1-2 \times-1 \\ -2 \times 1+2 \times-1+4 \times 1 & -2 \times 0+2 \times 1+4 \times-1 & -2 \times-1+2 \times 1+4 \times-1 \end{array}\right]\\ &=\left[\begin{array}{ccc} -1-1+2 & 0+1-2 & 1+1-2 \\ 1+1-2 & 0-1+2 & -1-1+2 \\ -2-2+4 & 0+2-4 & 2+2-4 \end{array}\right]\\ &=\left[\begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 0 \end{array}\right]\\ &|A \times B|=\left|\begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 0 \end{array}\right| \end{aligned}A×B=[112112224]×[101111111]=[1×1+1×1+2×11×0+1×1+2×11×1+1×1+2×11×11×12×11×01×12×11×11×12×12×1+2×1+4×12×0+2×1+4×12×1+2×1+4×1]=[11+20+121+121+1201+211+222+40+242+24]=[010010020]|A×B|=|010010020|A×B=[112112224]×[101111111]=[1×1+1×1+2×11×0+1×1+2×11×1+1×1+2×11×11×12×11×01×12×11×11×12×12×1+2×1+4×12×0+2×1+4×12×1+2×1+4×1]=[11+20+121+121+1201+211+222+40+242+24]=[010010020]|A×B|=|010010020|
| A × B | = | 0 1 0 0 1 0 0 2 0 | = 0 × | 1 0 2 0 | + 1 × | 0 0 0 0 | + 0 × | 0 1 0 2 | = 0 × ( 1 × 0 0 × ( 2 ) ) + 1 × ( 0 × 0 0 × 0 ) + 0 × ( 0 × ( 2 ) 1 × 0 ) = 0 × ( 0 + 0 ) + 1 × ( 0 + 0 ) + 0 × ( 0 + 0 ) = 0 × ( 0 ) + 1 × ( 0 ) + 0 × ( 0 ) = 0 + 0 + 0 = 0 Here, | A × B | = 0 , So ( A × B ) 1 is not possible. | A × B | = 0 1 0 0 1 0 0 2 0 = 0 × 1 0 2 0 + 1 × 0 0 0 0 + 0 × 0 1 0 2 = 0 × ( 1 × 0 0 × ( 2 ) ) + 1 × ( 0 × 0 0 × 0 ) + 0 × ( 0 × ( 2 ) 1 × 0 ) = 0 × ( 0 + 0 ) + 1 × ( 0 + 0 ) + 0 × ( 0 + 0 ) = 0 × ( 0 ) + 1 × ( 0 ) + 0 × ( 0 ) = 0 + 0 + 0 = 0  Here,  | A × B | = 0 , So  ( A × B ) 1  is not possible.  {:[{:[|A xx B|=|[0,-1,0],[0,1,0],[0,-2,0]|],[=0xx|[1,0],[-2,0]|+1xx|[0,0],[0,0]|+0xx|[0,1],[0,-2]|],[=0xx(1xx0-0xx(-2))+1xx(0xx0-0xx0)+0xx(0xx(-2)-1xx0)],[=0xx(0+0)+1xx(0+0)+0xx(0+0)],[=0xx(0)+1xx(0)+0xx(0)],[=0+0+0],[=0]:}],[” Here, “|A xx B|=0”, So “(A xx B)^(-1)” is not possible. “]:}\begin{aligned} &\begin{aligned} & |A \times B|=\left|\begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 0 \end{array}\right| \\ & =0 \times\left|\begin{array}{cc} 1 & 0 \\ -2 & 0 \end{array}\right|+1 \times\left|\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right|+0 \times\left|\begin{array}{cc} 0 & 1 \\ 0 & -2 \end{array}\right| \\ & =0 \times(1 \times 0-0 \times(-2))+1 \times(0 \times 0-0 \times 0)+0 \times(0 \times(-2)-1 \times 0) \\ & =0 \times(0+0)+1 \times(0+0)+0 \times(0+0) \\ & =0 \times(0)+1 \times(0)+0 \times(0) \\ & =0+0+0 \\ & =0 \end{aligned}\\ &\text { Here, }|A \times B|=0 \text {, So }(A \times B)^{-1} \text { is not possible. } \end{aligned}|A×B|=|010010020|=0×|1020|+1×|0000|+0×|0102|=0×(1×00×(2))+1×(0×00×0)+0×(0×(2)1×0)=0×(0+0)+1×(0+0)+0×(0+0)=0×(0)+1×(0)+0×(0)=0+0+0=0 Here, |A×B|=0, So (A×B)1 is not possible. 
(iii) Show that ( A 1 ) 1 = A ( A 1 ) 1 = A (A^(-1))^(-1)=A(A^{-1})^{-1}=A(A1)1=A
| A | = | 1 1 2 1 1 2 2 2 4 | = 1 × | 1 2 2 4 | 1 × | 1 2 2 4 | + 2 × | 1 1 2 2 | = 1 × ( 1 × 4 ( 2 ) × 2 ) 1 × ( 1 × 4 ( 2 ) × ( 2 ) ) + 2 × ( 1 × 2 ( 1 ) × ( 2 ) ) = 1 × ( 4 + 4 ) 1 × ( 4 4 ) + 2 × ( 2 2 ) = 1 × ( 0 ) 1 × ( 0 ) + 2 × ( 0 ) = 0 + 0 + 0 = 0 Here, | A | = 0 , So ( A ) 1 is not possible. | A | = 1 1 2 1 1 2 2 2 4 = 1 × 1 2 2 4 1 × 1 2 2 4 + 2 × 1 1 2 2 = 1 × ( 1 × 4 ( 2 ) × 2 ) 1 × ( 1 × 4 ( 2 ) × ( 2 ) ) + 2 × ( 1 × 2 ( 1 ) × ( 2 ) ) = 1 × ( 4 + 4 ) 1 × ( 4 4 ) + 2 × ( 2 2 ) = 1 × ( 0 ) 1 × ( 0 ) + 2 × ( 0 ) = 0 + 0 + 0 = 0  Here,  | A | = 0 , So  ( A ) 1  is not possible.  {:[{:[|A|=|[-1,1,2],[1,-1,-2],[-2,2,4]|],[=-1xx|[-1,-2],[2,4]|-1xx|[1,-2],[-2,4]|+2xx|[1,-1],[-2,2]|],[=-1xx(-1xx4-(-2)xx2)-1xx(1xx4-(-2)xx(-2))+2xx(1xx2-(-1)xx(-2))],[=-1xx(-4+4)-1xx(4-4)+2xx(2-2)],[=-1xx(0)-1xx(0)+2xx(0)],[=0+0+0],[=0]:}],[” Here, “|A|=0”, So “(A)^(-1)” is not possible. “]:}\begin{aligned} &\begin{aligned} & |A|=\left|\begin{array}{ccc} -1 & 1 & 2 \\ 1 & -1 & -2 \\ -2 & 2 & 4 \end{array}\right| \\ & =-1 \times\left|\begin{array}{cc} -1 & -2 \\ 2 & 4 \end{array}\right|-1 \times\left|\begin{array}{cc} 1 & -2 \\ -2 & 4 \end{array}\right|+2 \times\left|\begin{array}{cc} 1 & -1 \\ -2 & 2 \end{array}\right| \\ & =-1 \times(-1 \times 4-(-2) \times 2)-1 \times(1 \times 4-(-2) \times(-2))+2 \times(1 \times 2-(-1) \times(-2)) \\ & =-1 \times(-4+4)-1 \times(4-4)+2 \times(2-2) \\ & =-1 \times(0)-1 \times(0)+2 \times(0) \\ & =0+0+0 \\ & =0 \end{aligned}\\ &\text { Here, }|A|=0 \text {, So }(A)^{-1} \text { is not possible. } \end{aligned}|A|=|1121