MATH 334 - DIFFERENTIAL
EQUATIONS - Matrix methods and
eigenvalue problems
Question Bank - Set 2
Liberty University
Question 1
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 3,2,and −1. Given that the
eigenvector corresponding to λ= 3 is
1
2
−1
, find the eigenvectors corresponding
to the eigenvalues λ= 2 and λ=−1.
Solution
Step 1: Recall that for an n×nmatrix Awith eigenvalue λ, the eigenvector v
satisfies the equation (A−λI)v=0, where Iis the identity matrix of size n.
Step 2: We are given that the matrix Ahas eigenvalues λ= 3,2,and −1.
We will find the eigenvectors for each eigenvalue by solving the corresponding
system of equations for each λ.
Step 3: For λ= 2, we have the equation (A−2I)v=0. Substituting Aand
Iin, we get:
−1 0 0
000
0 0 −3
x
y
z
=
0
0
0
Step 4: The system of equations is:
(−x= 0
−3z= 0
Solving these equations, we find that x= 0 and z= 0. This gives us the
eigenvector corresponding to λ= 2:
0
1
0
.
Step 5: For λ=−1, we have the equation (A+I)v=0. Substituting A
and Iin, we get:
200
030
000
x
y
z
=
0
0
0
Step 6: The system of equations is:
(2x= 0
3y= 0
Solving these equations, we find that x= 0 and y= 0. This gives us the
eigenvector corresponding to λ=−1:
0
0
1
.
Question 2
Question
Let Abe a 3 ×3 matrix such that det(A) = 2 and tr(A) = 6, where det denotes
the determinant and tr denotes the trace of a matrix. If λ1,λ2, and λ3are the
eigenvalues of A, find the value of λ2
1+λ2
2+λ2
3.
Solution
Step 1: We know that the sum of the eigenvalues of a matrix is equal to the
trace of the matrix. Therefore, we have:
λ1+λ2+λ3= tr(A)=6
Step 2: We also know that the product of the eigenvalues is equal to the
determinant of the matrix. Thus, we have:
λ1·λ2·λ3= det(A)=2
Step 3: We can square the equation from Step 1 to find:
(λ1+λ2+λ3)2=λ2
1+λ2
2+λ2
3+ 2(λ1λ2+λ2λ3+λ3λ1)
36 = λ2
1+λ2
2+λ2
3+ 2(2) = λ2
1+λ2
2+λ2
3+ 4
Step 4: Substituting the values from Step 3 and Step 4, we find:
λ2
1+λ2
2+λ2
3+ 4 = 36
2
λ2
1+λ2
2+λ2
3= 36 −4
λ2
1+λ2
2+λ2
3= 32
Therefore, the value of λ2
1+λ2
2+λ2
3is 32.
Question 3
Question
Let Abe a 3×3 matrix with eigenvalues 2, −1, and 3. Determine the eigenvalues
of the matrix B= 2A2−A+I, where Iis the 3 ×3 identity matrix.
Solution
Given that Ahas eigenvalues 2, −1, and 3, we can find the eigenvalues of the
matrix Bby performing the following operations:
B= 2A2−A+I
= 2(AA)−A+I
= 2(A2)−A+I
Step 1: Find the eigenvalues of A2Since Ahas eigenvalues 2, −1, and
3, the matrix A2will have eigenvalues equal to the square of these values. So
the eigenvalues of A2are 22= 4, (−1)2= 1, and 32= 9.
Step 2: Find the eigenvalues of BNow we can find the eigenvalues of
Busing the expression B= 2A2−A+I. Applying our results from Step 1
and given that adding or multiplying a matrix by a scalar does not change its
eigenvalues, the eigenvalues of Bare:
Eigenvalues of B= 2 ·4−2+1,2·1−(−1) + 1,and 2 ·9−3+1
= 7,4,17
Therefore, the eigenvalues of the matrix Bare 7, 4, and 17.
Question 4
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Given
that the corresponding eigenvectors are
v1=
1
2
1
, v2=
0
1
1
,and v3=
1
0
1
,
find the matrix A.
3
Solution
To find the matrix A, we know that if λis an eigenvalue of Awith corresponding
eigenvector v, then
Av =λv.
Therefore, we have:
A
1
2
1
= 1
1
2
1
,
A
0
1
1
= 2
0
1
1
,
and
A
1
0
1
= 3
1
0
1
.
Step 1: Computing Av1=λ1v1:
A
1
2
1
=
a b c
d e f
g h i
1
2
1
=
a+ 2b+c
d+ 2e+f
g+ 2h+i
=
1
2
1
.
This gives us the system of equations:
a+ 2b+c= 1,
d+ 2e+f= 2,
g+ 2h+i= 1.
Step 2: Computing Av2=λ2v2:
A
0
1
1
=
a b c
d e f
g h i
0
1
1
=
b+c
e+f
h+i
=
0
2
2
.
This gives us the system of equations:
b+c= 0,
e+f= 2,
h+i= 2.
Step 3: Computing Av3=λ3v3:
A
1
0
1
=
a b c
d e f
g h i
1
0
1
=
a+c
d+f
g+i
=
3
0
3
.
4
This gives us the system of equations:
a+c= 3,
d+f= 0,
g+i= 3.
Step 4: Solve the system of equations to find the matrix A: From Step 1,
we have c= 0, a = 1, f =−2, d = 2, h = 2.
From Step 2, we have b= 0, e = 2, i = 2.
From Step 3, we have g= 0.
Therefore, the matrix Ais:
Question 5
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−3, and λ3= 1. Given
that Ahas an eigenvector corresponding to λ1= 2 of
1
1
1
, find the eigenvectors
corresponding to λ2and λ3.
Solution
Step 1: Recall that if vis an eigenvector of a matrix Acorresponding to an
eigenvalue λ, then we have the equation Av=λv.
Step 2: Given that λ1= 2 and an eigenvector corresponding to λ1is
1
1
1
,
we have that A
1
1
1
= 2
1
1
1
.
Step 3: Solving A
1
1
1
= 2
1
1
1
gives us the equation
A
1
1
1
=
2
2
2
.
Step 4: Subtracting
2
2
2
from both sides, we get
A
1
1
1
−
2
2
2
=
0
0
0
.
5
Step 5: To find the eigenvectors corresponding to λ2=−3 and λ3= 1, we
solve the equations Av2=−3v2and Av3=v3, respectively.
Step 6: Once the systems of equations Av2=−3v2and Av3=v3are set
up, they can be solved to find the eigenvectors corresponding to λ2and λ3.
Question 6
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 1,2,−3. If A2−6A+ 5I=0,
where Iis the identity matrix, find the inverse of A.
Solution
Step 1: First, we find the characteristic polynomial of matrix A, given that
its eigenvalues are λ= 1,2,−3. The characteristic polynomial is given by the
equation det(A−λI) = 0. So, we have:
det(A−λI) = (1 −λ)(2 −λ)(−3−λ)=0
Expanding this equation, we get:
(1 −λ)(2 −λ)(−3−λ) = −(λ3−2λ2+ 3λ−6) = 0
Therefore, the characteristic polynomial of Ais p(λ) = λ3−2λ2+ 3λ−6.
Step 2: Next, we use the equation provided: A2−6A+ 5I=0to simplify
the expression.
Since we know that Ahas eigenvalues 1,2,−3, we can rewrite A2and 6A:
A2=
1 0 0
0 2 0
0 0 −3
2
=
100
040
009
6A= 6
1 0 0
0 2 0
0 0 −3
=
6 0 0
0 12 0
0 0 −18
So the equation becomes:
100
040
009
−
6 0 0
0 12 0
0 0 −18
+5
100
010
001
=0
Step 3: Solving the simplified expression, we get:
−5 0 0
0−8 0
0 0 −9
=0
Now, we know that A−1exist when det(A)= 0.
Since det(A)=1×2× −3 = −6= 0, Ais invertible.
Therefore, A−1exists.
6
Question 7
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. Find
the eigenvectors corresponding to each eigenvalue.
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we need to solve the
system (A−2I)v1= 0, where v1is the eigenvector for λ1.
A−2I=
1−1 0
0 1 −1
0 0 2
Setting the matrix equation (A−2I)v1= 0, we get:
1−1 0
0 1 −1
0 0 2
x
y
z
=
0
0
0
This simplifies to the system of linear equations:
x−y= 0
y−z= 0
2z= 0
Step 2: Solving the system of equations, we find z= 0, y= 0, and x= 0.
Thus, the eigenvector corresponding to λ1= 2 is
0
0
0
.
Step 3: Similarly, for λ2=−1, we solve the system (A+I)v2= 0. This
gives us the matrix equation:
3−1 0
0 2 −1
0 0 5
x
y
z
=
0
0
0
Solving this system of equations, we find the eigenvector corresponding to λ2=
−1 is
1
2
0
.
Step 4: Finally, for λ3= 4, we solve the system (A−4I)v3= 0. This gives
us the matrix equation:
−1−1 0
0−2−1
001
x
y
z
=
0
0
0
7
Solving this system of equations, we find the eigenvector corresponding to λ3= 4
is
1
0
1
.
Question 8
Question
Let Abe a 3×3 matrix with eigenvalues λ=−2,1,3. Given that the eigenvector
corresponding to λ= 1 is v1=
1
0
0
and the eigenvector corresponding to λ= 3
is v2=
1
1
1
, determine the matrix A.
Solution
Step 1: Recall that any n×nmatrix Ais diagonalizable if it has nlinearly
independent eigenvectors. Since we are given three distinct eigenvalues with
corresponding eigenvectors, we can diagonalize A.
Step 2: The diagonal matrix Λ is defined as:
Λ =
−200
0 1 0
0 0 3
Step 3: Let Pbe the matrix formed by the eigenvectors of Aas its columns.
Thus, we have:
P=
11?
01?
01?
Step 4: To find the third column of P, we notice that the columns of Pmust
be linearly independent. Therefore, the third column of Pmust be a linear
combination of the given eigenvectors. Let the third column of Pbe v3=
a
b
c
.
Then, we have:
Av3=λv3
Av3= 3v3
A
a
b
c
= 3
a
b
c
8
Step 5: Solving A
a
b
c
= 3
a
b
c
gives:
−200
0 1 0
0 0 3
a
b
c
=
3a
3b
3c
Step 6: Solving the above equation, we get c= 0. Therefore, the third
column of Pand the matrix Abecomes:
P=
110
011
010
A=PΛP−1
Step 7: Compute P−1and then find A:
P−1=
1−1 0
0 1 −1
0 0 1
Step 8: Finally, we find:
A=
110
011
010
−200
0 1 0
0 0 3
1−1 0
0 1 −1
0 0 1
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 3, λ2=−2, and λ3= 1. Find a
basis for each eigenspace corresponding to these eigenvalues.
Solution
Step 1: To find the eigenvectors corresponding to each eigenvalue, we need to
solve the system of equations (A−λI)v=0.
For λ1= 3: Setting λ= 3 in (A−λI)v=0, we get:
(A−3I)v=
−1 1 2
000
1−1−1
x
y
z
=
0
0
0
Performing row operations, the reduced row-echelon form is:
1−1−1
0 0 0
0 0 0
9
So, the eigenvectors corresponding to λ= 3 are of the form
s
s
s
, where sis a
scalar.
Step 2: For λ2=−2: Setting λ=−2 in (A−λI)v=0, we get:
(A+ 2I)v=
512
020
1−1 1
x
y
z
=
0
0
0
Performing row operations, the reduced row-echelon form is:
101
010
000
So, the eigenvectors corresponding to λ=−2 are of the form
−s
0
s
, where sis
a scalar.
Step 3: For λ3= 1: Setting λ= 1 in (A−λI)v=0, we get:
(A−I)v=
−112
0−1 0
1−1 0
x
y
z
=
0
0
0
Performing row operations, the reduced row-echelon form is:
1−1 0
001
000
So, the eigenvectors corresponding to λ= 1 are of the form
s
s
0
, where sis a
scalar.
Thus, basis for eigenspace corresponding to λ1= 3:
1
1
1
Basis for eigenspace corresponding to λ2=−2:
−1
0
1
Basis for eigenspace corresponding to λ3= 1:
1
1
0
10
Question 10
Question
Let Abe a 3 ×3 matrix given by:
A=
2−1 0
−1 2 −1
0−1 2
Determine the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation:
det(A−λI)=0
where λis the eigenvalue we are solving for and Iis the identity matrix.
Step 2: Substitute the values into the equation and solve for λ:
det
2−λ−1 0
−1 2 −λ−1
0−1 2 −λ
= 0
Step 3: Expanding the determinant yields:
(2 −λ)3−2(2 −λ)−1=0
Step 4: Simplify the equation:
λ3−6λ2+ 11λ−6=0
Step 5: Solve the characteristic equation to find the eigenvalues. In this
case, the eigenvalues are λ= 1,2,3.
Step 6: We will now find the eigenvectors corresponding to each eigenvalue
by solving the system of equations (A−λI)x= 0.
For λ= 1: Substitute λ= 1 into A−λIand solve the system of equations:
1−1 0
−1 1 −1
0−1 1
x1
x2
x3
=
0
0
0
Step 7: The solutions to the system of equations for λ= 1 are:
x1=x2=x3
Thus, the eigenvector corresponding to λ= 1 is a scalar multiple of
1
1
1
.
For λ= 2 and λ= 3, follow the same steps to determine the eigenvectors.
11
Question 11
Question
Let Abe a 3 ×3 matrix given by
A=
1 2 1
0 2 0
−110
.
Find the eigenvectors and eigenvalues of A.
Solution
Step 1: To find the eigenvalues, we solve the characteristic equation |A−λI|= 0
where Iis the 3 ×3 identity matrix.
A−λI =
1−λ2 1
0 2 −λ0
−1 1 −λ
.
Step 2: Expanding the determinant gives
(1 −λ)((2 −λ)(−λ)−(0)(1)) −2(0 −(−1)(1)) + 1(0 −0) = 0.
Step 3: Simplifying the equation, we get
(1 −λ)(−2λ+λ2) + 2 = 0.
Step 4: Further simplification gives
λ3−3λ2+ 2λ+ 2 = 0.
Step 5: By solving the characteristic equation, we find that the eigenvalues
of Aare λ1= 2 and λ2= 1.
Step 6: To find the eigenvectors, we substitute each eigenvalue back into the
equation (A−λI)v=0.
Step 7: For λ= 2, we have
A−2I=
−1 2 1
000
−1 1 −2
.
Step 8: We row reduce to find that the eigenvector corresponding to λ= 2
is v1=
1
2
1
.
Step 9: For λ= 1, we have
A−I=
021
010
−1 1 −1
.
12
Step 10: We row reduce to find that the eigenvector corresponding to λ= 1
is v2=
−2
1
2
.
Therefore, the eigenvectors of Aare v1=
1
2
1
corresponding to λ= 2 and
v2=
−2
1
2
corresponding to λ= 1.
Question 12
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 2, and λ3= 3. Find the
eigenvectors corresponding to each eigenvalue.
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we need to solve the
equation (A−2I)v1=0, where v1is the eigenvector corresponding to λ1.
A−2I=
210
021
001
The augmented matrix formed by (A−2I)v1=0would be:
010|0
001|0
000|0
Step 2: Reduce the augmented matrix to row-echelon form:
010|0
001|0
000|0
→
010|0
001|0
000|0
Step 3: The solution to the system is x1= 0, x2= 0, and x3= 0. Therefore,
the eigenvector corresponding to λ1= 2 is v1=
0
0
0
.
Repeat Steps 1-3 for λ2= 2 and λ3= 3 to find the corresponding eigenvec-
tors.
13
Question 13
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 3, λ2= 1, and λ3=−2. If v1,
v2, and v3are the corresponding eigenvectors of A, find the matrix Pformed
by placing v1,v2, and v3as columns. Then, determine P−1AP .
Solution
Step 1: Form the matrix Pby arranging the eigenvectors v1,v2, and v3as
columns.
P=
| | |
v1v2v3
| | |
Step 2: Since v1,v2, and v3are eigenvectors of A, we know that Avi=λivi
for i= 1,2,3. Therefore, we have:
Av1= 3v1
Av2=v2
Av3=−2v3
Step 3: By substituting these relations into the definition of P−1AP , we
have:
P−1AP =
| | |
v1v2v3
| | |
−1
3 0 0
0 1 0
0 0 −2
| | |
v1v2v3
| | |
Step 4: We can calculate P−1by using the formula P−1=1
det(P)adj(P),
where adj(P) is the adjugate of P.
P−1=1
det(P)adj(P)
Step 5: Calculate the determinant of Pto find 1
det(P).
det(P) = v1·(v2×v3)
Step 6: Compute the adjugate of P.
adj(P) =
|||
v2×v3v3×v1v1×v2
|||
Step 7: Substitute the values of det(P) and adj(P) into P−1to find P−1.
Step 8: Finally, calculate P−1AP using the formula derived in Step 3.
14
Question 14
Question
Let Abe a 3 ×3 matrix given by
A=
211
131
112
.
Find the eigenvalues and eigenvectors of A. Normalize the eigenvectors.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0, where λis the eigenvalue and Iis the identity matrix.
det
211
131
112
−λ
100
010
001
= 0
det
2−λ1 1
1 3 −λ1
1 1 2 −λ
= 0
(2 −λ)[(3 −λ)(2 −λ)−1] −(1)[(1)(2 −λ)−(1)(1)] + (1)[(1)(1) −(1)(3 −λ)] = 0
Solving this equation gives us the eigenvalues.
Step 2: Now that we have found the eigenvalues, we can find the correspond-
ing eigenvectors by solving the equation (A−λI)x=0for each eigenvalue.
Step 3: Normalize the eigenvectors by dividing each by its magnitude.
Question 15
Question
Let A=4 6
3 5. Find the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we need to solve the characteristic equation
|A−λI|= 0, where λis the eigenvalue and Iis the identity matrix.
A−λI=4 6
3 5−λ1 0
0 1=4−λ6
3 5 −λ
Setting the determinant of A−λIto zero, we have:
4−λ6
3 5 −λ
= (4 −λ)(5 −λ)−18 = λ2−9λ+ 2 = 0
15
Step 2: Solve the quadratic equation λ2−9λ+ 2 = 0 to find the eigenvalues
λ. Using the quadratic formula λ=−(−9)±√(−9)2−4∗1∗2
2∗1, we get:
λ=9±√81 −8
2=9±√73
2
So, the eigenvalues are λ1=9+√73
2and λ2=9−√73
2.
Step 3: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute each eigenvalue back into the equation (A−λI)v=0, where vis the
eigenvector. For λ=9+√73
2, we have:
−√73
26
3−√73
2!x
y=0
0
Solving the system of equations, we get the eigenvector corresponding to λ1.
For λ=9−√73
2, we have:
√73
26
3√73
2!x
y=0
0
Solving the system of equations, we get the eigenvector corresponding to λ2.
Question 16
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3=−1. If
B= 3A−2I, where Iis the 3 ×3 identity matrix, find the eigenvalues of B.
Solution
Step 1: Find the eigenvalues of B. Given B= 3A−2I, the eigenvalues of B
are related to the eigenvalues of Aby the following equation:
det(B−λiI) = det(3A−2I−λiI)=0
where λiare the eigenvalues of A.
Step 2: Substitute the eigenvalues of Ainto the equation. Substitute the
eigenvalues of Ainto the equation to find the eigenvalues of B:
det(3A−2I−2I) = det(3A−4I) = 0
det(3A−2I−3I) = det(3A−5I) = 0
det(3A−2I+I) = det(3A−I)=0
16
Step 3: Find the eigenvalues of B. The eigenvalues of Bare the solutions to
the equations derived in Step 2:
λ′
1= 3(2) −1=5
λ′
2= 3(3) −1=8
λ′
3= 3(−1) −1 = −4
Therefore, the eigenvalues of matrix Bare λ′
1= 5, λ′
2= 8, and λ′
3=−4.
Question 17
Question
Let Abe a 3 ×3 matrix with eigenvalues −2,1,and 3. Determine the corre-
sponding eigenvectors for each eigenvalue.
Solution
Step 1: For eigenvalue λ=−2: Let v1be the eigenvector corresponding to
λ=−2. We solve the system (A+ 2I)v1=0, where Iis the 3 ×3 identity
matrix.
(A+ 2I)v1=
−200
0−1 0
0 0 1
v1=
0
0
0
Step 2: Solve for v1: This system simplifies to:
−2v11= 0
−v12= 0
v13= 0
So, the eigenvector corresponding to λ=−2 is v1=
0
0
1
.
Step 3: For eigenvalue λ= 1: Let v2be the eigenvector corresponding to
λ= 1. We solve the system (A−I)v2=0.
(A−I)v2=
−300
0 0 0
0 0 2
v2=
0
0
0
Step 4: Solve for v2: This system simplifies to:
−3v21= 0
2v23= 0
17
So, the eigenvector corresponding to λ= 1 is v2=
0
1
0
.
Step 5: For eigenvalue λ= 3: Let v3be the eigenvector corresponding to
λ= 3. We solve the system (A−3I)v3=0.
(A−3I)v3=
−5 0 0
0−2 0
0 0 −2
v3=
0
0
0
Step 6: Solve for v3: This system simplifies to:
−5v31= 0
−2v32= 0
−2v33= 0
So, the eigenvector corresponding to λ= 3 is v3=
0
0
1
.
Question 18
Question
Let Abe a 4 ×4 matrix given by
A=
2−1 0 0
1 2 −1 0
0 1 2 −1
0 0 1 2
.
Determine the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues, we need to solve the characteristic equation
det(A−λI) = 0 where λis the eigenvalue.
det(A−λI) = det
2−λ−1 0 0
1 2 −λ−1 0
0 1 2 −λ−1
0 0 1 2 −λ
= 0.
18
Step 2: Simplify the determinant using row operations.
= (2 −λ)det
2−λ−1 0
1 2 −λ−1
0 1 2 −λ
= (2 −λ)(2 −λ)det 2−λ−1
1 2 −λ−(−1)det 1−1
0 2 −λ
= (2 −λ)[(2 −λ)((2 −λ)(2 −λ)−(−1)(1)) −(−1)(2 −λ)]
= (2 −λ)[(2 −λ)((4 −4λ+λ2) + 1) −(−2 + λ)]
= (2 −λ)[(2 −λ)(5 + 4λ−4−4λ+λ2) + λ−2]
= (2 −λ)[(2 −λ)(λ2+ 1) + λ−2]
= (2 −λ)[(2 −λ)λ2+ 2 −λ2+λ−2]
= (2 −λ)(2λ2−2λ3+λ−λ2)
= 2λ2−2λ3+λ−λ2−2λ2+ 2λ3
=−λ2+λ.
Step 3: Set the characteristic equation −λ2+λ= 0 and solve for the eigen-
values.
−λ2+λ= 0 =⇒λ(−λ+ 1) = 0.
So, λ= 0 or λ= 1.
Step 4: To find the eigenvectors, we substitute each eigenvalue back into the
equation (A−λI)x= 0 and solve for x.
For λ= 0, we solve:
Ax= 0 =⇒
2−1 0 0
1 2 −1 0
0 1 2 −1
0 0 1 2
x1
x2
x3
x4
=
0
0
0
0
.
This system of equations leads to x1=x2=x3=x4, so the eigenvector
corresponding to λ= 0 is
1
1
1
1
.
For λ= 1, we solve:
Ax
Question 19
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 2,3,and 4. If B= 2A−3I, where
Iis the 3 ×3 identity matrix, find the eigenvalues of B.
19
Solution
Given that Ahas eigenvalues λ= 2, 3, and 4, we know that the characteristic
polynomial of Ais given by
det(A−λI) = 0
which can be expanded as
det
a11 −λ a12 a13
a21 a22 −λ a23
a31 a32 a33 −λ
= 0
Substitute λ= 2, λ= 3, and λ= 4 into the characteristic equation to get three
equations. By solving these equations, we can find the matrix A.
Step 1: Find the matrix A using the eigenvalues. For λ= 2:
det
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
= 0
For λ= 3:
det
a11 −3a12 a13
a21 a22 −3a23
a31 a32 a33 −3
= 0
For λ= 4:
det
a11 −4a12 a13
a21 a22 −4a23
a31 a32 a33 −4
= 0
Solving these equations will give us the matrix A.
Step 2: Find matrix B. Given B= 2A−3I, we can find Bonce we know
matrix A.
Step 3: Find the eigenvalues of matrix B. To find the eigenvalues of
matrix B, we can use the relationship between eigenvalues and similar matrices.
Since B= 2A−3I, the eigenvalues of Bwill be λB= 2λA−3, where λAare
the eigenvalues of matrix A.
Therefore, the eigenvalues of matrix Bwill be 2(2) −3 = 1, 2(3) −3 = 3,
and 2(4) −3 = 5.
Question 20
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 4, λ2= 2,and λ3=−1. If
B=A2−3A+ 5I, where Iis the 3 ×3 identity matrix, find the eigenvalues of
B.
20
Solution
Given that Ahas eigenvalues λ1= 4, λ2= 2,and λ3=−1, we have that
the eigenvalues of A2,−3A, and 5Iare (λi)2,−3λi, and 5 respectively, where
i∈ {1,2,3}.
Step 1: Find the eigenvalues of A2. Since Ahas eigenvalues λ1= 4, λ2= 2,
and λ3=−1, the eigenvalues of A2are λ2
1= 16, λ2
2= 4,and λ2
3= 1.
Step 2: Find the eigenvalues of −3A. Since Ahas eigenvalues λ1= 4, λ2=
2,and λ3=−1, the eigenvalues of −3Aare −3λ1=−12,−3λ2=−6,and
−3λ3= 3.
Step 3: Find the eigenvalues of 5I. Since Iis the identity matrix of size
3×3, the eigenvalues of 5Iare all equal to 5.
Step 4: Find the eigenvalues of B. To find the eigenvalues of B=A2−
3A+ 5I, we add the corresponding eigenvalues of A2,−3A, and 5I. Thus, the
eigenvalues of Bare 16 −12 + 5 = 9,4−6 + 5 = 3,and 1 + 3 + 5 = 9.
Therefore, the eigenvalues of Bare λ1= 9, λ2= 3,and λ3= 9.
Question 21
Question
Let A=3 2
2 6. Find the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation |A−λI|= 0 where Iis the identity matrix. The characteristic equation
is given by:
det 3−λ2
2 6 −λ= 0
Step 2: Calculating the determinant, we get:
(3 −λ)(6 −λ)−2·2 = 0
Step 3: Simplifying the equation, we have:
λ2−9λ+ 16 = 0
Step 4: Solving the quadratic equation, we find the eigenvalues:
λ=9±√92−4·1·16
2=9±√81 −64
2
λ=9±√17
2
21
Step 5: Therefore, the eigenvalues of matrix Aare λ1=9+√17
2and λ2=
9−√17
2.
Step 6: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute the eigenvalues back into the equation (A−λI)X= 0, where Xis the
eigenvector.
Step 7: For λ1=9+√17
2, subtract λ1from the diagonal of matrix Aand
form the system of equations:
−√17
22
2−√17
2!x
y=0
0
Step 8: Solving this system of equations yields the eigenvector corresponding
to λ1.
Step 9: For λ2=9−√17
2, follow the same process to find the corresponding
eigenvector.
Step 10: Therefore, the eigenvectors for λ1and λ2are the solutions obtained
in Step 8 and Step 9 respectively.
Question 22
Question
Let Abe a 3x3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. If
the eigenvectors corresponding to λ1and λ2are v1=
1
1
0
and v2=
1
0
1
respectively, find the matrix A.
Solution
Step 1: Recall that a matrix Acan be diagonalized by a matrix Pif A=P DP −1
where Dis a diagonal matrix with the eigenvalues of Aand the columns of P
are the eigenvectors of A.
Step 2: Since we are given the eigenvalues and eigenvectors of A, we can
construct the matrix Pas follows:
P= [v1,v2,v3] =
1 1 a
1 0 b
0 1 c
Step 3: To find the third eigenvector v3, we use the fact that the columns
of Pmust be linearly independent. Thus, v3is orthogonal to v1and v2and
satisfies Av3=λ3v3.
Step 4: Since v3is orthogonal to v1and v2, we can solve the system of
equations:
1 1 a
1 0 b
0 1 c
x
y
z
= 4
x
y
z
22
Step 5: Solving the system of equations yields a=−1, b= 1, and c= 1.
Therefore, the matrix Pis:
P=
1 1 −1
1 0 1
0 1 1
Step 6: Now, we can compute P−1using the inverse of a 3x3 matrix formula:
P−1=1
det(P)adj(P)
where det(P) is the determinant of Pand adj(P) is the adjugate of P.
Step 7: Calculate det(P) = −1 and adj(P) is found by taking the transpose
of the matrix of cofactors of P.
Step 8: Using the formula for P−1, we find:
P−1=
0 1 −1
1 1 −1
−1−1 2
Step 9: Finally, we can determine the matrix Aby the formula A=P DP −1.
Substituting the given eigenvalues into the diagonal matrix D, we have:
D=
200
0−1 0
004
Step 10: Calculate A=P DP −1:
A=
1 1 −1
1 0 1
0 1 1
200
0−1 0
004
0 1 −1
1 1 −1
−1−1 2
Question 23
Question
Consider the matrix A=3 1
1 3. Find the eigenvalues and corresponding
eigenvectors of matrix A.
Solution
To find the eigenvalues of matrix A, we need to solve the characteristic equation
|A−λI|= 0, where Iis the identity matrix.
Step 1: Find the characteristic equation The characteristic equation
is given by:
det 3−λ1
1 3 −λ= 0
23
Expanding the determinant, we get:
(3 −λ)2−1=0
Simplifying, we find:
λ2−6λ+ 8 = 0
Step 2: Find the eigenvalues Solving the characteristic equation λ2−
6λ+ 8 = 0, we get:
(λ−2)(λ−4) = 0
So the eigenvalues are λ1= 2 and λ2= 4.
Step 3: Find the eigenvectors To find the eigenvector corresponding to
λ= 2, substitute λ= 2 back into the equation (A−λI)v=0and solve for v:
3−2 1
1 3 −2x
y=0
0
This simplifies to the equations x+y= 0.
So, when x= 1, y=−1, and vice versa, the eigenvector corresponding to
λ= 2 is v1=1
−1.
Similarly, for λ= 4, we find the eigenvector v2=1
1.
Question 24
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 3, and λ3= 4. If v1,v2,
and v3are the corresponding eigenvectors, find the matrix A.
Solution
Step 1: We know that if Ais a square matrix with eigenvalues λ1, λ2, . . . , λn
and corresponding eigenvectors v1, v2, . . . , vn, then we can write Ain terms of
its eigenvectors and eigenvalues.
Step 2: The matrix Acan be expressed as A=P DP −1, where Pis the
matrix whose columns are the eigenvectors v1, v2, . . . , vnand Dis a diagonal
matrix with the eigenvalues λ1, λ2, . . . , λnon the diagonal.
Step 3: In this case, since Ais a 3 ×3 matrix, we will have P= [v1v2v3]
and D=
λ10 0
0λ20
0 0 λ3
.
Step 4: Substituting the given eigenvalues and eigenvectors into the expres-
sion for A, we have:
A=P DP −1=v1v2v3
100
030
004
v1v2v3−1
24
Step 5: Next, we compute P−1, which is the inverse of the matrix whose
columns are the eigenvectors v1, v2, v3. Let P−1= [w1w2w3].
Step 6: Therefore, the matrix Acan be calculated as:
A=P DP −1=v1v2v3
100
030
004
w1w2w3
Question 25
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 3.
If the eigenvector corresponding to λ1is v1=
1
0
−1
, find the eigenvectors
corresponding to the eigenvalues λ2and λ3.
Solution
1. Let v2and v3be the eigenvectors corresponding to the eigenvalues λ2and
λ3respectively. Since λ1= 2 is an eigenvalue with eigenvector v1=
1
0
−1
,
we have the following equation:
Av1=λ1v1
2. Substituting the values, we get:
A
1
0
−1
= 2
1
0
−1
This simplifies to:
a b c
d e f
g h i
1
0
−1
=
2
0
−2
3. Solving the matrix equation yields:
a−c= 2
d−f= 0
g+i=−2
25
4. Since λ2=−1 is an eigenvalue, we can set up a similar equation for v2:
Av2=λ2v2
Substituting the values and solving for v2will give the eigenvector corre-
sponding to λ2.
5. Similarly, since λ3= 3 is an eigenvalue, we can set up an equation for v3:
Av3=λ3v3
Substituting the values and solving for v3will give the eigenvector corre-
sponding to λ3.
Question 26
Question
Let’s consider the matrix A=4 1
−2−3. Find the eigenvalues and eigenvectors
of A. Then, use these eigenvalues and eigenvectors to diagonalize Aif possible.
Solution
Step 1: To find the eigenvalues of A, we need to solve the characteristic equation
|A−λI|= 0.
|A−λI|=
4−λ1
−2−3−λ
= (4 −λ)(−3−λ)−(−2)(1) = λ2−λ−14 = 0
Solving the quadratic equation, we find the eigenvalues:
λ=1±√1+4×14
2=1±√57
2
Step 2: Next, we find the corresponding eigenvectors. For each eigenvalue,
we solve the system of equations (A−λI)v= 0. For λ=1+√57
2:
"4−1+√57
21
−2−3−1+√57
2#x
y=0
0
"7−√57
21
−2−5−√57
2#x
y=0
0
Solving this system, we find the eigenvector corresponding to λ=1+√57
2.
Step 3: Similarly, for λ=1−√57
2, find the eigenvector.
26
Step 4: If the matrix Ais diagonalizable, we can form the matrix Pusing
the eigenvectors as columns.
P=eigenvector 1 eigenvector 2
Step 5: Then, the diagonal matrix Dis given by:
D=P−1AP
Step 6: If Ais diagonalizable, then A=P DP −1. Otherwise, Ais not
diagonalizable.
Question 27
Question
Let Abe a 3 ×3 matrix given by
A=
123
021
111
.
Determine the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues, we solve the characteristic equation |A−λI|= 0,
where Iis the identity matrix.
|A−λI|=
1−λ2 3
0 2 −λ1
1 1 1 −λ
= (1 −λ)
2−λ1
1 1 −λ−2
0 1
1 1 −λ
+ 3
0 2
1 1
= (1 −λ)((2 −λ)(1 −λ)−1) −2(−1) + 3(−2)
= (1 −λ)(λ2−3λ+ 1) + 2 −6
=−λ3+ 4λ2−2λ+ 1.
Step 2: Setting |A−λI|= 0, we find the eigenvalues of Aby solving the
equation −λ3+ 4λ2−2λ+ 1 = 0. This is a challenging polynomial to factor
by hand, so we will use numerical methods or software to find the roots. The
eigenvalues are approximately λ1≈0.277, λ2≈1.571, and λ3≈4.152.
Step 3: Next, we find the eigenvectors corresponding to each eigenvalue by
solving (A−λI)v= 0, where vis the eigenvector. For λ= 0.277: We solve
(A−0.277I)v1= 0:
0.723 2 3
0 1.723 1
1 1 0.723
x
y
z
=
0
0
0
.
27
Solving this system of equations, we find v1≈
−0.754
0.342
1
.
Repeat the above process for the other two eigenvalues to find their corre-
sponding eigenvectors.
Question 28
Question
Let Abe a 3 ×3 matrix with eigenvalues 2 and −1. If one of the eigenvectors
corresponding to the eigenvalue 2 is
1
0
1
, find the eigenvectors corresponding
to the eigenvalue −1.
Solution
Step 1: Recall that eigenvectors of a matrix Aare the non-zero vectors vsuch
that Av =λv, where λis the corresponding eigenvalue.
Step 2: Since λ=−1 is an eigenvalue of A, we have Av=−vfor some
non-zero vector v.
Step 3: Let v=
x
y
z
be the eigenvector corresponding to the eigenvalue
−1. Substituting into Av=λv, we get:
a b c
d e f
g h i
x
y
z
=−
x
y
z
Step 4: This equation can be expanded to form a system of equations:
ax +by +cz =−x
dx +ey +fz =−y
gx +hy +iz =−z
Step 5: Using the fact that
1
0
1
is an eigenvector corresponding to the
eigenvalue 2, we know A
1
0
1
= 2
1
0
1
. This gives us another equation to use.
28
Step 6: By substituting
1
0
1
into Av=λvfor the eigenvalue 2, we get:
a b c
d e f
g h i
1
0
1
= 2
1
0
1
Step 7: Solving this equation, we find further information about the matrix
Athat can be used to solve for the eigenvector corresponding to the eigenvalue
−1.
Step 8: Once we determine the matrix A, we can solve the system of equa-
tions from Step 4 to find the eigenvector corresponding to the eigenvalue −1.
Question 29
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2 and λ2= 3. If the eigenvector
corresponding to λ2= 3 is v2=
1
0
1
, find the eigenvector corresponding to
λ1= 2.
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we will use the eigen-
value equation Av=λv, where vis the eigenvector we want to find.
Step 2: In general, for a matrix Aand its eigenvalues λ1, λ2, . . . , λn, the
eigenvector corresponding to each eigenvalue can be found by solving (A−
λI)v=0, where Iis the identity matrix of the same size as A.
Step 3: Substituting the given eigenvalues into the equation, we have: (A−
2I)v=0.
Step 4: Given Ais a 3 ×3 matrix, we write the expression explicitly:
a11 a12 a13
a21 a22 a23
a31 a32 a33
−2
100
010
001
x
y
z
=
0
0
0
.
Step 5: Simplifying the above equation, we get:
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
x
y
z
=
0
0
0
.
29
Step 6: Since the matrix
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
is singular (has de-
terminant zero), the eigenvector
x
y
z
can be determined by solving the system
of equations formed by setting up the matrix equation equal to 0.
Step 7: Solving the system of equations obtained from Step 5 will yield the
eigenvector v1=
0
1
0
corresponding to the eigenvalue λ1= 2.
Question 30
Question
Let Abe a 3×3 matrix with eigenvalues λ= 1,2,3. We are given that the eigen-
vectors corresponding to λ= 1 and λ= 3 are
1
1
1
and
−1
0
1
, respectively.
Find the eigenvector corresponding to λ= 2.
Solution
The eigenvectors corresponding to distinct eigenvalues of a matrix are
linearly independent.
Since we are given that the eigenvectors corresponding to λ= 1 and
λ= 3 are linearly independent, we can use them to find the eigenvector
corresponding to λ= 2.
Step 1:
Let the eigenvector corresponding to λ= 2 be
a
b
c
. Since this eigenvector is
independent from the others, it must satisfy the equation A
a
b
c
= 2
a
b
c
.
Step 2:
Thus, we have the equation
1−1 1
100
111
a
b
c
= 2
a
b
c
which simplifies to
a−b+c
a
a+b+c
=
2a
2b
2c
30
Step 3:
Solving the system of equations gives us
a−b+c= 2a, a = 2b, a +b+c= 2c
Step 4:
From the second equation, we have a= 2b. Substituting this into the first and
third equations gives us b=−cand a= 2b=−2c. Hence, a possible eigenvector
corresponding to λ= 2 is
−2
1
−1
.
Therefore, the eigenvector corresponding to λ= 2 is
−2
1
−1
.
Question 31
Question
Let Abe a 3×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Determine
the eigenvalues of the matrix 2A−A2.
Solution
Given that Ais a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3, we
need to find the eigenvalues of the matrix 2A−A2.
Step 1: Calculate the matrix 2ATo find the matrix 2A, we multiply the
scalar 2 by each element of A:
2A= 2 ·
a b c
d e f
g h i
=
2a2b2c
2d2e2f
2g2h2i
Step 2: Calculate the matrix A2To find the matrix A2, we multiply A
by itself:
A2=A·A=
a b c
d e f
g h i
·
a b c
d e f
g h i
=
a2+bd +cg ab +be +ci ac +bf +ci
ad +de +fg bd +e2+fh cd +ef +hi
ag +dh +gi bg +eh +gi cg +fh +i2
Step 3: Calculate the matrix 2A−A2Subtracting A2from 2A, we get:
2A−A2=
2a2b2c
2d2e2f
2g2h2i
−
a2+bd +cg ab +be +ci ac +bf +ci
ad +de +fg bd +e2+fh cd +ef +hi
ag +dh +gi bg +eh +gi cg +fh +i2
=
2a−(a2+bd +cg) 2b−(ab +be +ci) 2c−(ac +bf +ci)
2d−(ad +de +fg) 2e−(bd +e2+f h) 2f−(cd +ef +hi)
2g−(ag +dh +gi) 2h−(bg +eh +gi) 2i−(cg +f h +i2)
Step 4: Determine the eigenvalues of 2A−A2From the expression
above, we see that the eigenvalues of 2A−A2are given by subtracting the
31
corresponding terms of A2from 2A. Therefore, the eigenvalues of 2A−A2are
2−12= 1, 2 −22=−2, and 2 −32=−7. Thus, the eigenvalues of 2A−A2are
1, −2, and −7.
Question 32
Question
Let Abe a 3 ×3 matrix given by
A=
210
120
001
.
Find the eigenvectors corresponding to the eigenvalues of A.
Solution
Step 1: Find the eigenvalues of the matrix Aby solving the characteristic equa-
tion |A−λI|= 0, where λis the eigenvalue and Iis the identity matrix.
|A−λI|=
2−λ1 0
1 2 −λ0
0 0 1 −λ
= (2 −λ)((2 −λ)(1 −λ)−0) −(1)((1)(1 −λ)−0)
= (2 −λ)(2 −3λ+λ2)−(1)(1 −λ)
= (2 −λ)(2 −3λ+λ2)−1 + λ
= 4 −6λ+ 2λ2−2λ+ 3λ2−λ3−1 + λ
=−λ3+ 5λ2−7λ+ 3
=−(λ−1)(λ−1)(λ−3) = 0
Solving λ−1 = 0, we get λ1= 1.
Solving λ−3 = 0, we get λ2= 3 (multiplicity 2).
Step 2: For each eigenvalue, find the eigenvectors by solving the system of
equations (A−λI)v=0. For λ= 1, we have (A−λI)v=
110
110
000
x
y
z
=
0
0
0
. Solving x+y= 0 gives x=−y. Let y=t, then the eigenvector
corresponding to λ= 1 is v1=t
−1
1
0
.
32
For λ= 3, we have (A−λI)v=
−1 1 0
1−1 0
0 0 −2
x
y
z
=
0
0
0
. Solving
−x+y= 0 and −x+y= 0 gives x=y. Let y=t, then the eigenvector
corresponding to λ= 3 is v2=t
1
1
0
.
Thus, the eigenvectors corresponding to the eigenvalues of matrix Aare
v1=t
−1
1
0
for λ= 1 and v2=t
1
1
0
for λ= 3.
Question 33
Question
Let Abe a 3×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Determine
the eigenvalues of the matrix B= 2A2−3A+5I, where Iis the identity matrix.
Solution
To find the eigenvalues of matrix B, we can use the properties of eigenvalues
and eigenvectors.
Step 1: Express Bin terms of A
First, express matrix Bin terms of matrix A:
B= 2A2−3A+ 5I
Step 2: Find the eigenvalues of B
We know that the eigenvalues of A2are the squares of the eigenvalues of A.
Therefore, the eigenvalues of A2are λ2
1= 1, λ2
2= 4, and λ2
3= 9. Similarly,
since Ahas eigenvalues λ1= 1, λ2= 2, and λ3= 3, we can find the eigenvalues
of B:
Eigenvalues of B= 2λ2
i−3λi+ 5, for i= 1,2,3
Calculating this for each eigenvalue λi, we get:
Eigenvalues of B={2(1)2−3(1) + 5,2(4) −3(2) + 5,2(9) −3(3) + 5}
Eigenvalues of B={4,6,14}
Therefore, the eigenvalues of matrix Bare 4, 6, and 14.
33
Question 34
Question
Let Abe a 3 ×3 matrix given by
A=
1 2 0
2−1 2
0 2 −1
.
Find the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation given by det(A−λI) = 0, where λis the eigenvalue and Iis the identity
matrix.
det
1−λ2 0
2−1−λ2
0 2 −1−λ
= 0.
Simplifying the determinant, we get
1−λ2 0
2−1−λ2
0 2 −1−λ
= (1−λ)((−1−λ)(−1−λ)−4)−2(−2(−1−λ)−0) = 0.
This leads to the characteristic equation −λ3−λ2+ 6λ−7 = 0.
Step 2: We need to find the roots of the characteristic equation, which will
give us the eigenvalues of matrix A. The roots can be found by inspection or by
using numerical methods. By inspection, we can see that λ= 1 is a root. By
polynomial division, we can find that (λ−1)(−λ2−2λ+ 7) = 0, which gives us
the other two roots λ=−1 and λ= 7.
Step 3: Now, we will find the eigenvectors associated with each eigenvalue.
For λ= 1: Solve the system (A−λI)X= 0 to find the eigenvector.
(A−I)X=
0 2 0
2−2 2
0 2 −2
X= 0.
By row-reducing the augmented matrix [A−I|0], we find the solution X1=
t
1
1
1
, where tis a scalar.
Step 4: For λ=−1: Solve the system (A+I)X= 0 to find the eigenvector.
(A+I)X=
220
202
020
X= 0.
34
By row-reducing the augmented matrix [A+I|0], we find the solution X−1=
s
−1
1
1
, where sis a scalar.
Step 5: For λ= 7: Solve the system (A−7I)X= 0 to find the eigenvector.
(A−7I)X=
−6 2 0
2−8 2
0 2 −8
X= 0.
By row-reducing the augmented matrix [A−7I|0], we find the solution X7=
r
1
2
1
, where ris a scalar.
Question 35
Question
Let Abe a 3 ×3 matrix defined as
A=
5−2 1
1 0 2
−114
Find the eigenvalues and eigenvectors of the matrix A.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation |A−λI|= 0, where λrepresents the eigenvalues and Iis the identity
matrix of the same size as A.
Starting with the matrix A:
A=
5−2 1
1 0 2
−114
Subtracting λtimes the identity matrix:
A−λI =
5−λ−2 1
1−λ2
−1 1 4 −λ
Calculating the determinant:
|A−λI|= (5−λ) [(−λ)(4 −λ)−(1)(1)]−(−2) [(1)(4 −λ)−(−1)(1)]+(1) [(1)(1) −(−λ)(−2)]
|A−λI|= (5 −λ)(λ2−4λ−1) + 2(4 −λ−1) + (1)(1 + 2λ)
35
Question 7
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. Find
the eigenvectors corresponding to each eigenvalue.
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we need to solve the
system (A−2I)v1= 0, where v1is the eigenvector for λ1.
A−2I=
1−1 0
0 1 −1
0 0 2
Setting the matrix equation (A−2I)v1= 0, we get:
1−1 0
0 1 −1
0 0 2
x
y
z
=
0
0
0
This simplifies to the system of linear equations:
x−y= 0
y−z= 0
2z= 0
Step 2: Solving the system of equations, we find z= 0, y= 0, and x= 0.
Thus, the eigenvector corresponding to λ1= 2 is
0
0
0
.
Step 3: Similarly, for λ2=−1, we solve the system (A+I)v2= 0. This
gives us the matrix equation:
3−1 0
0 2 −1
0 0 5
x
y
z
=
0
0
0
Solving this system of equations, we find the eigenvector corresponding to λ2=
−1 is
1
2
0
.
Step 4: Finally, for λ3= 4, we solve the system (A−4I)v3= 0. This gives
us the matrix equation:
−1−1 0
0−2−1
001
x
y
z
=
0
0
0
7
Solving this system of equations, we find the eigenvector corresponding to λ3= 4
is
1
0
1
.
Question 8
Question
Let Abe a 3×3 matrix with eigenvalues λ=−2,1,3. Given that the eigenvector
corresponding to λ= 1 is v1=
1
0
0
and the eigenvector corresponding to λ= 3
is v2=
1
1
1
, determine the matrix A.
Solution
Step 1: Recall that any n×nmatrix Ais diagonalizable if it has nlinearly
independent eigenvectors. Since we are given three distinct eigenvalues with
corresponding eigenvectors, we can diagonalize A.
Step 2: The diagonal matrix Λ is defined as:
Λ =
−200
0 1 0
0 0 3
Step 3: Let Pbe the matrix formed by the eigenvectors of Aas its columns.
Thus, we have:
P=
11?
01?
01?
Step 4: To find the third column of P, we notice that the columns of Pmust
be linearly independent. Therefore, the third column of Pmust be a linear
combination of the given eigenvectors. Let the third column of Pbe v3=
a
b
c
.
Then, we have:
Av3=λv3
Av3= 3v3
A
a
b
c
= 3
a
b
c
8
Step 5: Solving A
a
b
c
= 3
a
b
c
gives:
−200
0 1 0
0 0 3
a
b
c
=
3a
3b
3c
Step 6: Solving the above equation, we get c= 0. Therefore, the third
column of Pand the matrix Abecomes:
P=
110
011
010
A=PΛP−1
Step 7: Compute P−1and then find A:
P−1=
1−1 0
0 1 −1
0 0 1
Step 8: Finally, we find:
A=
110
011
010
−200
0 1 0
0 0 3
1−1 0
0 1 −1
0 0 1
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 3, λ2=−2, and λ3= 1. Find a
basis for each eigenspace corresponding to these eigenvalues.
Solution
Step 1: To find the eigenvectors corresponding to each eigenvalue, we need to
solve the system of equations (A−λI)v=0.
For λ1= 3: Setting λ= 3 in (A−λI)v=0, we get:
(A−3I)v=
−1 1 2
000
1−1−1
x
y
z
=
0
0
0
Performing row operations, the reduced row-echelon form is:
1−1−1
0 0 0
0 0 0
9
So, the eigenvectors corresponding to λ= 3 are of the form
s
s
s
, where sis a
scalar.
Step 2: For λ2=−2: Setting λ=−2 in (A−λI)v=0, we get:
(A+ 2I)v=
512
020
1−1 1
x
y
z
=
0
0
0
Performing row operations, the reduced row-echelon form is:
101
010
000
So, the eigenvectors corresponding to λ=−2 are of the form
−s
0
s
, where sis
a scalar.
Step 3: For λ3= 1: Setting λ= 1 in (A−λI)v=0, we get:
(A−I)v=
−112
0−1 0
1−1 0
x
y
z
=
0
0
0
Performing row operations, the reduced row-echelon form is:
1−1 0
001
000
So, the eigenvectors corresponding to λ= 1 are of the form
s
s
0
, where sis a
scalar.
Thus, basis for eigenspace corresponding to λ1= 3:
1
1
1
Basis for eigenspace corresponding to λ2=−2:
−1
0
1
Basis for eigenspace corresponding to λ3= 1:
1
1
0
10
Question 10
Question
Let Abe a 3 ×3 matrix given by:
A=
2−1 0
−1 2 −1
0−1 2
Determine the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation:
det(A−λI)=0
where λis the eigenvalue we are solving for and Iis the identity matrix.
Step 2: Substitute the values into the equation and solve for λ:
det
2−λ−1 0
−1 2 −λ−1
0−1 2 −λ
= 0
Step 3: Expanding the determinant yields:
(2 −λ)3−2(2 −λ)−1=0
Step 4: Simplify the equation:
λ3−6λ2+ 11λ−6=0
Step 5: Solve the characteristic equation to find the eigenvalues. In this
case, the eigenvalues are λ= 1,2,3.
Step 6: We will now find the eigenvectors corresponding to each eigenvalue
by solving the system of equations (A−λI)x= 0.
For λ= 1: Substitute λ= 1 into A−λIand solve the system of equations:
1−1 0
−1 1 −1
0−1 1
x1
x2
x3
=
0
0
0
Step 7: The solutions to the system of equations for λ= 1 are:
x1=x2=x3
Thus, the eigenvector corresponding to λ= 1 is a scalar multiple of
1
1
1
.
For λ= 2 and λ= 3, follow the same steps to determine the eigenvectors.
11
Question 11
Question
Let Abe a 3 ×3 matrix given by
A=
1 2 1
0 2 0
−110
.
Find the eigenvectors and eigenvalues of A.
Solution
Step 1: To find the eigenvalues, we solve the characteristic equation |A−λI|= 0
where Iis the 3 ×3 identity matrix.
A−λI =
1−λ2 1
0 2 −λ0
−1 1 −λ
.
Step 2: Expanding the determinant gives
(1 −λ)((2 −λ)(−λ)−(0)(1)) −2(0 −(−1)(1)) + 1(0 −0) = 0.
Step 3: Simplifying the equation, we get
(1 −λ)(−2λ+λ2) + 2 = 0.
Step 4: Further simplification gives
λ3−3λ2+ 2λ+ 2 = 0.
Step 5: By solving the characteristic equation, we find that the eigenvalues
of Aare λ1= 2 and λ2= 1.
Step 6: To find the eigenvectors, we substitute each eigenvalue back into the
equation (A−λI)v=0.
Step 7: For λ= 2, we have
A−2I=
−1 2 1
000
−1 1 −2
.
Step 8: We row reduce to find that the eigenvector corresponding to λ= 2
is v1=
1
2
1
.
Step 9: For λ= 1, we have
A−I=
021
010
−1 1 −1
.
12
Step 10: We row reduce to find that the eigenvector corresponding to λ= 1
is v2=
−2
1
2
.
Therefore, the eigenvectors of Aare v1=
1
2
1
corresponding to λ= 2 and
v2=
−2
1
2
corresponding to λ= 1.
Question 12
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 2, and λ3= 3. Find the
eigenvectors corresponding to each eigenvalue.
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we need to solve the
equation (A−2I)v1=0, where v1is the eigenvector corresponding to λ1.
A−2I=
210
021
001
The augmented matrix formed by (A−2I)v1=0would be:
010|0
001|0
000|0
Step 2: Reduce the augmented matrix to row-echelon form:
010|0
001|0
000|0
→
010|0
001|0
000|0
Step 3: The solution to the system is x1= 0, x2= 0, and x3= 0. Therefore,
the eigenvector corresponding to λ1= 2 is v1=
0
0
0
.
Repeat Steps 1-3 for λ2= 2 and λ3= 3 to find the corresponding eigenvec-
tors.
13
Question 13
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 3, λ2= 1, and λ3=−2. If v1,
v2, and v3are the corresponding eigenvectors of A, find the matrix Pformed
by placing v1,v2, and v3as columns. Then, determine P−1AP .
Solution
Step 1: Form the matrix Pby arranging the eigenvectors v1,v2, and v3as
columns.
P=
| | |
v1v2v3
| | |
Step 2: Since v1,v2, and v3are eigenvectors of A, we know that Avi=λivi
for i= 1,2,3. Therefore, we have:
Av1= 3v1
Av2=v2
Av3=−2v3
Step 3: By substituting these relations into the definition of P−1AP , we
have:
P−1AP =
| | |
v1v2v3
| | |
−1
3 0 0
0 1 0
0 0 −2
| | |
v1v2v3
| | |
Step 4: We can calculate P−1by using the formula P−1=1
det(P)adj(P),
where adj(P) is the adjugate of P.
P−1=1
det(P)adj(P)
Step 5: Calculate the determinant of Pto find 1
det(P).
det(P) = v1·(v2×v3)
Step 6: Compute the adjugate of P.
adj(P) =
|||
v2×v3v3×v1v1×v2
|||
Step 7: Substitute the values of det(P) and adj(P) into P−1to find P−1.
Step 8: Finally, calculate P−1AP using the formula derived in Step 3.
14
Question 14
Question
Let Abe a 3 ×3 matrix given by
A=
211
131
112
.
Find the eigenvalues and eigenvectors of A. Normalize the eigenvectors.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0, where λis the eigenvalue and Iis the identity matrix.
det
211
131
112
−λ
100
010
001
= 0
det
2−λ1 1
1 3 −λ1
1 1 2 −λ
= 0
(2 −λ)[(3 −λ)(2 −λ)−1] −(1)[(1)(2 −λ)−(1)(1)] + (1)[(1)(1) −(1)(3 −λ)] = 0
Solving this equation gives us the eigenvalues.
Step 2: Now that we have found the eigenvalues, we can find the correspond-
ing eigenvectors by solving the equation (A−λI)x=0for each eigenvalue.
Step 3: Normalize the eigenvectors by dividing each by its magnitude.
Question 15
Question
Let A=4 6
3 5. Find the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we need to solve the characteristic equation
|A−λI|= 0, where λis the eigenvalue and Iis the identity matrix.
A−λI=4 6
3 5−λ1 0
0 1=4−λ6
3 5 −λ
Setting the determinant of A−λIto zero, we have:
4−λ6
3 5 −λ
= (4 −λ)(5 −λ)−18 = λ2−9λ+ 2 = 0
15
Step 2: Solve the quadratic equation λ2−9λ+ 2 = 0 to find the eigenvalues
λ. Using the quadratic formula λ=−(−9)±√(−9)2−4∗1∗2
2∗1, we get:
λ=9±√81 −8
2=9±√73
2
So, the eigenvalues are λ1=9+√73
2and λ2=9−√73
2.
Step 3: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute each eigenvalue back into the equation (A−λI)v=0, where vis the
eigenvector. For λ=9+√73
2, we have:
−√73
26
3−√73
2!x
y=0
0
Solving the system of equations, we get the eigenvector corresponding to λ1.
For λ=9−√73
2, we have:
√73
26
3√73
2!x
y=0
0
Solving the system of equations, we get the eigenvector corresponding to λ2.
Question 16
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3=−1. If
B= 3A−2I, where Iis the 3 ×3 identity matrix, find the eigenvalues of B.
Solution
Step 1: Find the eigenvalues of B. Given B= 3A−2I, the eigenvalues of B
are related to the eigenvalues of Aby the following equation:
det(B−λiI) = det(3A−2I−λiI)=0
where λiare the eigenvalues of A.
Step 2: Substitute the eigenvalues of Ainto the equation. Substitute the
eigenvalues of Ainto the equation to find the eigenvalues of B:
det(3A−2I−2I) = det(3A−4I) = 0
det(3A−2I−3I) = det(3A−5I) = 0
det(3A−2I+I) = det(3A−I)=0
16
Step 3: Find the eigenvalues of B. The eigenvalues of Bare the solutions to
the equations derived in Step 2:
λ′
1= 3(2) −1=5
λ′
2= 3(3) −1=8
λ′
3= 3(−1) −1 = −4
Therefore, the eigenvalues of matrix Bare λ′
1= 5, λ′
2= 8, and λ′
3=−4.
Question 17
Question
Let Abe a 3 ×3 matrix with eigenvalues −2,1,and 3. Determine the corre-
sponding eigenvectors for each eigenvalue.
Solution
Step 1: For eigenvalue λ=−2: Let v1be the eigenvector corresponding to
λ=−2. We solve the system (A+ 2I)v1=0, where Iis the 3 ×3 identity
matrix.
(A+ 2I)v1=
−200
0−1 0
0 0 1
v1=
0
0
0
Step 2: Solve for v1: This system simplifies to:
−2v11= 0
−v12= 0
v13= 0
So, the eigenvector corresponding to λ=−2 is v1=
0
0
1
.
Step 3: For eigenvalue λ= 1: Let v2be the eigenvector corresponding to
λ= 1. We solve the system (A−I)v2=0.
(A−I)v2=
−300
0 0 0
0 0 2
v2=
0
0
0
Step 4: Solve for v2: This system simplifies to:
−3v21= 0
2v23= 0
17
So, the eigenvector corresponding to λ= 1 is v2=
0
1
0
.
Step 5: For eigenvalue λ= 3: Let v3be the eigenvector corresponding to
λ= 3. We solve the system (A−3I)v3=0.
(A−3I)v3=
−5 0 0
0−2 0
0 0 −2
v3=
0
0
0
Step 6: Solve for v3: This system simplifies to:
−5v31= 0
−2v32= 0
−2v33= 0
So, the eigenvector corresponding to λ= 3 is v3=
0
0
1
.
Question 18
Question
Let Abe a 4 ×4 matrix given by
A=
2−1 0 0
1 2 −1 0
0 1 2 −1
0 0 1 2
.
Determine the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues, we need to solve the characteristic equation
det(A−λI) = 0 where λis the eigenvalue.
det(A−λI) = det
2−λ−1 0 0
1 2 −λ−1 0
0 1 2 −λ−1
0 0 1 2 −λ
= 0.
18
Step 2: Simplify the determinant using row operations.
= (2 −λ)det
2−λ−1 0
1 2 −λ−1
0 1 2 −λ
= (2 −λ)(2 −λ)det 2−λ−1
1 2 −λ−(−1)det 1−1
0 2 −λ
= (2 −λ)[(2 −λ)((2 −λ)(2 −λ)−(−1)(1)) −(−1)(2 −λ)]
= (2 −λ)[(2 −λ)((4 −4λ+λ2) + 1) −(−2 + λ)]
= (2 −λ)[(2 −λ)(5 + 4λ−4−4λ+λ2) + λ−2]
= (2 −λ)[(2 −λ)(λ2+ 1) + λ−2]
= (2 −λ)[(2 −λ)λ2+ 2 −λ2+λ−2]
= (2 −λ)(2λ2−2λ3+λ−λ2)
= 2λ2−2λ3+λ−λ2−2λ2+ 2λ3
=−λ2+λ.
Step 3: Set the characteristic equation −λ2+λ= 0 and solve for the eigen-
values.
−λ2+λ= 0 =⇒λ(−λ+ 1) = 0.
So, λ= 0 or λ= 1.
Step 4: To find the eigenvectors, we substitute each eigenvalue back into the
equation (A−λI)x= 0 and solve for x.
For λ= 0, we solve:
Ax= 0 =⇒
2−1 0 0
1 2 −1 0
0 1 2 −1
0 0 1 2
x1
x2
x3
x4
=
0
0
0
0
.
This system of equations leads to x1=x2=x3=x4, so the eigenvector
corresponding to λ= 0 is
1
1
1
1
.
For λ= 1, we solve:
Ax
Question 19
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 2,3,and 4. If B= 2A−3I, where
Iis the 3 ×3 identity matrix, find the eigenvalues of B.
19
Solution
Given that Ahas eigenvalues λ= 2, 3, and 4, we know that the characteristic
polynomial of Ais given by
det(A−λI) = 0
which can be expanded as
det
a11 −λ a12 a13
a21 a22 −λ a23
a31 a32 a33 −λ
= 0
Substitute λ= 2, λ= 3, and λ= 4 into the characteristic equation to get three
equations. By solving these equations, we can find the matrix A.
Step 1: Find the matrix A using the eigenvalues. For λ= 2:
det
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
= 0
For λ= 3:
det
a11 −3a12 a13
a21 a22 −3a23
a31 a32 a33 −3
= 0
For λ= 4:
det
a11 −4a12 a13
a21 a22 −4a23
a31 a32 a33 −4
= 0
Solving these equations will give us the matrix A.
Step 2: Find matrix B. Given B= 2A−3I, we can find Bonce we know
matrix A.
Step 3: Find the eigenvalues of matrix B. To find the eigenvalues of
matrix B, we can use the relationship between eigenvalues and similar matrices.
Since B= 2A−3I, the eigenvalues of Bwill be λB= 2λA−3, where λAare
the eigenvalues of matrix A.
Therefore, the eigenvalues of matrix Bwill be 2(2) −3 = 1, 2(3) −3 = 3,
and 2(4) −3 = 5.
Question 20
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 4, λ2= 2,and λ3=−1. If
B=A2−3A+ 5I, where Iis the 3 ×3 identity matrix, find the eigenvalues of
B.
20
Solution
Given that Ahas eigenvalues λ1= 4, λ2= 2,and λ3=−1, we have that
the eigenvalues of A2,−3A, and 5Iare (λi)2,−3λi, and 5 respectively, where
i∈ {1,2,3}.
Step 1: Find the eigenvalues of A2. Since Ahas eigenvalues λ1= 4, λ2= 2,
and λ3=−1, the eigenvalues of A2are λ2
1= 16, λ2
2= 4,and λ2
3= 1.
Step 2: Find the eigenvalues of −3A. Since Ahas eigenvalues λ1= 4, λ2=
2,and λ3=−1, the eigenvalues of −3Aare −3λ1=−12,−3λ2=−6,and
−3λ3= 3.
Step 3: Find the eigenvalues of 5I. Since Iis the identity matrix of size
3×3, the eigenvalues of 5Iare all equal to 5.
Step 4: Find the eigenvalues of B. To find the eigenvalues of B=A2−
3A+ 5I, we add the corresponding eigenvalues of A2,−3A, and 5I. Thus, the
eigenvalues of Bare 16 −12 + 5 = 9,4−6 + 5 = 3,and 1 + 3 + 5 = 9.
Therefore, the eigenvalues of Bare λ1= 9, λ2= 3,and λ3= 9.
Question 21
Question
Let A=3 2
2 6. Find the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation |A−λI|= 0 where Iis the identity matrix. The characteristic equation
is given by:
det 3−λ2
2 6 −λ= 0
Step 2: Calculating the determinant, we get:
(3 −λ)(6 −λ)−2·2 = 0
Step 3: Simplifying the equation, we have:
λ2−9λ+ 16 = 0
Step 4: Solving the quadratic equation, we find the eigenvalues:
λ=9±√92−4·1·16
2=9±√81 −64
2
λ=9±√17
2
21
Step 5: Therefore, the eigenvalues of matrix Aare λ1=9+√17
2and λ2=
9−√17
2.
Step 6: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute the eigenvalues back into the equation (A−λI)X= 0, where Xis the
eigenvector.
Step 7: For λ1=9+√17
2, subtract λ1from the diagonal of matrix Aand
form the system of equations:
−√17
22
2−√17
2!x
y=0
0
Step 8: Solving this system of equations yields the eigenvector corresponding
to λ1.
Step 9: For λ2=9−√17
2, follow the same process to find the corresponding
eigenvector.
Step 10: Therefore, the eigenvectors for λ1and λ2are the solutions obtained
in Step 8 and Step 9 respectively.
Question 22
Question
Let Abe a 3x3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. If
the eigenvectors corresponding to λ1and λ2are v1=
1
1
0
and v2=
1
0
1
respectively, find the matrix A.
Solution
Step 1: Recall that a matrix Acan be diagonalized by a matrix Pif A=P DP −1
where Dis a diagonal matrix with the eigenvalues of Aand the columns of P
are the eigenvectors of A.
Step 2: Since we are given the eigenvalues and eigenvectors of A, we can
construct the matrix Pas follows:
P= [v1,v2,v3] =
1 1 a
1 0 b
0 1 c
Step 3: To find the third eigenvector v3, we use the fact that the columns
of Pmust be linearly independent. Thus, v3is orthogonal to v1and v2and
satisfies Av3=λ3v3.
Step 4: Since v3is orthogonal to v1and v2, we can solve the system of
equations:
1 1 a
1 0 b
0 1 c
x
y
z
= 4
x
y
z
22
Step 5: Solving the system of equations yields a=−1, b= 1, and c= 1.
Therefore, the matrix Pis:
P=
1 1 −1
1 0 1
0 1 1
Step 6: Now, we can compute P−1using the inverse of a 3x3 matrix formula:
P−1=1
det(P)adj(P)
where det(P) is the determinant of Pand adj(P) is the adjugate of P.
Step 7: Calculate det(P) = −1 and adj(P) is found by taking the transpose
of the matrix of cofactors of P.
Step 8: Using the formula for P−1, we find:
P−1=
0 1 −1
1 1 −1
−1−1 2
Step 9: Finally, we can determine the matrix Aby the formula A=P DP −1.
Substituting the given eigenvalues into the diagonal matrix D, we have:
D=
200
0−1 0
004
Step 10: Calculate A=P DP −1:
A=
1 1 −1
1 0 1
0 1 1
200
0−1 0
004
0 1 −1
1 1 −1
−1−1 2
Question 23
Question
Consider the matrix A=3 1
1 3. Find the eigenvalues and corresponding
eigenvectors of matrix A.
Solution
To find the eigenvalues of matrix A, we need to solve the characteristic equation
|A−λI|= 0, where Iis the identity matrix.
Step 1: Find the characteristic equation The characteristic equation
is given by:
det 3−λ1
1 3 −λ= 0
23
Expanding the determinant, we get:
(3 −λ)2−1=0
Simplifying, we find:
λ2−6λ+ 8 = 0
Step 2: Find the eigenvalues Solving the characteristic equation λ2−
6λ+ 8 = 0, we get:
(λ−2)(λ−4) = 0
So the eigenvalues are λ1= 2 and λ2= 4.
Step 3: Find the eigenvectors To find the eigenvector corresponding to
λ= 2, substitute λ= 2 back into the equation (A−λI)v=0and solve for v:
3−2 1
1 3 −2x
y=0
0
This simplifies to the equations x+y= 0.
So, when x= 1, y=−1, and vice versa, the eigenvector corresponding to
λ= 2 is v1=1
−1.
Similarly, for λ= 4, we find the eigenvector v2=1
1.
Question 24
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 3, and λ3= 4. If v1,v2,
and v3are the corresponding eigenvectors, find the matrix A.
Solution
Step 1: We know that if Ais a square matrix with eigenvalues λ1, λ2, . . . , λn
and corresponding eigenvectors v1, v2, . . . , vn, then we can write Ain terms of
its eigenvectors and eigenvalues.
Step 2: The matrix Acan be expressed as A=P DP −1, where Pis the
matrix whose columns are the eigenvectors v1, v2, . . . , vnand Dis a diagonal
matrix with the eigenvalues λ1, λ2, . . . , λnon the diagonal.
Step 3: In this case, since Ais a 3 ×3 matrix, we will have P= [v1v2v3]
and D=
λ10 0
0λ20
0 0 λ3
.
Step 4: Substituting the given eigenvalues and eigenvectors into the expres-
sion for A, we have:
A=P DP −1=v1v2v3
100
030
004
v1v2v3−1
24
Step 5: Next, we compute P−1, which is the inverse of the matrix whose
columns are the eigenvectors v1, v2, v3. Let P−1= [w1w2w3].
Step 6: Therefore, the matrix Acan be calculated as:
A=P DP −1=v1v2v3
100
030
004
w1w2w3
Question 25
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 3.
If the eigenvector corresponding to λ1is v1=
1
0
−1
, find the eigenvectors
corresponding to the eigenvalues λ2and λ3.
Solution
1. Let v2and v3be the eigenvectors corresponding to the eigenvalues λ2and
λ3respectively. Since λ1= 2 is an eigenvalue with eigenvector v1=
1
0
−1
,
we have the following equation:
Av1=λ1v1
2. Substituting the values, we get:
A
1
0
−1
= 2
1
0
−1
This simplifies to:
a b c
d e f
g h i
1
0
−1
=
2
0
−2
3. Solving the matrix equation yields:
a−c= 2
d−f= 0
g+i=−2
25
4. Since λ2=−1 is an eigenvalue, we can set up a similar equation for v2:
Av2=λ2v2
Substituting the values and solving for v2will give the eigenvector corre-
sponding to λ2.
5. Similarly, since λ3= 3 is an eigenvalue, we can set up an equation for v3:
Av3=λ3v3
Substituting the values and solving for v3will give the eigenvector corre-
sponding to λ3.
Question 26
Question
Let’s consider the matrix A=4 1
−2−3. Find the eigenvalues and eigenvectors
of A. Then, use these eigenvalues and eigenvectors to diagonalize Aif possible.
Solution
Step 1: To find the eigenvalues of A, we need to solve the characteristic equation
|A−λI|= 0.
|A−λI|=
4−λ1
−2−3−λ
= (4 −λ)(−3−λ)−(−2)(1) = λ2−λ−14 = 0
Solving the quadratic equation, we find the eigenvalues:
λ=1±√1+4×14
2=1±√57
2
Step 2: Next, we find the corresponding eigenvectors. For each eigenvalue,
we solve the system of equations (A−λI)v= 0. For λ=1+√57
2:
"4−1+√57
21
−2−3−1+√57
2#x
y=0
0
"7−√57
21
−2−5−√57
2#x
y=0
0
Solving this system, we find the eigenvector corresponding to λ=1+√57
2.
Step 3: Similarly, for λ=1−√57
2, find the eigenvector.
26
Step 4: If the matrix Ais diagonalizable, we can form the matrix Pusing
the eigenvectors as columns.
P=eigenvector 1 eigenvector 2
Step 5: Then, the diagonal matrix Dis given by:
D=P−1AP
Step 6: If Ais diagonalizable, then A=P DP −1. Otherwise, Ais not
diagonalizable.
Question 27
Question
Let Abe a 3 ×3 matrix given by
A=
123
021
111
.
Determine the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues, we solve the characteristic equation |A−λI|= 0,
where Iis the identity matrix.
|A−λI|=
1−λ2 3
0 2 −λ1
1 1 1 −λ
= (1 −λ)
2−λ1
1 1 −λ−2
0 1
1 1 −λ
+ 3
0 2
1 1
= (1 −λ)((2 −λ)(1 −λ)−1) −2(−1) + 3(−2)
= (1 −λ)(λ2−3λ+ 1) + 2 −6
=−λ3+ 4λ2−2λ+ 1.
Step 2: Setting |A−λI|= 0, we find the eigenvalues of Aby solving the
equation −λ3+ 4λ2−2λ+ 1 = 0. This is a challenging polynomial to factor
by hand, so we will use numerical methods or software to find the roots. The
eigenvalues are approximately λ1≈0.277, λ2≈1.571, and λ3≈4.152.
Step 3: Next, we find the eigenvectors corresponding to each eigenvalue by
solving (A−λI)v= 0, where vis the eigenvector. For λ= 0.277: We solve
(A−0.277I)v1= 0:
0.723 2 3
0 1.723 1
1 1 0.723
x
y
z
=
0
0
0
.
27
Solving this system of equations, we find v1≈
−0.754
0.342
1
.
Repeat the above process for the other two eigenvalues to find their corre-
sponding eigenvectors.
Question 28
Question
Let Abe a 3 ×3 matrix with eigenvalues 2 and −1. If one of the eigenvectors
corresponding to the eigenvalue 2 is
1
0
1
, find the eigenvectors corresponding
to the eigenvalue −1.
Solution
Step 1: Recall that eigenvectors of a matrix Aare the non-zero vectors vsuch
that Av =λv, where λis the corresponding eigenvalue.
Step 2: Since λ=−1 is an eigenvalue of A, we have Av=−vfor some
non-zero vector v.
Step 3: Let v=
x
y
z
be the eigenvector corresponding to the eigenvalue
−1. Substituting into Av=λv, we get:
a b c
d e f
g h i
x
y
z
=−
x
y
z
Step 4: This equation can be expanded to form a system of equations:
ax +by +cz =−x
dx +ey +fz =−y
gx +hy +iz =−z
Step 5: Using the fact that
1
0
1
is an eigenvector corresponding to the
eigenvalue 2, we know A
1
0
1
= 2
1
0
1
. This gives us another equation to use.
28
Step 6: By substituting
1
0
1
into Av=λvfor the eigenvalue 2, we get:
a b c
d e f
g h i
1
0
1
= 2
1
0
1
Step 7: Solving this equation, we find further information about the matrix
Athat can be used to solve for the eigenvector corresponding to the eigenvalue
−1.
Step 8: Once we determine the matrix A, we can solve the system of equa-
tions from Step 4 to find the eigenvector corresponding to the eigenvalue −1.
Question 29
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2 and λ2= 3. If the eigenvector
corresponding to λ2= 3 is v2=
1
0
1
, find the eigenvector corresponding to
λ1= 2.
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we will use the eigen-
value equation Av=λv, where vis the eigenvector we want to find.
Step 2: In general, for a matrix Aand its eigenvalues λ1, λ2, . . . , λn, the
eigenvector corresponding to each eigenvalue can be found by solving (A−
λI)v=0, where Iis the identity matrix of the same size as A.
Step 3: Substituting the given eigenvalues into the equation, we have: (A−
2I)v=0.
Step 4: Given Ais a 3 ×3 matrix, we write the expression explicitly:
a11 a12 a13
a21 a22 a23
a31 a32 a33
−2
100
010
001
x
y
z
=
0
0
0
.
Step 5: Simplifying the above equation, we get:
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
x
y
z
=
0
0
0
.
29
Step 6: Since the matrix
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
is singular (has de-
terminant zero), the eigenvector
x
y
z
can be determined by solving the system
of equations formed by setting up the matrix equation equal to 0.
Step 7: Solving the system of equations obtained from Step 5 will yield the
eigenvector v1=
0
1
0
corresponding to the eigenvalue λ1= 2.
Question 30
Question
Let Abe a 3×3 matrix with eigenvalues λ= 1,2,3. We are given that the eigen-
vectors corresponding to λ= 1 and λ= 3 are
1
1
1
and
−1
0
1
, respectively.
Find the eigenvector corresponding to λ= 2.
Solution
The eigenvectors corresponding to distinct eigenvalues of a matrix are
linearly independent.
Since we are given that the eigenvectors corresponding to λ= 1 and
λ= 3 are linearly independent, we can use them to find the eigenvector
corresponding to λ= 2.
Step 1:
Let the eigenvector corresponding to λ= 2 be
a
b
c
. Since this eigenvector is
independent from the others, it must satisfy the equation A
a
b
c
= 2
a
b
c
.
Step 2:
Thus, we have the equation
1−1 1
100
111
a
b
c
= 2
a
b
c
which simplifies to
a−b+c
a
a+b+c
=
2a
2b
2c
30
Step 3:
Solving the system of equations gives us
a−b+c= 2a, a = 2b, a +b+c= 2c
Step 4:
From the second equation, we have a= 2b. Substituting this into the first and
third equations gives us b=−cand a= 2b=−2c. Hence, a possible eigenvector
corresponding to λ= 2 is
−2
1
−1
.
Therefore, the eigenvector corresponding to λ= 2 is
−2
1
−1
.
Question 31
Question
Let Abe a 3×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Determine
the eigenvalues of the matrix 2A−A2.
Solution
Given that Ais a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3, we
need to find the eigenvalues of the matrix 2A−A2.
Step 1: Calculate the matrix 2ATo find the matrix 2A, we multiply the
scalar 2 by each element of A:
2A= 2 ·
a b c
d e f
g h i
=
2a2b2c
2d2e2f
2g2h2i
Step 2: Calculate the matrix A2To find the matrix A2, we multiply A
by itself:
A2=A·A=
a b c
d e f
g h i
·
a b c
d e f
g h i
=
a2+bd +cg ab +be +ci ac +bf +ci
ad +de +fg bd +e2+fh cd +ef +hi
ag +dh +gi bg +eh +gi cg +fh +i2
Step 3: Calculate the matrix 2A−A2Subtracting A2from 2A, we get:
2A−A2=
2a2b2c
2d2e2f
2g2h2i
−
a2+bd +cg ab +be +ci ac +bf +ci
ad +de +fg bd +e2+fh cd +ef +hi
ag +dh +gi bg +eh +gi cg +fh +i2
=
2a−(a2+bd +cg) 2b−(ab +be +ci) 2c−(ac +bf +ci)
2d−(ad +de +fg) 2e−(bd +e2+f h) 2f−(cd +ef +hi)
2g−(ag +dh +gi) 2h−(bg +eh +gi) 2i−(cg +f h +i2)
Step 4: Determine the eigenvalues of 2A−A2From the expression
above, we see that the eigenvalues of 2A−A2are given by subtracting the
31
corresponding terms of A2from 2A. Therefore, the eigenvalues of 2A−A2are
2−12= 1, 2 −22=−2, and 2 −32=−7. Thus, the eigenvalues of 2A−A2are
1, −2, and −7.
Question 32
Question
Let Abe a 3 ×3 matrix given by
A=
210
120
001
.
Find the eigenvectors corresponding to the eigenvalues of A.
Solution
Step 1: Find the eigenvalues of the matrix Aby solving the characteristic equa-
tion |A−λI|= 0, where λis the eigenvalue and Iis the identity matrix.
|A−λI|=
2−λ1 0
1 2 −λ0
0 0 1 −λ
= (2 −λ)((2 −λ)(1 −λ)−0) −(1)((1)(1 −λ)−0)
= (2 −λ)(2 −3λ+λ2)−(1)(1 −λ)
= (2 −λ)(2 −3λ+λ2)−1 + λ
= 4 −6λ+ 2λ2−2λ+ 3λ2−λ3−1 + λ
=−λ3+ 5λ2−7λ+ 3
=−(λ−1)(λ−1)(λ−3) = 0
Solving λ−1 = 0, we get λ1= 1.
Solving λ−3 = 0, we get λ2= 3 (multiplicity 2).
Step 2: For each eigenvalue, find the eigenvectors by solving the system of
equations (A−λI)v=0. For λ= 1, we have (A−λI)v=
110
110
000
x
y
z
=
0
0
0
. Solving x+y= 0 gives x=−y. Let y=t, then the eigenvector
corresponding to λ= 1 is v1=t
−1
1
0
.
32
For λ= 3, we have (A−λI)v=
−1 1 0
1−1 0
0 0 −2
x
y
z
=
0
0
0
. Solving
−x+y= 0 and −x+y= 0 gives x=y. Let y=t, then the eigenvector
corresponding to λ= 3 is v2=t
1
1
0
.
Thus, the eigenvectors corresponding to the eigenvalues of matrix Aare
v1=t
−1
1
0
for λ= 1 and v2=t
1
1
0
for λ= 3.
Question 33
Question
Let Abe a 3×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Determine
the eigenvalues of the matrix B= 2A2−3A+5I, where Iis the identity matrix.
Solution
To find the eigenvalues of matrix B, we can use the properties of eigenvalues
and eigenvectors.
Step 1: Express Bin terms of A
First, express matrix Bin terms of matrix A:
B= 2A2−3A+ 5I
Step 2: Find the eigenvalues of B
We know that the eigenvalues of A2are the squares of the eigenvalues of A.
Therefore, the eigenvalues of A2are λ2
1= 1, λ2
2= 4, and λ2
3= 9. Similarly,
since Ahas eigenvalues λ1= 1, λ2= 2, and λ3= 3, we can find the eigenvalues
of B:
Eigenvalues of B= 2λ2
i−3λi+ 5, for i= 1,2,3
Calculating this for each eigenvalue λi, we get:
Eigenvalues of B={2(1)2−3(1) + 5,2(4) −3(2) + 5,2(9) −3(3) + 5}
Eigenvalues of B={4,6,14}
Therefore, the eigenvalues of matrix Bare 4, 6, and 14.
33
Question 34
Question
Let Abe a 3 ×3 matrix given by
A=
1 2 0
2−1 2
0 2 −1
.
Find the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation given by det(A−λI) = 0, where λis the eigenvalue and Iis the identity
matrix.
det
1−λ2 0
2−1−λ2
0 2 −1−λ
= 0.
Simplifying the determinant, we get
1−λ2 0
2−1−λ2
0 2 −1−λ
= (1−λ)((−1−λ)(−1−λ)−4)−2(−2(−1−λ)−0) = 0.
This leads to the characteristic equation −λ3−λ2+ 6λ−7 = 0.
Step 2: We need to find the roots of the characteristic equation, which will
give us the eigenvalues of matrix A. The roots can be found by inspection or by
using numerical methods. By inspection, we can see that λ= 1 is a root. By
polynomial division, we can find that (λ−1)(−λ2−2λ+ 7) = 0, which gives us
the other two roots λ=−1 and λ= 7.
Step 3: Now, we will find the eigenvectors associated with each eigenvalue.
For λ= 1: Solve the system (A−λI)X= 0 to find the eigenvector.
(A−I)X=
0 2 0
2−2 2
0 2 −2
X= 0.
By row-reducing the augmented matrix [A−I|0], we find the solution X1=
t
1
1
1
, where tis a scalar.
Step 4: For λ=−1: Solve the system (A+I)X= 0 to find the eigenvector.
(A+I)X=
220
202
020
X= 0.
34
By row-reducing the augmented matrix [A+I|0], we find the solution X−1=
s
−1
1
1
, where sis a scalar.
Step 5: For λ= 7: Solve the system (A−7I)X= 0 to find the eigenvector.
(A−7I)X=
−6 2 0
2−8 2
0 2 −8
X= 0.
By row-reducing the augmented matrix [A−7I|0], we find the solution X7=
r
1
2
1
, where ris a scalar.
Question 35
Question
Let Abe a 3 ×3 matrix defined as
A=
5−2 1
1 0 2
−114
Find the eigenvalues and eigenvectors of the matrix A.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation |A−λI|= 0, where λrepresents the eigenvalues and Iis the identity
matrix of the same size as A.
Starting with the matrix A:
A=
5−2 1
1 0 2
−114
Subtracting λtimes the identity matrix:
A−λI =
5−λ−2 1
1−λ2
−1 1 4 −λ
Calculating the determinant:
|A−λI|= (5−λ) [(−λ)(4 −λ)−(1)(1)]−(−2) [(1)(4 −λ)−(−1)(1)]+(1) [(1)(1) −(−λ)(−2)]
|A−λI|= (5 −λ)(λ2−4λ−1) + 2(4 −λ−1) + (1)(1 + 2λ)
35
|A−λI|=λ3−9λ2+ 22λ−13
Setting |A−λI|= 0 and solving for λ:
λ3−9λ2+ 22λ−13 = 0
Solving this cubic equation, we find the eigenvalues.
Step 2: To find the eigenvectors associated with each eigenvalue, we sub-
stitute the eigenvalues back into the equation (A−λI)v= 0 and solve for
v.
Let’s say the eigenvalues are λ1, λ2, λ3. For each λi, we solve (A−λiI)v= 0
to find the corresponding eigenvector vi.
36
