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Department of Aerospace Engineering
AE602 Mathematics for Aerospace Engineers
Assignment No. 4
4.1 Decide whether or not the following vectors are linearly independent, by solving
𝑐1𝑣1 + 𝑐2𝑣2 + 𝑐3𝑣3 + 𝑐4𝑣4 = 0:
𝑣1 =
1100
, 𝑣2 =
1010
, 𝑣3 =
0011
, 𝑣4 =
0101
.
Decide also if they span 𝑹𝟒, by trying to solve 𝑐1𝑣1 + ⋯ + 𝑐4𝑣4 = 0, 0, 0, 1 .
SOLUTION:
To check for dependency or independency, we use
𝑐1𝑣1 + 𝑐2𝑣2 + 𝑐3𝑣3 + 𝑐4𝑣4 = 0 .............(1)
If 𝒄𝟏 = 𝒄𝟐 = 𝒄𝟑 = 𝒄𝟒 = 𝟎 is the only way to solve the above equation, then the
vectors are linearly independent. If any of the 𝒄𝒊 ≠ 0, then the given vectors are linearly
dependent. Solving the equation (1)
𝑐1
1100
+ 𝑐2
1010
+ 𝑐3
0011
+ 𝑐4
0101
= 0
we get four equations with four unknowns as follows
𝑐1 + 𝑐2 = 0
𝑐1 + 𝑐4 = 0
𝑐2 + 𝑐3 = 0
𝑐3 + 𝑐4 = 0
From the above set of equations we get 𝒄𝟐 = 𝒄𝟒 = −𝒄𝟑 = −𝒄𝟏
Hence with 𝒄𝟏 = 𝟏; 𝒄𝟐 = −𝟏; 𝒄𝟑 = 𝟏; 𝒄𝟒 = −𝟏 , the equation (1) goes to zero.
That is 𝑐1𝑣1 + 𝑐2𝑣2 + 𝑐3𝑣3 + 𝑐4𝑣4 = 1
1100
+ −1
1010
+ 1
0011
+ −1
0101
= 0
𝑁𝑜𝑡𝑒 𝑡𝑎𝑡 𝑐1, 𝑐2, 𝑐3, 𝑐4 can take any other values also.
Therefore the given vectors 𝒗𝟏, 𝒗𝟐, 𝒗𝟑, 𝒗𝟒 are linearly independent.
2
Now trying to solve
𝑐1𝑣1 + ⋯ + 𝑐4𝑣4 = 0, 0, 0, 1
We have
𝑐1 + 𝑐2 = 0.....(2)
𝑐1 + 𝑐4 = 0.....(3)
𝑐2 + 𝑐3 = 0.....(4)
𝑐3 + 𝑐4 = 1.....(5)
From (2) and (3), we get
𝑐2 = −𝑐1 𝑎𝑛𝑑 𝑐2 = −𝑐3 𝑎𝑛𝑑 𝑒𝑛𝑐𝑒 𝑐1 = 𝑐3
Substituting 𝑐1 = 𝑐3 in equation (3) , we get
𝑐3 + 𝑐4 = 0.....(6)
Equations (5) and (6) contradict each other and hence the vectors do not span 𝑹𝟒 .
4.2 Decide the dependence or independence of
(a) 1, 1, 2 , 1, 2, 1 , 3, 1,1 ;
SOLUTION:
Equation to be solved is,
𝑐1𝑣1 + 𝑐2𝑣2 + 𝑐3𝑣3 = 0
𝑐1 112 + 𝑐2
121 + 𝑐3
311 = 0
This is like solving Ac = 0
Solving Ac = 0 is equivalent to solving Uc = 0 where "U" is the upper triangular matrix form of
"A"
Let A = 1 1 31 2 12 1 1
R2→R2 - R1 ; R3→R3 - 2*R1
3
→ 1 1 30 1 −20 −1 −5
R3→R3 + R2
→ 1 1 30 1 −20 0 −7
= U
Now solving Uc = 0
𝑐1 100 + 𝑐2
110 + 𝑐3
3−2−7
= 0
-7. 𝑐3 = 0
𝒄𝟑 = 𝟎
𝑐2 − 2𝑐3 = 0
𝒄𝟐 = 𝟎
𝑐1 + 𝑐2 + 3𝑐3 = 0
𝒄𝟏 = 𝟎
𝒄𝟏 = 𝒄𝟐 = 𝒄𝟑 = 𝟎 and hence the given vectors are independent.
(b) 𝑣1 − 𝑣2 , 𝑣2−𝑣3, 𝑣3 − 𝑣4, 𝑣4 − 𝑣1 for any vectors 𝑣1 , 𝑣2,𝑣3, 𝑣4;
SOLUTION:
Solving the equation,
𝑐1 𝑣1 − 𝑣2 + 𝑐2 𝑣2−𝑣3 ) + 𝑐3(𝑣3 − 𝑣4 + 𝑐4( 𝑣4 − 𝑣1) = 0.........(1)
If 𝒄𝟏 = 𝒄𝟐 = 𝒄𝟑 = 𝒄𝟒 = 𝟏, then equation (1) gets satisfied
That is 𝑣1 − 𝑣2 + 𝑣2−𝑣3 + 𝑣3 − 𝑣4 + 𝑣4 − 𝑣1 = 0
Hence the vectors 𝒗𝟏 − 𝒗𝟐, 𝒗𝟐−𝒗𝟑, 𝒗𝟑 − 𝒗𝟒, 𝒗𝟒 − 𝒗𝟏 are linearly dependent for any vectors
𝒗𝟏, 𝒗𝟐,𝒗𝟑, 𝒗𝟒
4
(c) 1, 1, 0 , 1, 0, 0 , 0, 1,1 , 𝑥, 𝑦, 𝑧 , for any numbers 𝑥, 𝑦, 𝑧.
Solution:
To find dependency or independency we have to solve
𝑐1 110 + 𝑐2
100 + 𝑐3
011 + 𝑐4
𝑥𝑦𝑧 = 0
𝑐1 + 𝑐2 + 𝑐4𝑥 = 0.........(1)
𝑐1 + 𝑐3 + 𝑐4𝑦 = 0.........(2)
𝑐3 + 𝑐4𝑧 = 0.........(3)
Case 1: When x=y=z=0.
Then from equations (1), (2) and (3), we have 𝒄𝟏 = 𝒄𝟐 = 𝒄𝟑 = 𝟎 but 𝒄𝟒 can assume
any value and hence the vectors are not linearly independent.
Case 2: When either x or y or z not equal to zero (or) x, y & z all are not equal to zero
From equations (1), (2) and (3),
𝑧 =−𝑐3
𝑐4 .........(4)
𝑦 =−(𝑐1+𝑐3)
𝑐4 ...(5)
𝑥 =−(𝑐1+𝑐2)
𝑐4 ....(6)
If either x or y or z is not equal to zero, then for x or y or z to have a finite value (which is not
equal to zero), 𝒄𝟒 must not be equal to zero. If it is zero then the value tends to infinity.
Hence for the case 2 also the given vectors are not linearly dependent.
Hence the given set of vectors are always linearly dependent for any existing non-infinite
𝒙, 𝒚, 𝒛.
5
4.3 Prove that if any diagonal element of
𝑇 = 𝑎 𝑏 𝑐0 𝑑 𝑒0 0 𝑓
is zero, the rows are linearly dependent.
SOLUTION:
The equation to be solved for determining dependency or independency is,
𝑐1 𝑎𝑏𝑐 + 𝑐2
0𝑑𝑒 + 𝑐3
00𝑓 = 0
𝑐1𝑎 = 0.........(1)
𝑐1𝑏 + 𝑐2𝑑 = 0.........(2)
𝑐1𝑐 + 𝑐2𝑒 + 𝑐3𝑓 = 0.........(3)
CASE 1: a = 0, d ≠ 0, f ≠ 0
From equation (1),
If a = 0, 𝒄𝟏 need not be zero (it can assume any value). Hence the
rows are linearly dependent.
CASE 2: a ≠ 0, d = 0, f ≠ 0
From (1), we have 𝒄𝟏 = 𝟎
From (2), we have 𝑐2𝑑 = 0
Now since d = 0, 𝒄𝟐 need not be zero which proves that the rows are again linearly
dependent.
CASE 3: a ≠ 0, d ≠ 0, f = 0
From (1), we have 𝒄𝟏 = 𝟎
From (2), since d ≠ 0 , we have 𝒄𝟐 = 𝟎
From (3), since f = 0, 𝒄𝟑 can assume any value.
Hence for the present case the rows are again linearly dependent.
6
4.4 Is it true that if 𝑣1 , 𝑣2 ,𝑣3 are linearly dependent, then also the vectors 𝑤1 = 𝑣1 + 𝑣2, 𝑤2 =
𝑣1 + 𝑣3, 𝑤3 = 𝑣2 + 𝑣3, are linearly independent? (Hint: Assume some combination 𝑐1𝑤1 +𝑐2𝑤2 + 𝑐3𝑤3 = 0, and find which 𝑐𝑖 are possible.)
SOLUTION:
If 𝑣1, 𝑣2 ,𝑣3 are linearly dependent, then in the equation
𝑐1𝑣1 + 𝑐2𝑣 + 𝑐3𝑣3 = 0
there is a 𝒄𝒊 which is not equal to zero...........(1)
To check the dependency of vectors 𝑤1 = 𝑣1 + 𝑣2, 𝑤2 = 𝑣1 + 𝑣3, 𝑤3 = 𝑣2 + 𝑣3 , we consider
the equation
𝑐1𝑤1 + 𝑐2𝑤2 + 𝑐3𝑤3 = 0
𝑐1(𝑣1 + 𝑣2) + 𝑐2(𝑣1 + 𝑣3) + 𝑐3(𝑣2 + 𝑣3) = 0
(𝑐1 + 𝑐2)𝑣1 + (𝑐1 + 𝑐3)𝑣2 + (𝑐2 + 𝑐3)𝑣3 = 0
𝐶1𝑣1 + 𝐶2𝑣 + 𝐶3𝑣3 = 0
From (1), if there is a non-zero 𝒄𝒊 , then there should definitely be a non-zero 𝑪𝒊 since 𝑪𝒊 is
composed of 𝒄𝒊 . If there is such a non-zero 𝑪𝒊 , then vectors 𝒘𝟏, 𝒘𝟐, 𝒘𝟑 are also linearly
dependent.
Hence it is not true that if 𝑣1, 𝑣2 ,𝑣3 are linearly dependent, then the vectors 𝑤1 = 𝑣1 + 𝑣2,
𝑤2 = 𝑣1 + 𝑣3, 𝑤3 = 𝑣2 + 𝑣3, are linearly independent.
4.5 Describe geometrically the subspace of 𝑹𝟑 spanned by
(a) 0, 0, 0 , 0, 1, 0 , 0, 2,0 ;
SOLUTION:
Here the x and z components are zero and only the y component is present. Hence the given
vectors span a line (y-axis) in 𝑹𝟑.
(b) 0, 0, 1 , 0, 1, 1 , 0, 2,1 ;
SOLUTION:
Here x component is zero but y and z components are present. Hence these vectors span the yz
-plane in 𝑹𝟑.
7
(c) all six of these vectors. Which two form a basis?
SOLUTION:
In all the six vectors, the x-component is zero. Only the y and z components are present. Hence
these six vectors span the yz -plane in 𝑹𝟑.
The vectors (0,1,0) and (0,0,1) form the basis because the other vectors in the column space
can be expressed as a linear combination of these two vectors.
For example 2 010 + 1
001 =
021
(d) all vectors with positive components.
SOLUTION:
The vectors with positive components can be multiplied by any scalar (positive or negative)
so that they span the whole of 𝑹𝟑.
4.6 To decide whether 𝑏 is in the subspace spanned by 𝑤1,… , 𝑤𝑙 , let the vectors 𝑤 be the
columns of 𝐴 and try to solve 𝐴𝑥 = 𝑏 . What is it the result for
(a) 𝑤1 = 1, 1, 0 , 𝑤2 = 2, 2, 1 , 𝑤3 = 0, 0, 2 , 𝑏 = 3, 4, 5 ;
SOLUTION:
For the above system 𝑨𝒙 = 𝒃 to have a solution, the vector "b" must lie in the column
space of "A". Otherwise the system will not have a solution. "A" matrix is formed by the
vectors "w" as follows
A = 1 2 01 2 00 1 2
Hence 𝑨𝒙 = 𝒃 is 1 2 01 2 00 1 2
𝑥1
𝑥2
𝑥3
= 345
Applying Gauss elimination to the "A" matrix
R2 → R2-R1
→ 1 2 00 0 00 1 2
8
Interchanging rows 2 and 3
→ 1 2 00 1 20 0 0
= U
Here the third pivot is equal to zero and hence the system does not have a solution. As already
stated if there is no solution then it means that the vector "b" does not lie in the column space of
"A". Hence 𝒃 is not in the subspace spanned by 𝒘𝟏, 𝒘𝟐, 𝒘𝟑.
(b) 𝑤1 = 1, 2, 0 , 𝑤2 = 2, 5, 0 , 𝑤3 = 0, 0, 2 , 𝑤4 = 0, 0, 0 , and any 𝑏?
SOLUTION:
Ax = 120
2 0 05 0 00 2 0
𝑥1
𝑥2
𝑥3
𝑥4
=
𝑏1
𝑏2
𝑏3
Applying Gauss elimination
R2 → R2-2*R1
→ 100
2 0 01 0 00 2 0
𝑥1
𝑥2
𝑥3
𝑥4
=
𝑏1
𝑏2
𝑏3
− 2𝑏1
There are pivots in column 1, 2 and 3. Hence 𝒙𝟏, 𝒙𝟐𝒂𝒏𝒅 𝒙𝟑 are basic variables whereas 𝒙𝟒
is a free variable.
2𝑥3 = 𝑏3
𝒙𝟑 = 𝒃𝟑/𝟐
𝒙𝟐 = 𝒃𝟐 − 𝟐𝒃𝟏
𝑥1 + 2𝑥2 = 𝑏1
𝒙𝟏 = 𝟓𝒃𝟏 − 𝟐𝒃𝟐
9
𝑥1
𝑥2
𝑥3
𝑥4
=
5𝑏1 − 2𝑏2
𝑏2 − 2𝑏1
𝑏3/20
Whatever be the values of 𝑏1, 𝑏2, 𝑏3, there is definitely going to be a solution available as we can
clearly see from the above equation. Hence definitely vector "b" is in the subspace spanned
by vectors 𝒘𝟏, 𝒘𝟐, 𝒘𝟑, 𝒘𝟒 (which is the column space in 𝑹𝟑).
4.7 By locating the pivots, find a basis for the column space of
𝑈 =
0000
1000
4200
3200
Express each column that is not the basis as a combination of the basic columns. Find also a
matrix 𝐴 with this echelon form 𝑈, but a different column space.
SOLUTION:
Given
𝑈 =
0000
1000
4200
3200
which is in the echelon form. Here the columns 2 and 3 contain the pivots (1 and 2) respectively.
Hence columns 2 and 3 are the basis for the column space of "A".
Basis for column space are
1000
and
4200
....................(1)
Now columns that are not the basis can be expressed as a linear combination of the basis
columns as follows:
𝒗𝟒 = 𝒗𝟑 − 𝒗𝟐
That is
3200
=
4200
−
1000
Also 𝒗𝟏 = 𝟎𝒗𝟑 − 𝟎𝒗𝟐
By looking into "U" we can see that A =
𝟎𝟎𝟎𝟎
𝟏𝟎𝟎𝟐
𝟒𝟐𝟒𝟖
𝟑𝟐𝟒𝟔
for which when we apply the Gaussian
10
elimination, we get back the given "U" matrix. Again here columns 2 and 3 are the basis
for the column space as the remaining two columns (1 and 4) can be expressed as a linear
combination of the basis columns.
Now the basis for column space are
1002
and
4248
which are different from the basis obtained in
(1). Hence the above matrix 𝐴 is a matrix with this echelon form 𝑈, but a different column space.
4.8 Find the dimension and construct a basis for the four subspaces associated with each of the
matrices
𝐴 = 0 1 40 2 8
00 and 𝑈 =
0 1 40 0 0
00 .
SOLUTION :
First let us consider the matrix 𝐴 = 0 1 40 2 8
00
Here m=2; n=4; r=1 (where r is the rank of the matrix A)
Column space R(A):
Dimension of R(A) is "r" which is equal to 1.
Basis for R(A) is 12 because in the echelon form of "A", only the second
column contains the pivot (1 is the pivot).
Row space R(𝐴𝑇):
Dimension of R(𝐴𝑇) is "r" which is equal to 1.
Basis for R(𝐴𝑇) is 0 1 4 0 because it is the only non-zero row in
the echelon form of "A".
Null space N(A):
Dimension of N(A) is "n-r" which is equal to 3.
Basis for null space are found by solving Ax=0 where the solution vector
"x" forms the null space of "A".
Solving Ax = 0 is the same as solving Ux = 0
0 1 40 0 0
00
𝑥1
𝑥2
𝑥3
𝑥4
= 00
11
𝑥2 + 4𝑥3 = 0
𝑥2 = −4𝑥3
𝑥1
𝑥2
𝑥3
𝑥4
=
𝑥1
−4𝑥3
𝑥3
𝑥4
where 𝒙𝟏, 𝒙𝟑, 𝒙𝟒 are the free variables
When 𝒙𝟏 = 𝟏, 𝒙𝟑 = 𝟎, 𝒙𝟒 = 𝟎
𝑥1
𝑥2
𝑥3
𝑥4
=
1000
When 𝒙𝟏 = 𝟎, 𝒙𝟑 = 𝟎, 𝒙𝟒 = 𝟏
𝑥1
𝑥2
𝑥3
𝑥4
=
0001
When 𝒙𝟏 = 𝟎, 𝒙𝟑 = 𝟏, 𝒙𝟒 = 𝟎
𝑥1
𝑥2
𝑥3
𝑥4
=
0−410
Basis of null space of "A" are
1000
,
0001
and
0−410
Left Null space N(𝐴𝑇):
Dimension of N(𝐴𝑇) is "m-r" which is equal to 1
For finding the left null space of "A", we solve 𝑨𝑻𝒚 = 𝟎
0140
0280
𝑦1
𝑦2 =
0000
𝒚𝟏 = −𝟐𝒚𝟐
12
Here 𝒚𝟐 is the free variable because in the echelon form of "𝑨𝑻", only the first column
contains a pivot.
U =
1000
2000
is the echelon form
When 𝒚𝟐 = 𝟏, 𝒚𝟏 = −𝟐
Therefore basis for left null space is −21
Now considering the matrix 𝑈 = 0 1 40 0 0
00
Column space:
Dimension is 1.
Basis is the column containing the pivots which is 10
Row space:
Dimension is 1.
Basis is the non-zero row which is 0 1 4 0
Null space:
Dimension is 3.
Basis are
1000
,
0001
and
0−410
Left Null space:
Dimension is 1.
Basis is obtained by solving 𝑼𝑻𝒚 = 𝟎
0140
0000
𝑦1
𝑦2 =
0000
Here again 𝒚𝟐 is the free variable because in the echelon form of "𝑼𝑻", only the first column
contains a pivot.
13
When 𝒚𝟐 = 𝟏, basis of left null space is 01
4.9 Find the dimension and a basis for the four fundamental subspaces for both
𝐴 = 1 2 00 1 11 2 0
101 and 𝑈 =
1 2 00 1 10 0 0
100 .
SOLUTION:
First considering the matrix 𝐴 = 1 2 00 1 11 2 0
101
Here m=3; n=4; r=2 (since in the echelon form, the last row becomes zero
and there are two non-zero rows)
Column space:
Dimension is 2.
Basis are 101 𝑎𝑛𝑑
212 since these are the columns which contains pivots in
"U" matrix.
Row space:
Dimension is 2.
Basis are 1 2 0 1 and 0 1 1 0
Null space:
dimension is 2.
Basis are obtained by solving Ax = 0 or Ux= 0
1 2 00 1 10 0 0
100
𝑥1
𝑥2
𝑥3
𝑥4
= 000
Here 𝒙𝟏 & 𝑥𝟐𝒂𝒓𝒆 𝒃𝒂𝒔𝒊𝒄 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒂𝒏𝒅 𝒙𝟑 & 𝒙𝟒 𝒂𝒓𝒆 𝒕𝒉𝒆 𝒇𝒓𝒆𝒆 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔
𝒙𝟐 = −𝒙𝟑 and 𝒙𝟏 = 𝟐𝒙𝟑 − 𝒙𝟒
14
𝑥1
𝑥2
𝑥3
𝑥4
=
2𝑥3 − 𝑥4
−𝑥3
𝑥3
𝑥4
= 𝑥3
2−110
+ 𝑥4
−1001
When 𝒙𝟑 = 𝟎, 𝒙𝟒 = 𝟏
𝑥1
𝑥2
𝑥3
𝑥4
=
−1001
When 𝒙𝟑 = 𝟏, 𝒙𝟒 = 𝟎
𝑥1
𝑥2
𝑥3
𝑥4
=
2−110
Therefore the basis are
−𝟏𝟎𝟎𝟏
𝒂𝒏𝒅
𝟐−𝟏𝟏𝟎
Left null space:
dimension is 1.
Solving 𝑨𝑻𝒚 = 𝟎
1 0 12 1 20 1 01 0 1
𝑦1
𝑦2
𝑦3
=
0000
Here 𝑦3 is the free variable and 𝒚𝟏 = −𝒚𝟑 & 𝒚𝟐 = 𝟎
When 𝒚𝟑 = 𝟏, the basis becomes −101
Now considering the matrix 𝑈 = 1 2 00 1 10 0 0
100
Column space:
Dimension is 2.
Basis are 100 𝑎𝑛𝑑
210
15
Row space:
Dimension is 2.
Basis are 1 2 0 1 and 0 1 1 0
Null space:
Dimension is 2.
Basis are
−1001
𝑎𝑛𝑑
2−110
Left null space:
Dimension is 1.
Solving 𝑼𝑻𝒚 = 𝟎
1 0 02 1 00 1 01 0 0
𝑦1
𝑦2
𝑦3
=
0000
Here 𝑦3 is the free variable ; 𝒚𝟏 = 𝟎, 𝒚𝟐 = 𝟎,
Hence assigning 𝑦3 = 1, 𝑤𝑒 𝑔𝑒𝑡 𝑡𝑒 𝑏𝑎𝑠𝑖𝑠 𝑎𝑠 001
4.10 Describe the four subspaces in 3-dimensional space associated with
𝐴 = 0 1 00 0 10 0 0
.
SOLUTION:
Here m=3; n=3; r=2
Column space:
Dimension is 2.
Basis are 100 𝑎𝑛𝑑
010
The column space spans the x-y plane in the space 𝑹𝟑.
16
Row space:
Dimension is 2.
Basis are 0 1 0 and 0 0 1
The row space spans the y-z plane in the vector space 𝑹𝟑
Null space:
Dimension is 1.
Solving Ax = 0
0 1 00 0 10 0 0
𝑥1
𝑥2
𝑥3
= 000
Basis for null space is 100
The null space spans the line x-axis in the vector space 𝑹𝟑.
Left null space:
Dimension is 1.
Solving 𝑨𝑻𝒚 = 𝟎
0 0 01 0 00 1 0
𝑦1
𝑦2
𝑦3
= 000
𝑦1 = 0, 𝑦2 = 0 (𝑦3 𝑖𝑠 𝑡𝑒 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒)
Therefore Basis is 001
The left null space spans the line (z-axis) in the vector space 𝑹𝟑
4.11 Find the rank of 𝐴 and write the matrix as 𝐴 = 𝑢𝑣𝑇:
𝐴 = 1 0 00 0 02 0 0
306 and 𝐴 =
2 −22 −2
.
17
SOLUTION:
a) 𝐴 = 1 0 00 0 02 0 0
306
R3 → R3-2*R1
→ 1 0 00 0 00 0 0
300 which is the echelon form
Rank of "A" is 1. All matrices with rank 1 can be expressed in the
form 𝑨 = 𝒖𝒗𝑻
Therefore A = 102 1 0 0 3
b) 𝐴 = 2 −22 −2
= 2 1 −11 −1
R2 → R2-R1
→ 1 −10 0
which is the echelon form.
Hence rank is 1. All matrices with rank 1 can be expressed in the
form 𝑨 = 𝒖𝒗𝑻
A = (2) 11 1 −1
4.12 Find a left-inverse and/or a right –inverse (when they exist) for
𝐴 = 1 1 00 1 1
and 𝑀 = 1 01 10 1
and 𝑇 = 𝑎 𝑏0 𝑎
.
SOLUTION:
a) 𝐴 = 1 1 00 1 1
Here r=2; m=2; n=3
r=m implies right inverse exists for "A"
AC = I where C is the right inverse and C = 𝑨𝑻(𝑨𝑨𝑻)−𝟏
18
𝐴𝐴𝑇 = 2 11 2
(𝑨𝑨𝑻)−𝟏 = 2
3 −1
3
−13
23
C = 𝑨𝑻(𝑨𝑨𝑻)−𝟏 =
23
13
−13
−13
13
23
is the right inverse of "A"
b) 𝑀 = 1 01 10 1
R2 → R2-R1
→ 1 00 10 1
R3 → R3-R2
→ 1 00 10 0
which is the echelon form
Here r=2; m=3; n=2
r=n implies there exists a left inverse for "M"
BM = I where B is the left inverse and is equal to
B = (𝑴𝑻𝑴)−𝟏𝑴𝑻
𝑀𝑇𝑀 = 2 11 2
(𝑀𝑇𝑀)−1 = 2
3 −1
3
−13
23
B = (𝑴𝑻𝑴)−𝟏𝑴𝑻 = 2
3 1
3 −1
3
−13
13
23 is the left
inverse of "M"
19
c) 𝑇 = 𝑎 𝑏0 𝑎
Here r=m=n=2. Hence both left and right inverse exist
and they both are the same.
det(T) = 𝒂𝟐
𝑻−𝟏 = 𝟏𝒂𝟐
𝑎 −𝑏0 𝑎
𝑻−𝟏 = 1
𝑎 −𝑏𝑎2
0 1𝑎
is the left and right inverse of "T"