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} 2 0 marks. Word limit will not apply in the case of numerical questions.
2
×
20
=
40
2
×
20
=
40
2xx20=40 2 \times 20=40 2 × 20 = 40
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 = ( − 1 1 2 1 − 1 − 2 − 2 2 4 )
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 = ( 1 0 − 1 − 1 1 1 1 − 1 − 1 )
(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 = B − 1 A − 1
(iii) Show that
(
A
−
1
)
−
1
=
A
A
−
1
−
1
=
A
(A^(-1))^(-1)=A \left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A} ( A − 1 ) − 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} ( B − 1 ) − 1 = B
An individual consumer consumes two commodities
X
1
&
X
2
X
1
&
X
2
X_(1)&X_(2) X_1 \& X_2 X 1 & X 2 . 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 = X 1 0.4 X 2 0.6
The price of commodity one is
P
1
=
P
1
=
P_(1)= \mathrm{P}_1= P 1 = Rs. 3.00 , the price of commodity two is
P
2
=
P
2
=
P_(2)= \mathrm{P}_2= P 2 = 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}_2 X 1 & X 2 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} 1 0 marks. Word limit will not apply in the case of numerical questions.
3
×
10
=
30
3
×
10
=
30
3xx10=30 3 \times 10=30 3 × 10 = 30
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+23 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 .
(i) Find the stationary points of
z
z
z z z .
(ii) Determine if at these points the function is at a relative maximum, relative minimum, infixion point, or saddle point.
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=0 d 2 y d x 2 − 2 d y d x + 10 y = 0
given
Y
(
0
)
=
4
Y
(
0
)
=
4
Y(0)=4 Y(0)=4 Y ( 0 ) = 4
d
y
d
x
(
0
)
=
1
d
y
d
x
(
0
)
=
1
(dy)/(dx)(0)=1 \frac{d y}{d x}(0)=1 d y d x ( 0 ) = 1
If
Z
=
f
(
x
,
y
)
=
x
y
Z
=
f
(
x
,
y
)
=
x
y
Z=f(x,y)=xy Z=f(x, y)=x y Z = f ( x , y ) = x y
Find the maximum value for
f
(
x
,
y
)
f
(
x
,
y
)
f(x,y) f(x, y) f ( x , y ) if
x
x
x x x and
y
y
y y y are constrained to sum to 1 (That is,
x
+
y
=
1
x
+
y
=
1
x+y=1 x+y=1 x + 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 ∫ ( 7 x − 2 ) 3 x + 2 d x
8. Explain the concept of maximum value function.
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^b Q = A L a K b . Find the elasticity of production with respect to labour (L).
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 = ( 1 2 3 ) , b = ( − 2 1 − 3 ) , c = ( − 2 − 1 1 )
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^(‘)*b 2 \mathbf{a}-5 \mathbf{b}, \mathbf{2 a}-5 \mathbf{b}+\mathbf{c}, \mathbf{a}^{\prime} \cdot \mathbf{b} 2 a − 5 b , 2 a − 5 b + c , a ′ ⋅ b
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 = ( − 1 1 2 1 − 1 − 2 − 2 2 4 )
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 = ( 1 0 − 1 − 1 1 1 1 − 1 − 1 )
(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} ( A B ) − 1 = B − 1 A − 1
(iii) Show that
(
A
−
1
)
−
1
=
A
(
A
−
1
)
−
1
=
A
(A^(-1))^(-1)=A (A^{-1})^{-1}=A ( A − 1 ) − 1 = A
(iv) Show that
(
B
−
1
)
−
1
=
B
(
B
−
1
)
−
1
=
B
(B^(-1))^(-1)=B (B^{-1})^{-1}=B ( B − 1 ) − 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 [ − 1 1 2 1 − 1 − 2 − 2 2 4 ]
Now, reduce this matrix interchanging rows
R
1
↔
R
3
R
1
↔
R
3
R_(1)harrR_(3) R_1 \leftrightarrow R_3 R 1 ↔ R 3
=
[
−
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} = [ − 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
=
[
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] = [ 1 − 1 − 2 0 0 0 − 1 1 2 ]
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} R 3 ← R 3 + R 1 = [ 1 − 1 − 2 0 0 0 0 0 0 ]
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 [ 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
=
[
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] = [ 1 0 − 1 0 1 0 0 0 0 ]
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} ( A B ) − 1 = B − 1 A − 1
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 = [ − 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
|
=
|
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 | = | 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.
(iii) Show that
(
A
−
1
)
−
1
=
A
(
A
−
1
)
−
1
=
A
(A^(-1))^(-1)=A (A^{-1})^{-1}=A ( A − 1 ) − 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 | = | − 1 1 2 1 −