MATH 334 - DIFFERENTIAL
EQUATIONS - Matrix methods and
eigenvalue problems
Question Bank - Set 3
Liberty University
Question 1
Question
Let Abe a 3×3 matrix with eigenvalues −1, 2, and 4. Given that the eigenvector
corresponding to the eigenvalue −1 is
1
0
1
, find the eigenvectors corresponding
to the eigenvalues 2 and 4.
Solution
To find the eigenvectors corresponding to the eigenvalues 2 and 4, we can use
the eigenvalue equation Av=λv, where λis an eigenvalue of Aand vis the
corresponding eigenvector.
Step 1: Find eigenvector corresponding to eigenvalue 2 Let v2=
x
y
z
be the eigenvector corresponding to the eigenvalue 2. Substituting into
the eigenvalue equation gives:
Av2= 2v2
−101
2 0 1
1 0 0
x
y
z
= 2
x
y
z
Solving the system of equations, we get:
−x+z= 2x
2x+z= 2y
x= 2z
Solving this system of equations gives x= 2, y= 0, and z= 1. Therefore,
the eigenvector corresponding to the eigenvalue 2 is
2
0
1
.
Step 2: Find eigenvector corresponding to eigenvalue 4 Let v4=
p
q
r
be the eigenvector corresponding to the eigenvalue 4. Substituting into
the eigenvalue equation gives:
Av4= 4v4
−101
2 0 1
1 0 0
p
q
r
= 4
p
q
r
Solving the system of equations gives p=−2qand r= 3q. Therefore, the
eigenvector corresponding to the eigenvalue 4 is
−2
1
3
.
Question 2
Question
Let Abe a 3 ×3 matrix given by
A=
210
121
012
.
Find the eigenvalues and corresponding eigenvectors of matrix A.
Solution
To find the eigenvalues of matrix A, we need to solve the characteristic equation
det(A−λI) = 0, where Iis the identity matrix.
Step 1: Find the characteristic equation We have:
det(A−λI) = det
2−λ1 0
1 2 −λ1
0 1 2 −λ
= 0.
Expanding this determinant gives us the characteristic equation:
(2 −λ)((2 −λ)2−1) −(1)((2 −λ)−0) = 0.
Simplifying this equation gives us:
λ3−6λ2+ 9λ−3=0.
2
Step 2: Find the eigenvalues To find the eigenvalues, we solve the char-
acteristic equation λ3−6λ2+ 9λ−3 = 0. By inspection, we can see that λ= 1
is a root of this polynomial.
Dividing the characteristic equation by (λ−1) gives us:
λ2−5λ+ 3 = 0.
Using the quadratic formula to solve this equation, we get λ=5±√13
2.
Therefore, the eigenvalues of matrix Aare λ= 1,5+√13
2,5−√13
2.
Step 3: Find the eigenvectors Now, we find the eigenvectors correspond-
ing to each eigenvalue.
For λ= 1, we solve the system (A−I)v= 0:
110
111
011
v= 0.
Solving this system gives us one independent eigenvector v1=
1
−1
1
.
For λ=5+√13
2, we solve the system (A−5+√13
2I)v= 0 to find another
eigenvector v2.
For λ=5−√13
2, we solve the system (A−5−√13
2I)v= 0 to find the third
eigenvector v3.
Therefore, the eigenvalues and eigenvectors of matrix Aare: λ= 1 with
eigenvector v1=
1
−1
1
,λ=5+√13
2with eigenvector v2, and λ=5−√13
2with
eigenvector v3.
Question 3
Question
Let Abe a 3 ×3 matrix given by
A=
5 2 −1
222
−1 2 5
.
Find the eigenvalues and eigenvectors of matrix A.
3
Solution
Step 1: To find the eigenvalues of matrix A, we solve the characteristic equation
|A−λI|= 0, where λis the eigenvalue and Iis the 3 ×3 identity matrix.
|A−λI|=
5−λ2−1
2 2 −λ2
−1 2 5 −λ
.
Expanding the determinant, we get
(5 −λ)((2 −λ)(5 −λ)−4) −2(2(5 −λ)−2) + (−1)(2 ·2−2) = 0.
Simplifying the equation, we have
λ3−12λ2+ 39λ−28 = 0.
This equation can be factored as (λ−4)(λ−2)(λ−7) = 0.
Thus, the eigenvalues are λ1= 4, λ2= 2,and λ3= 7.
Step 2: Next, we find the eigenvectors corresponding to each eigenvalue by
substituting them back into the equation (A−λI)v= 0 and solving for the
vector v.
For λ1= 4, we solve (A−4I)v= 0:
1 2 −1
2−2 2
−1 2 1
v=
0
0
0
.
By row reducing the augmented matrix, the solution is v1=
1
1
1
.
For λ2= 2, we solve (A−2I)v= 0:
3 2 −1
202
−1 2 3
v=
0
0
0
.
By row reducing the augmented matrix, the solution is v2=
1
−1
1
.
For λ3= 7, we solve (A−7I)v= 0:
−2 2 −1
2−5 2
−1 2 −2
v=
0
0
0
.
By row reducing the augmented matrix, the solution is v3=
1
1
−1
.
Therefore, the eigenvalues of matrix Aare 4, 2, and 7, with corresponding
eigenvectors
1
1
1
,
1
−1
1
, and
1
1
−1
, respectively.
4
Question 4
Question
Let Abe a 3 ×3 matrix given by
A=
2 0 1
1 2 1
−101
.
Find the eigenvalues and eigenvectors of A.
Solution
Step 1: Find the characteristic equation The characteristic equation of a
square matrix Ais given by det(A−λI) = 0, where Iis the identity matrix.
For matrix Ain this case,
A−λI =
2−λ0 1
1 2 −λ1
−1 0 1 −λ
.
Thus, the characteristic equation becomes
det(A−λI) = det
2−λ0 1
1 2 −λ1
−1 0 1 −λ
= 0.
Step 2: Solve the characteristic equation Expanding the determinant
above, we get
(1 −λ)((2 −λ)(1 −λ)−0·0) −0−1(−1(1 −λ)−0) = 0
⇒(1 −λ)((2 −λ)(1 −λ) + 1(1 −λ)) = 0
⇒(1 −λ)(2 −λ)(1 −λ+ 1) = 0
⇒(1 −λ)(2 −λ)(2 −λ)=0.
Solving this equation gives the eigenvalues λ1= 1 and λ2= 2 with algebraic
multiplicity 2 and 1, respectively.
Step 3: Find the eigenvectors To find the eigenvector corresponding to
λ= 1: Substitute λ= 1 into the matrix A−λI,
A−λI =
1 0 1
1 1 1
−100
.
Reducing the matrix to row echelon form gives the eigenvector corresponding
to λ= 1 as
1
−1
1
.
5
For the eigenvalue λ= 2: Substitute λ= 2 into the matrix A−λI,
A−λI =
001
101
−1 0 −1
.
Reducing the matrix to row echelon form results in the eigenvector correspond-
ing to λ= 2 as
1
1
−1
.
Question 5
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 4, and λ3=−1. If v1,
v2, and v3are corresponding eigenvectors, find an orthogonal matrix Psuch
that P−1AP is a diagonal matrix.
Solution
Given that Ahas eigenvalues λ1= 2, λ2= 4, and λ3=−1, and corresponding
eigenvectors v1,v2, and v3, we can form the orthogonal matrix Pby normalizing
the eigenvectors.
P=v1v2v3
First, let’s normalize each eigenvector:
v1=
1
1
1
,v2=
2
−1
1
,v3=
0
1
−1
Next, let’s form the orthogonal matrix P:
P=
1
√3
2
√60
1
√3−1
√6
1
√2
1
√3
1
√6−1
√2
Now, we find P−1by using the property that for any orthogonal matrix Q,
Q−1=QT:
P−1=PT
Finally, to diagonalize A, we compute P−1AP :
P−1AP =
2 0 0
0 4 0
0 0 −1
Therefore, P−1AP is a diagonal matrix.
6
Question 6
Question
Let Abe a 3 ×3 matrix with eigenvalues 1,2,and 3. Determine the eigenvalues
of the matrix A2.
Solution
To determine the eigenvalues of A2, we first recall that if λis an eigenvalue of
a matrix B, then λ2is an eigenvalue of B2.
Step 1: Find the eigenvalues of A2using the eigenvalues of A. Since the
eigenvalues of Aare 1,2,and 3, the eigenvalues of A2are 12= 1,22= 4,and
32= 9.
Therefore, the eigenvalues of the matrix A2are 1,4,and 9.
Question 7
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If v1,
v2, and v3are the corresponding eigenvectors of A, find the eigenvalues and
eigenvectors of the matrix B=A3−2A2+ 3A.
Solution
Given that Ahas eigenvalues λ1= 1, λ2= 2, and λ3= 3, we also have
eigenvectors v1,v2, and v3for A.
Step 1: Find the eigenvalues of matrix B.The eigenvalues of Bcan
be found by evaluating the expression B=A3−2A2+ 3Ain terms of the
eigenvalues of A.
Let µbe an eigenvalue of Band wbe the corresponding eigenvector. Then,
we have: Bw = (A3−2A2+ 3A)w
=A3w−2A2w+ 3Aw
= (λ3
1−2λ2
1+ 3λ1)w
= (13−2·12+ 3 ·1)w= 2w
Therefore, the eigenvalues of Bare 2, 2, and 2.
Step 2: Find the corresponding eigenvectors of matrix B.Since
all the eigenvalues of Bare the same, the eigenvectors corresponding to the
eigenvalue of 2 can be found by solving the system (B−2I)w= 0, where Iis
the identity matrix.
7
Substitute B=A3−2A2+ 3Ainto the equation:
(A3−2A2+ 3A−2I)w= 0
(A3−2A2+ 3A−2I)w= 0
(A3−2A2+ 3A−2I)w= 0
Solving this system of equations will give us the eigenvectors corresponding
to the eigenvalue 2 of matrix B.
Question 8
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. Given
that the eigenvectors corresponding to λ1and λ2are v1=
1
1
1
and v2=
1
−1
0
,
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: Recall that for a matrix Aand an eigenvalue λwith eigenvector v, we
have the relationship Av=λv.
Step 2: Let v3=
x
y
z
be the eigenvector corresponding to the eigenvalue
λ3= 4. We need to solve the equation Av3= 4v3.
Step 3: Substitute v3into the equation Av3= 4v3:
Av3=
2 1 3
1 0 1
−115
x
y
z
= 4
x
y
z
Step 4: This gives us the system of equations:
2x+y+ 3z= 4x
x+ 1z= 4y
−x+y+ 5z= 4z
Step 5: Simplifying each equation, we get:
−2x+y+ 3z= 0
x−4y+z= 0
−x+y−z= 0
Step 6: Solving this system of equations, we find that x= 1, y= 1, and
z= 0.
8
Therefore, the eigenvector corresponding to the eigenvalue λ3= 4 is v3=
1
1
0
.
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 1,2,and 3. If the eigenvectors
corresponding to λ= 1 and λ= 3 are v1=
1
0
1
and v3=
0
1
1
respectively,
find the eigenvector corresponding to λ= 2.
Solution
Step 1: Recall that the eigenvectors of a matrix Acorresponding to distinct
eigenvalues are linearly independent. Thus, we can use the eigenvectors corre-
sponding to λ= 1 and λ= 3 to find the eigenvector corresponding to λ= 2.
Step 2: Let v2=
x
y
z
be the eigenvector corresponding to λ= 2. We know
that Av2=λv2.
Step 3: We have Av2= 2v2. Thus,
Av2=
101
020
103
x
y
z
=
2x
2y
2z
Step 4: Simplifying the matrix multiplication, we get the system of equa-
tions: x+z= 2x,
2y= 2y,
x+ 3z= 2z.
Step 5: Solving the system of equations, we find that x=z. Therefore, the
eigenvector corresponding to λ= 2 is v2=
1
y
1
for any y= 0.
Question 10
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 4, and λ3=−1. If
v1,v2, and v3are the corresponding eigenvectors, find a matrix Psuch that
A=P DP −1, where Dis a diagonal matrix.
9
Solution
Step 1: The matrix Dis a diagonal matrix with the eigenvalues of Aalong the
diagonal. So,
D=
2 0 0
0 4 0
0 0 −1
Step 2: The matrix Pis formed by placing the eigenvectors of Aas columns
in the order corresponding to the eigenvalues. So,
P=v1v2v3=v1v2v3
Step 3: To find P, we use the equation Av =λv for each eigenvector v. This
gives us a system of equations to solve for each eigenvector. Once we have found
the eigenvectors, we can construct Pas shown in Step 2.
Step 4: To find P−1, we can use the formula P−1=PT, as Pis an orthogonal
matrix.
Step 5: Therefore, the matrix Psuch that A=P DP −1is:
P=v1v2v3
Remember to normalize the eigenvectors if necessary to make Pan orthog-
onal matrix.
Question 11
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4.
Determine the characteristic polynomial of Aand find an expression for the
matrix inverse of Ausing its eigenvalues.
Solution
1. We know that the characteristic polynomial of a square matrix Ais given
by det(A−λI), where λis an eigenvalue and Iis the identity matrix of
the same size. Therefore, the characteristic polynomial of Ais:
det(A−λI)=(λ−2)(λ+ 1)(λ−4)
2. To find the matrix inverse of Ausing its eigenvalues, we can use the
formula:
A−1=
n
X
i=1
1
λi
projvi
where λiare eigenvalues and projvidenotes the projection matrix onto
the eigenvector corresponding to eigenvalue λi.
10
3. Let v1, v2, v3be the eigenvectors corresponding to the eigenvalues 2,−1,4
respectively. Then, we have:
A−1=1
2projv1+1
−1projv2+1
4projv3
=1
2v1vT
1+1
−1v2vT
2+1
4v3vT
3
Question 12
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3 and λ3= 3. If the
eigenvectors corresponding to λ2= 3 are v2=
1
2
−1
and v3=
1
3
−2
, find
the matrix A.
Solution
Step 1: Recall that for each eigenvalue λi, the corresponding eigenvectors vi
satisfy the equation (A−λiI)vi=0.
Step 2: For λ2= 3, we have:
(A−3I)v2=0
Substitute the given values:
a b c
d e f
g h i
−
300
030
003
1
2
−1
=
0
0
0
Step 3: This gives us the following system of equations:
a−3=0
d= 0
g+ 3 = 0
b+ 2e= 0
c−f= 0
h= 0
i−3=0
Step 4: Solving the system of equations, we find a= 3, b=−2, c= 3, d= 0,
e= 1, f= 3, g=−3, h= 0, i= 3.
Therefore, the matrix Ais:
A=
3−2 3
0 1 3
−303
11
Question 13
Question
Let Abe a 3×3 matrix with eigenvalues λ= 2,1,−3. If det(2A2−A+3I) = 60,
where Iis the identity matrix, find det(A).
Solution
Step 1: We know that the determinant of a matrix is equal to the product of
its eigenvalues. Let Ahave eigenvalues λ1, λ2, λ3. Therefore, we have:
det(A) = λ1·λ2·λ3
Step 2: From the given information, we know the eigenvalues of Aare λ1= 2,
λ2= 1, and λ3=−3. Therefore,
det(A)=2·1·(−3) = −6
Step 3: Next, we are given that det(2A2−A+ 3I) = 60. We can use the
fact that det(cA) = cn·det(A), where cis a scalar and nis the dimension of the
matrix A. Using this property, we can simplify det(2A2−A+ 3I) as follows:
det(2A2−A+ 3I) = 23·det(A2)−det(A) + 33·det(I) = 8det(A2)−6 + 27 ·1
Step 4: Since det(A2) = (det(A))2, we can rewrite the expression as:
det(2A2−A+ 3I) = 8(det(A))2−6 + 27
Step 5: Given that the expression equals 60, we have:
8(det(A))2−6 + 27 = 60 =⇒8(det(A))2= 39 =⇒(det(A))2=39
8
Step 6: Finally, we find det(A) by taking the square root of 39
8:
det(A) = r39
8=√39
2√2=√39
2√2·√2
√2=√78
4=√78
4
Question 14
Question
Let Abe a 3 ×3 matrix given by
A=
310
020
112
.
Find the eigenvalues and eigenvectors of matrix A.
12
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation det(A−λI) = 0, where λis the eigenvalue and Iis the identity matrix.
So we have:
A−λI =
3−λ1 0
0 2 −λ0
1 1 2 −λ
.
Finding the determinant and setting it equal to zero gives us:
det(A−λI) =
3−λ1 0
0 2 −λ0
1 1 2 −λ
= 0.
Step 2: Expanding the determinant, we get:
(3 −λ)((2 −λ)(2 −λ)−0) −1(0 −0) + 0 = 0.
(3 −λ)(λ2−4λ+ 4) = 0.
(3 −λ)(λ−2)2= 0.
Step 3: Solving (3 −λ)(λ−2)2= 0 gives us the eigenvalues:
λ1= 3 multiplicity 1
λ2= 2 multiplicity 2
Step 4: To find the eigenvectors corresponding to each eigenvalue, we solve
the system of equations (A−λI)v= 0 for each eigenvalue.
For λ= 3, solving (A−3I)v= 0 gives the eigenvector
v1=
1
0
−1
.
For λ= 2, solving (A−2I)v= 0 gives the eigenvector
v2=
−1
0
1
.
Therefore, the eigenvalues of matrix Aare 3 with eigenvector
1
0
−1
and 2
with eigenvector
−1
0
1
.
13
Question 15
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3=−1.
Determine the eigenvalues of the matrix 2A−3A2.
Solution
Step 1: Find the eigenvalues of A2.
Since Ahas eigenvalues λ1= 2, λ2= 3, and λ3=−1, the eigenvalues of A2
will be λ2
1= 4, λ2
2= 9, and λ2
3= 1.
Step 2: Calculate the eigenvalues of 2A−3A2.
The matrix 2A−3A2will have eigenvalues 2λi−3λ2
ifor i= 1,2,3.
Substituting the given eigenvalues, we have: - For λ1= 2: 2(2) −3(4) =
4−12 = −8 - For λ2= 3: 2(3) −3(9) = 6 −27 = −21 - For λ3=−1:
2(−1) −3(1) = −2−3 = −5
Therefore, the eigenvalues of 2A−3A2are −8, −21, and −5.
Question 16
Question
Let
A=2 1
1 3
be a 2x2 matrix. Find the eigenvalues of Aand corresponding eigenvectors.
Solution
Step 1: To find the eigenvalues of matrix A, we solve the characteristic equation
|A−λI|= 0, where λrepresents the eigenvalue and Iis the identity matrix.
A−λI =2−λ1
1 3 −λ
|A−λI|=
2−λ1
1 3 −λ
= (2 −λ)(3 −λ)−1=0
λ2−5λ+ 5 = 0
Step 2: Solving the quadratic equation gives the eigenvalues:
λ=5±√5
2
14
Step 3: Now, we find the eigenvectors corresponding to each eigenvalue. For
λ=5+√5
2, we solve the system (A−λI)v= 0:
"2−5+√5
21
1 3 −5+√5
2#v1
v2=0
0
Solving the equations gives v1=p5 + √5 and v2= 1.
Therefore, the eigenvector corresponding to λ=5+√5
2is p5 + √5
1.
Step 4: Similarly, for λ=5−√5
2, we solve the system (A−λI)v= 0 to find
the eigenvector:
v=p5−√5
−1
Therefore, the eigenvector corresponding to λ=5−√5
2is p5−√5
−1.
Question 17
Question
Let Aand Bbe 2×2 matrices with eigenvalues λ1= 2 and λ2=−1, respectively.
Consider the matrix C=A2+ 3B.
Find the eigenvalues of matrix C.
Solution
Let’s first find the eigenvalues of matrix C=A2+ 3B.
Step 1: Find the eigenvalues of A2.
Since Ahas eigenvalues λ1= 2 and λ2=−1, the eigenvalues of A2are the
squares of the eigenvalues of A:
λA2= 22= 4 and (−1)2= 1.
Step 2: Find the eigenvalues of 3B.
Since Bhas eigenvalues λ1= 2 and λ2=−1, the eigenvalues of 3Bare
simply 3 times the eigenvalues of B:
λ3B= 3 ·2 = 6 and 3 ·(−1) = −3.
Step 3: Find the eigenvalues of C=A2+ 3B.
The eigenvalues of Care the sums of the eigenvalues of A2and 3B:
λC={4+6,4−3,1+6,1−3}={10,1,7,−2}.
Therefore, the eigenvalues of matrix Care λ1= 10, λ2= 1, λ3= 7, λ4=−2.
15
Question 18
Question
Let A=5 2
4 3. Determine the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation det(A−λI) = 0, where Iis the identity matrix.
Step 2: The characteristic equation is det 5 2
4 3−λ1 0
0 1= 0.
Step 3: Simplifying, we have det 5−λ2
4 3 −λ= 0.
Step 4: Expanding the determinant, we get (5 −λ)(3 −λ)−2·4 = 0.
Step 5: Simplifying further, we obtain 15 −8λ+λ2−8 = 0.
Step 6: This equation simplifies to λ2−8λ+ 7 = 0.
Step 7: Solving the quadratic equation, we find two eigenvalues: λ1= 1 and
λ2= 7.
Step 8: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute them back into (A−λI)v=0, where vis the eigenvector.
Step 9: For λ= 1, we have (A−1I)v=4 2
4 2x
y=0
0.
Step 10: This gives us the equations 4x+ 2y= 0 and 4x+ 2y= 0. Solving,
we find that x=−1
2y. Letting y= 2, we get the eigenvector v1=−1
2.
Step 11: For λ= 7, we have (A−7I)v=−2 2
4−4x
y=0
0.
Step 12: This gives us the equations −2x+ 2y= 0 and 4x−4y= 0. Solving,
we find that x=y. Letting y= 1, we get the eigenvector v2=1
1.
Step 13: Therefore, the eigenvalues of matrix Aare λ1= 1 with eigenvector
v1=−1
2and λ2= 7 with eigenvector v2=1
1.
Question 19
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2,and λ3= 4. If
B=A2−3A+ 2I, find the eigenvalues of B.
Solution
Given that Ais a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2,and λ3= 4. We
want to find the eigenvalues of matrix B=A2−3A+ 2I.
16
Step 1: Find the eigenvalues of B.
First, let’s express Bin terms of A:
B=A2−3A+ 2I
We can write this as:
B=A2−3A+ 2I=A·A−3A+ 2I
Now, let’s substitute the eigenvalues of Ainto the expression:
B=A2−3A+ 2I=
100
020
004
2
−3
100
020
004
+ 2
100
010
001
Calculating the expression, we get:
B=
1 0 0
0 4 0
0 0 16
−
3 0 0
0 6 0
0 0 12
+
200
020
002
B=
000
000
006
Step 2: Find the eigenvalues of B.
The eigenvalues of Bare the values of λsuch that |B−λI|= 0. Let’s
calculate this determinant.
|B−λI|=
−λ0 0
0−λ0
0 0 6 −λ
= (−λ)(−λ)(6 −λ)
=−λ2(6 −λ) = −λ2(6 −λ) = −λ3+ 6λ2
To find the eigenvalues of B, we solve the characteristic equation −λ3+6λ2=
0. Factoring out λ2:
λ2(−λ+ 6) = 0
Setting each factor to 0 gives us the eigenvalues:
λ2= 0 =⇒λ= 0
−λ+ 6 = 0 =⇒λ= 6
Therefore, the eigenvalues of matrix Bare λ= 0 and λ= 6.
Question 20
Question
Let Abe a 3 ×3 matrix with eigenvalues 1, −2, and 4. If B=A3−2A2+ 3A,
find the eigenvalues of B.
17
Solution
Step 1: First, let’s find the eigenvalues of A3,A2, and A.
For the matrix Awith eigenvalues 1, −2, and 4, the eigenvalues of A3,
A2, and Aare simply the eigenvalues of Araised to the power of 3, 2, and 1
respectively. Thus, the eigenvalues of A3are 13= 1, (−2)3=−8, and 43= 64,
the eigenvalues of A2are 12= 1, (−2)2= 4, and 42= 16, and the eigenvalues
of Aare 1, −2, and 4.
Step 2: Next, let’s find the matrix B.
We have B=A3−2A2+ 3A. Substituting the eigenvalues of A3,A2, and
Afound in step 1, we get the eigenvalues of Bas follows:
Eigenvalues of B= 1 −2·1+3·1=2,
Eigenvalues of B=−8−2·4+3·(−2) = −21,
Eigenvalues of B= 64 −2·16 + 3 ·4 = 24.
Therefore, the eigenvalues of matrix Bare 2, −21, and 24.
Question 21
Question
Let Abe a 3 ×3 matrix with characteristic polynomial λ3−6λ2+ 11λ−6.
If v1, v2,and v3are eigenvectors corresponding to eigenvalues λ1, λ2,and λ3
respectively, find vT
1Av2.
Solution
Step 1: First, we need to find the eigenvalues of the matrix Aby solving the
characteristic equation det(A−λI) = 0.
(A−λI) =
a11 −λ a12 a13
a21 a22 −λ a23
a31 a32 a33 −λ
The characteristic polynomial p(λ) = det(A−λI) is given as λ3−6λ2+ 11λ−6.
Thus, the eigenvalues are the roots of this polynomial, which are λ= 1,2,3.
Step 2: Next, we find the eigenvectors by solving the system of equations
(A−λiI)vi= 0 for each eigenvalue λi. For λ= 1: Solve (A−I)v1= 0. For
λ= 2: Solve (A−2I)v2= 0. For λ= 3: Solve (A−3I)v3= 0.
Step 3: Since v1, v2,and v3are eigenvectors, they are linearly independent
and form a basis of R3. Thus, vT
1Av2= 0 as any two different eigenvectors are
orthogonal.
18
Question 22
Question
Let Abe a 3 ×3 matrix given by
A=
−2 1 0
0−2 1
0 0 −2
.
Find the eigenvalues and eigenvectors of A.
Solution
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 of the same size
as A.
|A−λI|=
−2−λ1 0
0−2−λ1
0 0 −2−λ
= (−2−λ)((−2−λ)2)
= (−2−λ)3= 0.
This implies that the eigenvalue is λ=−2 with algebraic multiplicity of 3.
To find the eigenvectors corresponding to the eigenvalue λ=−2, we need to
solve the system of equations (A−λI)v= 0. Substitute λ=−2 back into the
equation A−λI = 0:
A+ 2I=
010
001
000
.
Solving the system (A+ 2I)v= 0 leads to the following eigenvectors:
v=t
1
0
0
,
where tis a scalar. Thus, the eigenvalue -2 has a geometric multiplicity of 1.
Question 23
Question
Let Abe a 3x3 matrix with eigenvalues λ1= 2, λ2= 4, and λ3= 6. Find the
eigenvectors corresponding to each eigenvalue.
19
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we solve the system of
equations (A−2I)v1=0where v1=
x
y
z
.
A−2I=
−1 2 −3
101
2−2−4
Using row reduction, we get:
−1 2 −3
101
2−2−4
→
101
011
000
So, one possible eigenvector is v1=
−1
−1
1
.
Step 2: To find the eigenvector corresponding to λ2= 4, we solve the system
of equations (A−4I)v2=0where v2=
x
y
z
.
A−4I=
−2 2 −3
1−4 1
2−2−6
Using row reduction, we get:
−2 2 −3
1−4 1
2−2−6
→
101
011
000
So, one possible eigenvector is v2=
−1
−1
1
.
Step 3: To find the eigenvector corresponding to λ3= 6, we solve the system
of equations (A−6I)v3=0where v3=
x
y
z
.
A−6I=
−4 2 −3
1−6 1
2−2−8
Using row reduction, we get:
−4 2 −3
1−6 1
2−2−8
→
101
011
000
20
So, one possible eigenvector is v3=
−1
−1
1
.
Question 24
Question
Let Abe a 3×3 matrix with eigenvalues 2, −1, and 3. Given that the eigenvector
corresponding to the eigenvalue 2 is
1
0
1
, find the eigenvectors corresponding
to the eigenvalues −1 and 3.
Solution
Step 1: Finding the eigenvector corresponding to eigenvalue −1
Let vbe the eigenvector corresponding to the eigenvalue −1. We know that
for an eigenvector vcorresponding to an eigenvalue λ, the matrix Asatisfies the
equation (A−λI)v= 0, where Iis the identity matrix.
Therefore, for λ=−1:
(A+I)v= 0
Substitute Aand λ:
200
0−1 0
003
+
100
010
001
x
y
z
=
0
0
0
Simplify:
300
000
004
x
y
z
=
0
0
0
This gives the following system of equations:
3x= 0,0y= 0,4z= 0
Since ycan be any value, let y=t. We can choose x=z= 0 and y= 1
to get a non-trivial solution. Therefore, an eigenvector corresponding to the
eigenvalue −1 is
0
1
0
.
Step 2: Finding the eigenvector corresponding to eigenvalue 3
Let wbe the eigenvector corresponding to the eigenvalue 3. Using the same
method as above, we have:
(A−3I)w= 0
21
Substitute Aand λ:
200
0−1 0
003
−
300
030
003
x
y
z
=
0
0
0
This gives the following system of equations:
−x= 0,−y= 0,0=0
So x=yand zis arbitrary. Let x=y= 1 and z= 0, we get an eigenvector
corresponding to the eigenvalue 3 as
1
1
0
.
Question 25
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−3, and λ3= 4. Given
that the corresponding eigenvectors are v1=
1
1
1
,v2=
1
0
1
, and v3=
0
1
1
,
find an invertible matrix Psuch that P−1AP is a diagonal matrix.
Solution
Step 1: Form the matrix Pusing the eigenvectors of A.
P= [v1,v2,v3] =
110
101
111
Step 2: Calculate the inverse of P. Firstly, find the determinant of P:
det(P) = 1(0 −1) −1(1 −0) + 0(1 −1) = −1
Next, calculate the cofactor matrix of Pby finding the matrix of minors,
then appending the correct signs:
Cof(P) =
0−1 1
1−1−1
−1 1 0
Then, find the adjugate of Pby taking the transpose of the cofactor matrix:
Adj(P) =
0 1 −1
−1−1 1
1−1 0
22
Finally, calculate P−1using the formula P−1=1
det(P)·Adj(P).
P−1=1
−1
0 1 −1
−1−1 1
1−1 0
=
0−1 1
1 1 −1
−1 1 0
Step 3: Calculate P−1AP to obtain a diagonal matrix.
P−1AP =
0−1 1
1 1 −1
−1 1 0
200
0−3 0
004
110
101
111
=
200
0−3 0
004
Therefore, the matrix P−1AP is a diagonal matrix as desired.
Question 26
Question
Let Abe a 3 ×3 matrix with eigenvalues 1,2,and 3. If B= 2A2−3A+I,
where Iis the identity matrix, determine the eigenvalues of B.
Solution
To find the eigenvalues of B, we need to find the roots of the characteristic
polynomial of B.
Step 1: Find the characteristic polynomial of BThe characteristic
polynomial of a matrix Bis given by det(B−λI) = 0, where λis the eigenvalue
we are trying to find. Therefore,
det(B−λI) = det(2A2−3A+I−λI) = 0.
Step 2: Substitute the expression for BSubstitute B= 2A2−3A+I
into the equation above:
det(2A2−3A+I−λI)=0.
Step 3: Factor out the terms Since Ahas eigenvalues 1,2,and 3, we can
rewrite Aas a diagonal matrix:
A=
100
020
003
.
Substitute this into the equation:
det
2
100
020
003
2
−3
100
020
003
+
100
010
001
−λ
100
010
001
= 0.
Step 4: Calculate the characteristic polynomial Calculate the deter-
minant of the given matrix expression to obtain the characteristic polynomial
of B. Solving this characteristic polynomial will give us the eigenvalues of B.
23
Question 27
Question
Let Abe a 3 ×3 matrix such that A3−6A2+ 11A−6I=O, where Iis the
3×3 identity matrix. Find all the eigenvalues of A.
Solution
To find the eigenvalues of A, we will first find the characteristic polynomial of
Ausing the given equation.
Step 1: Define the characteristic polynomial p(λ). Let p(λ) = det(A−λI) be
the characteristic polynomial of A. The given equation A3−6A2+11A−6I=O
can be rewritten as:
A(A2−6A+ 11I) = 6I
Multiplying by A−λI on both sides, we get:
A(A2−6A+ 11I)=6I=⇒det(A−λI) = 0
Thus, p(λ) = λ3−6λ2+ 11λ−6.
Step 2: Find the eigenvalues of A. To find the eigenvalues of A, we solve
p(λ) = 0. By inspection, we find that λ= 1 is a root of p(λ). Using polynomial
division or synthetic division, we find that:
p(λ)=(λ−1)(λ2−5λ+ 6) = (λ−1)2(λ−6)
Therefore, the eigenvalues of Aare λ= 1 with multiplicity 2 and λ= 6.
Question 28
Question
Let Abe a square matrix such that A2= 5A−6I, where Iis the identity matrix.
If vis an eigenvector of Acorresponding to the eigenvalue 3, find A−1v.
Solution
Step 1: First, let’s find the eigenvalues of matrix A. Given that A2= 5A−6I, we
can rewrite the equation as A2−5A+6I= 0. This means that the characteristic
polynomial of Ais detA2−5A+ 6I= 0. Expanding this determinant gives us
detA2−5A+ 6I= det(A−3I) det(A−2I) = 0. Therefore, the eigenvalues
of Aare 3 and 2.
Step 2: Next, find the eigenvectors corresponding to the eigenvalues. For
eigenvalue 3, we solve the system (A−3I)v= 0 where vis the eigenvector
corresponding to eigenvalue 3. This gives us (A−3I)v= 0 which implies
(A−3I)v= 0v. Solving the equation (A−3I)v= 0 gives us the eigenvector
corresponding to eigenvalue 3.
24
Step 3: Now that we have the eigenvector vcorresponding to the eigenvalue
3, we can find A−1v. Recall that if vis an eigenvector of Awith eigenvalue
λ, then A−1vis an eigenvector of A−1with eigenvalue 1/λ. Since vis an
eigenvector of Acorresponding to the eigenvalue 3, A−1vis an eigenvector of
A−1with eigenvalue 1/3.
Therefore, A−1vis an eigenvector of A−1with eigenvalue 1/3.
Question 29
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, λ3= 3. Given that
the eigenvectors corresponding to λ1and λ3are
1
0
2
and
1
1
1
, respectively, find
the eigenvector corresponding to λ2.
Solution
Step 1: Recall that the eigenvectors of a matrix Aare the solutions to the
equation (A−λI)x=0, where λis an eigenvalue of A.
Step 2: Let’s first find the eigenvector corresponding to λ2= 2. Using the
equation (A−λ2I)v=0, we have:
a b c
d e f
g h i
−2
100
010
001
x
y
z
=
0
0
0
Simplifying, we get:
a−2b c
d e −2f
g h i −2
x
y
z
=
0
0
0
Step 3: So, we have the following system of equations:
(a−2)x+by +cz = 0
dx + (e−2)y+fz = 0
gx +hy + (i−2)z= 0
Step 4: Since we are given the eigenvectors corresponding to λ1and λ3, we
can substitute them into the system of equations to obtain additional informa-
tion. Let’s substitute
1
0
2
and
1
1
1
into the system.
Step 5: Substituting
1
0
2
into the system gives us the equations:
25
(a−2) + 0 + 2c= 0
d+ (e−2) ·0 + f·2=0
g+h+ (i−2) ·2=0
Solving this system we can find values for a, b, c, d, e, f, g, h, i.
Step 6: Next, substituting
1
1
1
into the system gives us another set of
equations, which we can solve to find the eigenvector corresponding to λ2= 2.
Question 30
Question
Let A=4 1
1 4. Find the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0, where Iis the identity matrix.
det 4−λ1
1 4 −λ= 0
(4 −λ)2−1=0
λ2−8λ+ 15 = 0
(λ−3)(λ−5) = 0
So, the eigenvalues are λ1= 3 and λ2= 5.
Step 2: Now, we find the eigenvector associated with λ1= 3. Let v1=x
y
be the eigenvector. Then, we have:
(A−3I)v1= 0
1 1
1 1x
y=0
0
x+y= 0
Thus, the eigenvector corresponding to λ1= 3 is any non-zero scalar multiple
of 1
−1.
26
Step 3: Next, we find the eigenvector associated with λ2= 5. Let v2=x
y
be the eigenvector. Then, we have:
(A−5I)v2= 0
−1 1
1−1x
y=0
0
−x+y= 0
Thus, the eigenvector corresponding to λ2= 5 is any non-zero scalar multiple
of 1
1.
Question 31
Question
Let Abe a 3 ×3 matrix with eigenvalues −2,1,and 3. Find the eigenvectors
corresponding to each eigenvalue.
Solution
Step 1: To find the eigenvectors, we need to solve the system of equations
(A−λI)v=0for each eigenvalue λ.
For λ=−2:
A−λI =
000
030
005
−(−2)
100
010
001
=
200
050
007
The system of equations becomes:
200
050
007
x
y
z
=
0
0
0
Solving this system, we get x=y=z= 0. Therefore, the eigenvector
corresponding to λ=−2 is
0
0
0
.
Step 2: Repeat the same process for λ= 1:
A−λI =
000
030
005
−1
100
010
001
=
−100
0 2 0
0 0 4
27
The system of equations becomes:
−100
0 2 0
0 0 4
x
y
z
=
0
0
0
Solving this system, we get x=y= 0 and zcan be any non-zero scalar.
Therefore, an eigenvector corresponding to λ= 1 is any vector of the form
0
0
1
.
Step 3: Lastly, repeat the process for λ= 3:
A−λI =
000
030
005
−3
100
010
001
=
−300
0 0 0
0 0 2
The system of equations becomes:
−300
0 0 0
0 0 2
x
y
z
=
0
0
0
Solving this system, we get y=z= 0 and xcan be any non-zero scalar.
Therefore, an eigenvector corresponding to λ= 3 is any vector of the form
1
0
0
.
Question 32
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 3. Given
that corresponding eigenvectors are
v1=
1
0
1
,v2=
1
1
0
,and v3=
1
0
−1
,
write the matrix A.
Solution
To write the matrix A, we can use the formula for diagonalizing a matrix:
A=P DP −1where Dis a diagonal matrix with eigenvalues of Aon the diagonal
sorted according to their respective eigenvectors in P.
Step 1: Compute the matrix Pusing the given eigenvectors.
P=v1v2v3=
1 1 1
0 1 0
1 0 −1
28
Step 2: Compute the inverse of matrix Pto get P−1.To find P−1,
we need to solve the equation P P −1=I, where Iis the identity matrix.
1 1 1
0 1 0
1 0 −1
a d g
b e h
c f i
=
100
010
001
Solving this system of equations gives P−1.
P−1=
1−1 1
010
−1 0 −1
Step 3: Compute the diagonal matrix Dwith eigenvalues on the
diagonal.
D=
200
0−1 0
003
Step 4: Write the matrix Ausing the formula A=P DP −1.Substi-
tuting the values of P,D, and P−1into the formula:
A=P DP −1=
1 1 1
0 1 0
1 0 −1
200
0−1 0
003
1−1 1
010
−1 0 −1
Calculating the matrix product gives us the matrix A.
Question 33
Question
Let A be a 3x3 matrix with eigenvalues 2, 3, and 4. If B = A2−3A+ 5I, find
the eigenvalues of B.
Solution
To find the eigenvalues of B, we will first find the matrix A and then use the
properties of eigenvalues to find the eigenvalues of B.
Step 1: Finding A From the given eigenvalues of A, we know that the
characteristic equation of A is given by:
det(A−λI) = 0
where λrepresents an eigenvalue.
Substitute the given eigenvalues into the characteristic equation to find the
matrix A.
29
Step 2: Finding B Now, we can find B using the equation stated in the
question: B = A2−3A+ 5I.
Substitute the matrix A into the expression for B to calculate the matrix B.
Step 3: Finding the eigenvalues of B The eigenvalues of B are equal to
the eigenvalues of the matrix B.
Calculate the eigenvalues of B by finding the roots of the characteristic
equation of B, given by:
det(B−λI)=0
where λrepresents an eigenvalue.
Solve the characteristic equation to find the eigenvalues of B.
Therefore, the eigenvalues of matrix B are the roots of the characteristic
equation.
Question 34
Question
Let A=3 1
3 1. Determine the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0.
det 3−λ1
3 1 −λ= 0.
Step 2: Expanding the determinant, we get:
(3 −λ)(1 −λ)−3=0.
Step 3: Simplifying the equation gives:
λ2−4λ= 0.
Step 4: Factoring out λ, we have:
λ(λ−4) = 0.
Step 5: This equation has two solutions: λ= 0 and λ= 4. Therefore, the
eigenvalues of Aare 0 and 4.
Step 6: To find the eigenvector corresponding to λ= 0, we substitute λ= 0
back into the equation (A−λI)v= 0.
(A−0I)v=3 1
3 1v=0
0.
30
Step 7: Solving the system of equations, we get v=t−1
3for some scalar
t= 0.
Step 8: Therefore, the eigenvector corresponding to λ= 0 is v1=−1
3.
Step 9: To find the eigenvector corresponding to λ= 4, we substitute λ= 4
back into the equation (A−λI)v= 0.
(A−4I)v=−1 1
3−3v=0
0.
Step 10: Solving the system of equations, we get v=t1
1for some scalar
t= 0.
Step 11: Therefore, the eigenvector corresponding to λ= 4 is v2=1
1.
Question 35
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 5. Find the
eigenvalues of the matrix A2−4A+ 3I, where Iis the 3 ×3 identity matrix.
Solution
Given that Ahas eigenvalues λ1= 2, λ2= 3, and λ3= 5, we want to find the
eigenvalues of the matrix A2−4A+ 3I.
Step 1: Find the eigenvalues of A2The eigenvalues of A2are simply the
squares of the eigenvalues of A. Therefore, the eigenvalues of A2are λ2
1= 22= 4,
λ2
2= 32= 9, and λ2
3= 52= 25.
Step 2: Find the eigenvalues of −4AThe eigenvalues of −4Aare simply
−4 times the eigenvalues of A. Therefore, the eigenvalues of −4Aare −4·2 = −8,
−4·3 = −12, and −4·5 = −20.
Step 3: Find the eigenvalues of A2−4A+ 3ITo find the eigenvalues of
A2−4A+ 3I, we add the corresponding eigenvalues of A2,−4A, and 3I. Thus,
the eigenvalues of A2−4A+ 3Iare:
4−8 + 3 = −1,9−12 + 3 = 0,and 25 −20 + 3 = 8.
Therefore, the eigenvalues of the matrix A2−4A+ 3Iare −1, 0, and 8.
31
Solving this system of equations gives x= 2, y= 0, and z= 1. Therefore,
the eigenvector corresponding to the eigenvalue 2 is
2
0
1
.
Step 2: Find eigenvector corresponding to eigenvalue 4 Let v4=
p
q
r
be the eigenvector corresponding to the eigenvalue 4. Substituting into
the eigenvalue equation gives:
Av4= 4v4
−101
2 0 1
1 0 0
p
q
r
= 4
p
q
r
Solving the system of equations gives p=−2qand r= 3q. Therefore, the
eigenvector corresponding to the eigenvalue 4 is
−2
1
3
.
Question 2
Question
Let Abe a 3 ×3 matrix given by
A=
210
121
012
.
Find the eigenvalues and corresponding eigenvectors of matrix A.
Solution
To find the eigenvalues of matrix A, we need to solve the characteristic equation
det(A−λI) = 0, where Iis the identity matrix.
Step 1: Find the characteristic equation We have:
det(A−λI) = det
2−λ1 0
1 2 −λ1
0 1 2 −λ
= 0.
Expanding this determinant gives us the characteristic equation:
(2 −λ)((2 −λ)2−1) −(1)((2 −λ)−0) = 0.
Simplifying this equation gives us:
λ3−6λ2+ 9λ−3=0.
2
Step 2: Find the eigenvalues To find the eigenvalues, we solve the char-
acteristic equation λ3−6λ2+ 9λ−3 = 0. By inspection, we can see that λ= 1
is a root of this polynomial.
Dividing the characteristic equation by (λ−1) gives us:
λ2−5λ+ 3 = 0.
Using the quadratic formula to solve this equation, we get λ=5±√13
2.
Therefore, the eigenvalues of matrix Aare λ= 1,5+√13
2,5−√13
2.
Step 3: Find the eigenvectors Now, we find the eigenvectors correspond-
ing to each eigenvalue.
For λ= 1, we solve the system (A−I)v= 0:
110
111
011
v= 0.
Solving this system gives us one independent eigenvector v1=
1
−1
1
.
For λ=5+√13
2, we solve the system (A−5+√13
2I)v= 0 to find another
eigenvector v2.
For λ=5−√13
2, we solve the system (A−5−√13
2I)v= 0 to find the third
eigenvector v3.
Therefore, the eigenvalues and eigenvectors of matrix Aare: λ= 1 with
eigenvector v1=
1
−1
1
,λ=5+√13
2with eigenvector v2, and λ=5−√13
2with
eigenvector v3.
Question 3
Question
Let Abe a 3 ×3 matrix given by
A=
5 2 −1
222
−1 2 5
.
Find the eigenvalues and eigenvectors of matrix A.
3
Solution
Step 1: To find the eigenvalues of matrix A, we solve the characteristic equation
|A−λI|= 0, where λis the eigenvalue and Iis the 3 ×3 identity matrix.
|A−λI|=
5−λ2−1
2 2 −λ2
−1 2 5 −λ
.
Expanding the determinant, we get
(5 −λ)((2 −λ)(5 −λ)−4) −2(2(5 −λ)−2) + (−1)(2 ·2−2) = 0.
Simplifying the equation, we have
λ3−12λ2+ 39λ−28 = 0.
This equation can be factored as (λ−4)(λ−2)(λ−7) = 0.
Thus, the eigenvalues are λ1= 4, λ2= 2,and λ3= 7.
Step 2: Next, we find the eigenvectors corresponding to each eigenvalue by
substituting them back into the equation (A−λI)v= 0 and solving for the
vector v.
For λ1= 4, we solve (A−4I)v= 0:
1 2 −1
2−2 2
−1 2 1
v=
0
0
0
.
By row reducing the augmented matrix, the solution is v1=
1
1
1
.
For λ2= 2, we solve (A−2I)v= 0:
3 2 −1
202
−1 2 3
v=
0
0
0
.
By row reducing the augmented matrix, the solution is v2=
1
−1
1
.
For λ3= 7, we solve (A−7I)v= 0:
−2 2 −1
2−5 2
−1 2 −2
v=
0
0
0
.
By row reducing the augmented matrix, the solution is v3=
1
1
−1
.
Therefore, the eigenvalues of matrix Aare 4, 2, and 7, with corresponding
eigenvectors
1
1
1
,
1
−1
1
, and
1
1
−1
, respectively.
4
Question 4
Question
Let Abe a 3 ×3 matrix given by
A=
2 0 1
1 2 1
−101
.
Find the eigenvalues and eigenvectors of A.
Solution
Step 1: Find the characteristic equation The characteristic equation of a
square matrix Ais given by det(A−λI) = 0, where Iis the identity matrix.
For matrix Ain this case,
A−λI =
2−λ0 1
1 2 −λ1
−1 0 1 −λ
.
Thus, the characteristic equation becomes
det(A−λI) = det
2−λ0 1
1 2 −λ1
−1 0 1 −λ
= 0.
Step 2: Solve the characteristic equation Expanding the determinant
above, we get
(1 −λ)((2 −λ)(1 −λ)−0·0) −0−1(−1(1 −λ)−0) = 0
⇒(1 −λ)((2 −λ)(1 −λ) + 1(1 −λ)) = 0
⇒(1 −λ)(2 −λ)(1 −λ+ 1) = 0
⇒(1 −λ)(2 −λ)(2 −λ)=0.
Solving this equation gives the eigenvalues λ1= 1 and λ2= 2 with algebraic
multiplicity 2 and 1, respectively.
Step 3: Find the eigenvectors To find the eigenvector corresponding to
λ= 1: Substitute λ= 1 into the matrix A−λI,
A−λI =
1 0 1
1 1 1
−100
.
Reducing the matrix to row echelon form gives the eigenvector corresponding
to λ= 1 as
1
−1
1
.
5
For the eigenvalue λ= 2: Substitute λ= 2 into the matrix A−λI,
A−λI =
001
101
−1 0 −1
.
Reducing the matrix to row echelon form results in the eigenvector correspond-
ing to λ= 2 as
1
1
−1
.
Question 5
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 4, and λ3=−1. If v1,
v2, and v3are corresponding eigenvectors, find an orthogonal matrix Psuch
that P−1AP is a diagonal matrix.
Solution
Given that Ahas eigenvalues λ1= 2, λ2= 4, and λ3=−1, and corresponding
eigenvectors v1,v2, and v3, we can form the orthogonal matrix Pby normalizing
the eigenvectors.
P=v1v2v3
First, let’s normalize each eigenvector:
v1=
1
1
1
,v2=
2
−1
1
,v3=
0
1
−1
Next, let’s form the orthogonal matrix P:
P=
1
√3
2
√60
1
√3−1
√6
1
√2
1
√3
1
√6−1
√2
Now, we find P−1by using the property that for any orthogonal matrix Q,
Q−1=QT:
P−1=PT
Finally, to diagonalize A, we compute P−1AP :
P−1AP =
2 0 0
0 4 0
0 0 −1
Therefore, P−1AP is a diagonal matrix.
6
Question 6
Question
Let Abe a 3 ×3 matrix with eigenvalues 1,2,and 3. Determine the eigenvalues
of the matrix A2.
Solution
To determine the eigenvalues of A2, we first recall that if λis an eigenvalue of
a matrix B, then λ2is an eigenvalue of B2.
Step 1: Find the eigenvalues of A2using the eigenvalues of A. Since the
eigenvalues of Aare 1,2,and 3, the eigenvalues of A2are 12= 1,22= 4,and
32= 9.
Therefore, the eigenvalues of the matrix A2are 1,4,and 9.
Question 7
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If v1,
v2, and v3are the corresponding eigenvectors of A, find the eigenvalues and
eigenvectors of the matrix B=A3−2A2+ 3A.
Solution
Given that Ahas eigenvalues λ1= 1, λ2= 2, and λ3= 3, we also have
eigenvectors v1,v2, and v3for A.
Step 1: Find the eigenvalues of matrix B.The eigenvalues of Bcan
be found by evaluating the expression B=A3−2A2+ 3Ain terms of the
eigenvalues of A.
Let µbe an eigenvalue of Band wbe the corresponding eigenvector. Then,
we have: Bw = (A3−2A2+ 3A)w
=A3w−2A2w+ 3Aw
= (λ3
1−2λ2
1+ 3λ1)w
= (13−2·12+ 3 ·1)w= 2w
Therefore, the eigenvalues of Bare 2, 2, and 2.
Step 2: Find the corresponding eigenvectors of matrix B.Since
all the eigenvalues of Bare the same, the eigenvectors corresponding to the
eigenvalue of 2 can be found by solving the system (B−2I)w= 0, where Iis
the identity matrix.
7
Substitute B=A3−2A2+ 3Ainto the equation:
(A3−2A2+ 3A−2I)w= 0
(A3−2A2+ 3A−2I)w= 0
(A3−2A2+ 3A−2I)w= 0
Solving this system of equations will give us the eigenvectors corresponding
to the eigenvalue 2 of matrix B.
Question 8
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. Given
that the eigenvectors corresponding to λ1and λ2are v1=
1
1
1
and v2=
1
−1
0
,
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: Recall that for a matrix Aand an eigenvalue λwith eigenvector v, we
have the relationship Av=λv.
Step 2: Let v3=
x
y
z
be the eigenvector corresponding to the eigenvalue
λ3= 4. We need to solve the equation Av3= 4v3.
Step 3: Substitute v3into the equation Av3= 4v3:
Av3=
2 1 3
1 0 1
−115
x
y
z
= 4
x
y
z
Step 4: This gives us the system of equations:
2x+y+ 3z= 4x
x+ 1z= 4y
−x+y+ 5z= 4z
Step 5: Simplifying each equation, we get:
−2x+y+ 3z= 0
x−4y+z= 0
−x+y−z= 0
Step 6: Solving this system of equations, we find that x= 1, y= 1, and
z= 0.
8
Therefore, the eigenvector corresponding to the eigenvalue λ3= 4 is v3=
1
1
0
.
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 1,2,and 3. If the eigenvectors
corresponding to λ= 1 and λ= 3 are v1=
1
0
1
and v3=
0
1
1
respectively,
find the eigenvector corresponding to λ= 2.
Solution
Step 1: Recall that the eigenvectors of a matrix Acorresponding to distinct
eigenvalues are linearly independent. Thus, we can use the eigenvectors corre-
sponding to λ= 1 and λ= 3 to find the eigenvector corresponding to λ= 2.
Step 2: Let v2=
x
y
z
be the eigenvector corresponding to λ= 2. We know
that Av2=λv2.
Step 3: We have Av2= 2v2. Thus,
Av2=
101
020
103
x
y
z
=
2x
2y
2z
Step 4: Simplifying the matrix multiplication, we get the system of equa-
tions: x+z= 2x,
2y= 2y,
x+ 3z= 2z.
Step 5: Solving the system of equations, we find that x=z. Therefore, the
eigenvector corresponding to λ= 2 is v2=
1
y
1
for any y= 0.
Question 10
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 4, and λ3=−1. If
v1,v2, and v3are the corresponding eigenvectors, find a matrix Psuch that
A=P DP −1, where Dis a diagonal matrix.
9
Solution
Step 1: The matrix Dis a diagonal matrix with the eigenvalues of Aalong the
diagonal. So,
D=
2 0 0
0 4 0
0 0 −1
Step 2: The matrix Pis formed by placing the eigenvectors of Aas columns
in the order corresponding to the eigenvalues. So,
P=v1v2v3=v1v2v3
Step 3: To find P, we use the equation Av =λv for each eigenvector v. This
gives us a system of equations to solve for each eigenvector. Once we have found
the eigenvectors, we can construct Pas shown in Step 2.
Step 4: To find P−1, we can use the formula P−1=PT, as Pis an orthogonal
matrix.
Step 5: Therefore, the matrix Psuch that A=P DP −1is:
P=v1v2v3
Remember to normalize the eigenvectors if necessary to make Pan orthog-
onal matrix.
Question 11
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4.
Determine the characteristic polynomial of Aand find an expression for the
matrix inverse of Ausing its eigenvalues.
Solution
1. We know that the characteristic polynomial of a square matrix Ais given
by det(A−λI), where λis an eigenvalue and Iis the identity matrix of
the same size. Therefore, the characteristic polynomial of Ais:
det(A−λI)=(λ−2)(λ+ 1)(λ−4)
2. To find the matrix inverse of Ausing its eigenvalues, we can use the
formula:
A−1=
n
X
i=1
1
λi
projvi
where λiare eigenvalues and projvidenotes the projection matrix onto
the eigenvector corresponding to eigenvalue λi.
10
3. Let v1, v2, v3be the eigenvectors corresponding to the eigenvalues 2,−1,4
respectively. Then, we have:
A−1=1
2projv1+1
−1projv2+1
4projv3
=1
2v1vT
1+1
−1v2vT
2+1
4v3vT
3
Question 12
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3 and λ3= 3. If the
eigenvectors corresponding to λ2= 3 are v2=
1
2
−1
and v3=
1
3
−2
, find
the matrix A.
Solution
Step 1: Recall that for each eigenvalue λi, the corresponding eigenvectors vi
satisfy the equation (A−λiI)vi=0.
Step 2: For λ2= 3, we have:
(A−3I)v2=0
Substitute the given values:
a b c
d e f
g h i
−
300
030
003
1
2
−1
=
0
0
0
Step 3: This gives us the following system of equations:
a−3=0
d= 0
g+ 3 = 0
b+ 2e= 0
c−f= 0
h= 0
i−3=0
Step 4: Solving the system of equations, we find a= 3, b=−2, c= 3, d= 0,
e= 1, f= 3, g=−3, h= 0, i= 3.
Therefore, the matrix Ais:
A=
3−2 3
0 1 3
−303
11
Question 13
Question
Let Abe a 3×3 matrix with eigenvalues λ= 2,1,−3. If det(2A2−A+3I) = 60,
where Iis the identity matrix, find det(A).
Solution
Step 1: We know that the determinant of a matrix is equal to the product of
its eigenvalues. Let Ahave eigenvalues λ1, λ2, λ3. Therefore, we have:
det(A) = λ1·λ2·λ3
Step 2: From the given information, we know the eigenvalues of Aare λ1= 2,
λ2= 1, and λ3=−3. Therefore,
det(A)=2·1·(−3) = −6
Step 3: Next, we are given that det(2A2−A+ 3I) = 60. We can use the
fact that det(cA) = cn·det(A), where cis a scalar and nis the dimension of the
matrix A. Using this property, we can simplify det(2A2−A+ 3I) as follows:
det(2A2−A+ 3I) = 23·det(A2)−det(A) + 33·det(I) = 8det(A2)−6 + 27 ·1
Step 4: Since det(A2) = (det(A))2, we can rewrite the expression as:
det(2A2−A+ 3I) = 8(det(A))2−6 + 27
Step 5: Given that the expression equals 60, we have:
8(det(A))2−6 + 27 = 60 =⇒8(det(A))2= 39 =⇒(det(A))2=39
8
Step 6: Finally, we find det(A) by taking the square root of 39
8:
det(A) = r39
8=√39
2√2=√39
2√2·√2
√2=√78
4=√78
4
Question 14
Question
Let Abe a 3 ×3 matrix given by
A=
310
020
112
.
Find the eigenvalues and eigenvectors of matrix A.
12
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation det(A−λI) = 0, where λis the eigenvalue and Iis the identity matrix.
So we have:
A−λI =
3−λ1 0
0 2 −λ0
1 1 2 −λ
.
Finding the determinant and setting it equal to zero gives us:
det(A−λI) =
3−λ1 0
0 2 −λ0
1 1 2 −λ
= 0.
Step 2: Expanding the determinant, we get:
(3 −λ)((2 −λ)(2 −λ)−0) −1(0 −0) + 0 = 0.
(3 −λ)(λ2−4λ+ 4) = 0.
(3 −λ)(λ−2)2= 0.
Step 3: Solving (3 −λ)(λ−2)2= 0 gives us the eigenvalues:
λ1= 3 multiplicity 1
λ2= 2 multiplicity 2
Step 4: To find the eigenvectors corresponding to each eigenvalue, we solve
the system of equations (A−λI)v= 0 for each eigenvalue.
For λ= 3, solving (A−3I)v= 0 gives the eigenvector
v1=
1
0
−1
.
For λ= 2, solving (A−2I)v= 0 gives the eigenvector
v2=
−1
0
1
.
Therefore, the eigenvalues of matrix Aare 3 with eigenvector
1
0
−1
and 2
with eigenvector
−1
0
1
.
13
Question 15
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3=−1.
Determine the eigenvalues of the matrix 2A−3A2.
Solution
Step 1: Find the eigenvalues of A2.
Since Ahas eigenvalues λ1= 2, λ2= 3, and λ3=−1, the eigenvalues of A2
will be λ2
1= 4, λ2
2= 9, and λ2
3= 1.
Step 2: Calculate the eigenvalues of 2A−3A2.
The matrix 2A−3A2will have eigenvalues 2λi−3λ2
ifor i= 1,2,3.
Substituting the given eigenvalues, we have: - For λ1= 2: 2(2) −3(4) =
4−12 = −8 - For λ2= 3: 2(3) −3(9) = 6 −27 = −21 - For λ3=−1:
2(−1) −3(1) = −2−3 = −5
Therefore, the eigenvalues of 2A−3A2are −8, −21, and −5.
Question 16
Question
Let
A=2 1
1 3
be a 2x2 matrix. Find the eigenvalues of Aand corresponding eigenvectors.
Solution
Step 1: To find the eigenvalues of matrix A, we solve the characteristic equation
|A−λI|= 0, where λrepresents the eigenvalue and Iis the identity matrix.
A−λI =2−λ1
1 3 −λ
|A−λI|=
2−λ1
1 3 −λ
= (2 −λ)(3 −λ)−1=0
λ2−5λ+ 5 = 0
Step 2: Solving the quadratic equation gives the eigenvalues:
λ=5±√5
2
14
Step 3: Now, we find the eigenvectors corresponding to each eigenvalue. For
λ=5+√5
2, we solve the system (A−λI)v= 0:
"2−5+√5
21
1 3 −5+√5
2#v1
v2=0
0
Solving the equations gives v1=p5 + √5 and v2= 1.
Therefore, the eigenvector corresponding to λ=5+√5
2is p5 + √5
1.
Step 4: Similarly, for λ=5−√5
2, we solve the system (A−λI)v= 0 to find
the eigenvector:
v=p5−√5
−1
Therefore, the eigenvector corresponding to λ=5−√5
2is p5−√5
−1.
Question 17
Question
Let Aand Bbe 2×2 matrices with eigenvalues λ1= 2 and λ2=−1, respectively.
Consider the matrix C=A2+ 3B.
Find the eigenvalues of matrix C.
Solution
Let’s first find the eigenvalues of matrix C=A2+ 3B.
Step 1: Find the eigenvalues of A2.
Since Ahas eigenvalues λ1= 2 and λ2=−1, the eigenvalues of A2are the
squares of the eigenvalues of A:
λA2= 22= 4 and (−1)2= 1.
Step 2: Find the eigenvalues of 3B.
Since Bhas eigenvalues λ1= 2 and λ2=−1, the eigenvalues of 3Bare
simply 3 times the eigenvalues of B:
λ3B= 3 ·2 = 6 and 3 ·(−1) = −3.
Step 3: Find the eigenvalues of C=A2+ 3B.
The eigenvalues of Care the sums of the eigenvalues of A2and 3B:
λC={4+6,4−3,1+6,1−3}={10,1,7,−2}.
Therefore, the eigenvalues of matrix Care λ1= 10, λ2= 1, λ3= 7, λ4=−2.
15
Question 18
Question
Let A=5 2
4 3. Determine the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation det(A−λI) = 0, where Iis the identity matrix.
Step 2: The characteristic equation is det 5 2
4 3−λ1 0
0 1= 0.
Step 3: Simplifying, we have det 5−λ2
4 3 −λ= 0.
Step 4: Expanding the determinant, we get (5 −λ)(3 −λ)−2·4 = 0.
Step 5: Simplifying further, we obtain 15 −8λ+λ2−8 = 0.
Step 6: This equation simplifies to λ2−8λ+ 7 = 0.
Step 7: Solving the quadratic equation, we find two eigenvalues: λ1= 1 and
λ2= 7.
Step 8: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute them back into (A−λI)v=0, where vis the eigenvector.
Step 9: For λ= 1, we have (A−1I)v=4 2
4 2x
y=0
0.
Step 10: This gives us the equations 4x+ 2y= 0 and 4x+ 2y= 0. Solving,
we find that x=−1
2y. Letting y= 2, we get the eigenvector v1=−1
2.
Step 11: For λ= 7, we have (A−7I)v=−2 2
4−4x
y=0
0.
Step 12: This gives us the equations −2x+ 2y= 0 and 4x−4y= 0. Solving,
we find that x=y. Letting y= 1, we get the eigenvector v2=1
1.
Step 13: Therefore, the eigenvalues of matrix Aare λ1= 1 with eigenvector
v1=−1
2and λ2= 7 with eigenvector v2=1
1.
Question 19
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2,and λ3= 4. If
B=A2−3A+ 2I, find the eigenvalues of B.
Solution
Given that Ais a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2,and λ3= 4. We
want to find the eigenvalues of matrix B=A2−3A+ 2I.
16
Step 1: Find the eigenvalues of B.
First, let’s express Bin terms of A:
B=A2−3A+ 2I
We can write this as:
B=A2−3A+ 2I=A·A−3A+ 2I
Now, let’s substitute the eigenvalues of Ainto the expression:
B=A2−3A+ 2I=
100
020
004
2
−3
100
020
004
+ 2
100
010
001
Calculating the expression, we get:
B=
1 0 0
0 4 0
0 0 16
−
3 0 0
0 6 0
0 0 12
+
200
020
002
B=
000
000
006
Step 2: Find the eigenvalues of B.
The eigenvalues of Bare the values of λsuch that |B−λI|= 0. Let’s
calculate this determinant.
|B−λI|=
−λ0 0
0−λ0
0 0 6 −λ
= (−λ)(−λ)(6 −λ)
=−λ2(6 −λ) = −λ2(6 −λ) = −λ3+ 6λ2
To find the eigenvalues of B, we solve the characteristic equation −λ3+6λ2=
0. Factoring out λ2:
λ2(−λ+ 6) = 0
Setting each factor to 0 gives us the eigenvalues:
λ2= 0 =⇒λ= 0
−λ+ 6 = 0 =⇒λ= 6
Therefore, the eigenvalues of matrix Bare λ= 0 and λ= 6.
Question 20
Question
Let Abe a 3 ×3 matrix with eigenvalues 1, −2, and 4. If B=A3−2A2+ 3A,
find the eigenvalues of B.
17
Solution
Step 1: First, let’s find the eigenvalues of A3,A2, and A.
For the matrix Awith eigenvalues 1, −2, and 4, the eigenvalues of A3,
A2, and Aare simply the eigenvalues of Araised to the power of 3, 2, and 1
respectively. Thus, the eigenvalues of A3are 13= 1, (−2)3=−8, and 43= 64,
the eigenvalues of A2are 12= 1, (−2)2= 4, and 42= 16, and the eigenvalues
of Aare 1, −2, and 4.
Step 2: Next, let’s find the matrix B.
We have B=A3−2A2+ 3A. Substituting the eigenvalues of A3,A2, and
Afound in step 1, we get the eigenvalues of Bas follows:
Eigenvalues of B= 1 −2·1+3·1=2,
Eigenvalues of B=−8−2·4+3·(−2) = −21,
Eigenvalues of B= 64 −2·16 + 3 ·4 = 24.
Therefore, the eigenvalues of matrix Bare 2, −21, and 24.
Question 21
Question
Let Abe a 3 ×3 matrix with characteristic polynomial λ3−6λ2+ 11λ−6.
If v1, v2,and v3are eigenvectors corresponding to eigenvalues λ1, λ2,and λ3
respectively, find vT
1Av2.
Solution
Step 1: First, we need to find the eigenvalues of the matrix Aby solving the
characteristic equation det(A−λI) = 0.
(A−λI) =
a11 −λ a12 a13
a21 a22 −λ a23
a31 a32 a33 −λ
The characteristic polynomial p(λ) = det(A−λI) is given as λ3−6λ2+ 11λ−6.
Thus, the eigenvalues are the roots of this polynomial, which are λ= 1,2,3.
Step 2: Next, we find the eigenvectors by solving the system of equations
(A−λiI)vi= 0 for each eigenvalue λi. For λ= 1: Solve (A−I)v1= 0. For
λ= 2: Solve (A−2I)v2= 0. For λ= 3: Solve (A−3I)v3= 0.
Step 3: Since v1, v2,and v3are eigenvectors, they are linearly independent
and form a basis of R3. Thus, vT
1Av2= 0 as any two different eigenvectors are
orthogonal.
18
Question 22
Question
Let Abe a 3 ×3 matrix given by
A=
−2 1 0
0−2 1
0 0 −2
.
Find the eigenvalues and eigenvectors of A.
Solution
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 of the same size
as A.
|A−λI|=
−2−λ1 0
0−2−λ1
0 0 −2−λ
= (−2−λ)((−2−λ)2)
= (−2−λ)3= 0.
This implies that the eigenvalue is λ=−2 with algebraic multiplicity of 3.
To find the eigenvectors corresponding to the eigenvalue λ=−2, we need to
solve the system of equations (A−λI)v= 0. Substitute λ=−2 back into the
equation A−λI = 0:
A+ 2I=
010
001
000
.
Solving the system (A+ 2I)v= 0 leads to the following eigenvectors:
v=t
1
0
0
,
where tis a scalar. Thus, the eigenvalue -2 has a geometric multiplicity of 1.
Question 23
Question
Let Abe a 3x3 matrix with eigenvalues λ1= 2, λ2= 4, and λ3= 6. Find the
eigenvectors corresponding to each eigenvalue.
19
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we solve the system of
equations (A−2I)v1=0where v1=
x
y
z
.
A−2I=
−1 2 −3
101
2−2−4
Using row reduction, we get:
−1 2 −3
101
2−2−4
→
101
011
000
So, one possible eigenvector is v1=
−1
−1
1
.
Step 2: To find the eigenvector corresponding to λ2= 4, we solve the system
of equations (A−4I)v2=0where v2=
x
y
z
.
A−4I=
−2 2 −3
1−4 1
2−2−6
Using row reduction, we get:
−2 2 −3
1−4 1
2−2−6
→
101
011
000
So, one possible eigenvector is v2=
−1
−1
1
.
Step 3: To find the eigenvector corresponding to λ3= 6, we solve the system
of equations (A−6I)v3=0where v3=
x
y
z
.
A−6I=
−4 2 −3
1−6 1
2−2−8
Using row reduction, we get:
−4 2 −3
1−6 1
2−2−8
→
101
011
000
20
So, one possible eigenvector is v3=
−1
−1
1
.
Question 24
Question
Let Abe a 3×3 matrix with eigenvalues 2, −1, and 3. Given that the eigenvector
corresponding to the eigenvalue 2 is
1
0
1
, find the eigenvectors corresponding
to the eigenvalues −1 and 3.
Solution
Step 1: Finding the eigenvector corresponding to eigenvalue −1
Let vbe the eigenvector corresponding to the eigenvalue −1. We know that
for an eigenvector vcorresponding to an eigenvalue λ, the matrix Asatisfies the
equation (A−λI)v= 0, where Iis the identity matrix.
Therefore, for λ=−1:
(A+I)v= 0
Substitute Aand λ:
200
0−1 0
003
+
100
010
001
x
y
z
=
0
0
0
Simplify:
300
000
004
x
y
z
=
0
0
0
This gives the following system of equations:
3x= 0,0y= 0,4z= 0
Since ycan be any value, let y=t. We can choose x=z= 0 and y= 1
to get a non-trivial solution. Therefore, an eigenvector corresponding to the
eigenvalue −1 is
0
1
0
.
Step 2: Finding the eigenvector corresponding to eigenvalue 3
Let wbe the eigenvector corresponding to the eigenvalue 3. Using the same
method as above, we have:
(A−3I)w= 0
21
Substitute Aand λ:
200
0−1 0
003
−
300
030
003
x
y
z
=
0
0
0
This gives the following system of equations:
−x= 0,−y= 0,0=0
So x=yand zis arbitrary. Let x=y= 1 and z= 0, we get an eigenvector
corresponding to the eigenvalue 3 as
1
1
0
.
Question 25
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−3, and λ3= 4. Given
that the corresponding eigenvectors are v1=
1
1
1
,v2=
1
0
1
, and v3=
0
1
1
,
find an invertible matrix Psuch that P−1AP is a diagonal matrix.
Solution
Step 1: Form the matrix Pusing the eigenvectors of A.
P= [v1,v2,v3] =
110
101
111
Step 2: Calculate the inverse of P. Firstly, find the determinant of P:
det(P) = 1(0 −1) −1(1 −0) + 0(1 −1) = −1
Next, calculate the cofactor matrix of Pby finding the matrix of minors,
then appending the correct signs:
Cof(P) =
0−1 1
1−1−1
−1 1 0
Then, find the adjugate of Pby taking the transpose of the cofactor matrix:
Adj(P) =
0 1 −1
−1−1 1
1−1 0
22
Finally, calculate P−1using the formula P−1=1
det(P)·Adj(P).
P−1=1
−1
0 1 −1
−1−1 1
1−1 0
=
0−1 1
1 1 −1
−1 1 0
Step 3: Calculate P−1AP to obtain a diagonal matrix.
P−1AP =
0−1 1
1 1 −1
−1 1 0
200
0−3 0
004
110
101
111
=
200
0−3 0
004
Therefore, the matrix P−1AP is a diagonal matrix as desired.
Question 26
Question
Let Abe a 3 ×3 matrix with eigenvalues 1,2,and 3. If B= 2A2−3A+I,
where Iis the identity matrix, determine the eigenvalues of B.
Solution
To find the eigenvalues of B, we need to find the roots of the characteristic
polynomial of B.
Step 1: Find the characteristic polynomial of BThe characteristic
polynomial of a matrix Bis given by det(B−λI) = 0, where λis the eigenvalue
we are trying to find. Therefore,
det(B−λI) = det(2A2−3A+I−λI) = 0.
Step 2: Substitute the expression for BSubstitute B= 2A2−3A+I
into the equation above:
det(2A2−3A+I−λI)=0.
Step 3: Factor out the terms Since Ahas eigenvalues 1,2,and 3, we can
rewrite Aas a diagonal matrix:
A=
100
020
003
.
Substitute this into the equation:
det
2
100
020
003
2
−3
100
020
003
+
100
010
001
−λ
100
010
001
= 0.
Step 4: Calculate the characteristic polynomial Calculate the deter-
minant of the given matrix expression to obtain the characteristic polynomial
of B. Solving this characteristic polynomial will give us the eigenvalues of B.
23
Question 27
Question
Let Abe a 3 ×3 matrix such that A3−6A2+ 11A−6I=O, where Iis the
3×3 identity matrix. Find all the eigenvalues of A.
Solution
To find the eigenvalues of A, we will first find the characteristic polynomial of
Ausing the given equation.
Step 1: Define the characteristic polynomial p(λ). Let p(λ) = det(A−λI) be
the characteristic polynomial of A. The given equation A3−6A2+11A−6I=O
can be rewritten as:
A(A2−6A+ 11I) = 6I
Multiplying by A−λI on both sides, we get:
A(A2−6A+ 11I)=6I=⇒det(A−λI) = 0
Thus, p(λ) = λ3−6λ2+ 11λ−6.
Step 2: Find the eigenvalues of A. To find the eigenvalues of A, we solve
p(λ) = 0. By inspection, we find that λ= 1 is a root of p(λ). Using polynomial
division or synthetic division, we find that:
p(λ)=(λ−1)(λ2−5λ+ 6) = (λ−1)2(λ−6)
Therefore, the eigenvalues of Aare λ= 1 with multiplicity 2 and λ= 6.
Question 28
Question
Let Abe a square matrix such that A2= 5A−6I, where Iis the identity matrix.
If vis an eigenvector of Acorresponding to the eigenvalue 3, find A−1v.
Solution
Step 1: First, let’s find the eigenvalues of matrix A. Given that A2= 5A−6I, we
can rewrite the equation as A2−5A+6I= 0. This means that the characteristic
polynomial of Ais detA2−5A+ 6I= 0. Expanding this determinant gives us
detA2−5A+ 6I= det(A−3I) det(A−2I) = 0. Therefore, the eigenvalues
of Aare 3 and 2.
Step 2: Next, find the eigenvectors corresponding to the eigenvalues. For
eigenvalue 3, we solve the system (A−3I)v= 0 where vis the eigenvector
corresponding to eigenvalue 3. This gives us (A−3I)v= 0 which implies
(A−3I)v= 0v. Solving the equation (A−3I)v= 0 gives us the eigenvector
corresponding to eigenvalue 3.
24
Step 3: Now that we have the eigenvector vcorresponding to the eigenvalue
3, we can find A−1v. Recall that if vis an eigenvector of Awith eigenvalue
λ, then A−1vis an eigenvector of A−1with eigenvalue 1/λ. Since vis an
eigenvector of Acorresponding to the eigenvalue 3, A−1vis an eigenvector of
A−1with eigenvalue 1/3.
Therefore, A−1vis an eigenvector of A−1with eigenvalue 1/3.
Question 29
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, λ3= 3. Given that
the eigenvectors corresponding to λ1and λ3are
1
0
2
and
1
1
1
, respectively, find
the eigenvector corresponding to λ2.
Solution
Step 1: Recall that the eigenvectors of a matrix Aare the solutions to the
equation (A−λI)x=0, where λis an eigenvalue of A.
Step 2: Let’s first find the eigenvector corresponding to λ2= 2. Using the
equation (A−λ2I)v=0, we have:
a b c
d e f
g h i
−2
100
010
001
x
y
z
=
0
0
0
Simplifying, we get:
a−2b c
d e −2f
g h i −2
x
y
z
=
0
0
0
Step 3: So, we have the following system of equations:
(a−2)x+by +cz = 0
dx + (e−2)y+fz = 0
gx +hy + (i−2)z= 0
Step 4: Since we are given the eigenvectors corresponding to λ1and λ3, we
can substitute them into the system of equations to obtain additional informa-
tion. Let’s substitute
1
0
2
and
1
1
1
into the system.
Step 5: Substituting
1
0
2
into the system gives us the equations:
25
(a−2) + 0 + 2c= 0
d+ (e−2) ·0 + f·2=0
g+h+ (i−2) ·2=0
Solving this system we can find values for a, b, c, d, e, f, g, h, i.
Step 6: Next, substituting
1
1
1
into the system gives us another set of
equations, which we can solve to find the eigenvector corresponding to λ2= 2.
Question 30
Question
Let A=4 1
1 4. Find the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0, where Iis the identity matrix.
det 4−λ1
1 4 −λ= 0
(4 −λ)2−1=0
λ2−8λ+ 15 = 0
(λ−3)(λ−5) = 0
So, the eigenvalues are λ1= 3 and λ2= 5.
Step 2: Now, we find the eigenvector associated with λ1= 3. Let v1=x
y
be the eigenvector. Then, we have:
(A−3I)v1= 0
1 1
1 1x
y=0
0
x+y= 0
Thus, the eigenvector corresponding to λ1= 3 is any non-zero scalar multiple
of 1
−1.
26
Step 3: Next, we find the eigenvector associated with λ2= 5. Let v2=x
y
be the eigenvector. Then, we have:
(A−5I)v2= 0
−1 1
1−1x
y=0
0
−x+y= 0
Thus, the eigenvector corresponding to λ2= 5 is any non-zero scalar multiple
of 1
1.
Question 31
Question
Let Abe a 3 ×3 matrix with eigenvalues −2,1,and 3. Find the eigenvectors
corresponding to each eigenvalue.
Solution
Step 1: To find the eigenvectors, we need to solve the system of equations
(A−λI)v=0for each eigenvalue λ.
For λ=−2:
A−λI =
000
030
005
−(−2)
100
010
001
=
200
050
007
The system of equations becomes:
200
050
007
x
y
z
=
0
0
0
Solving this system, we get x=y=z= 0. Therefore, the eigenvector
corresponding to λ=−2 is
0
0
0
.
Step 2: Repeat the same process for λ= 1:
A−λI =
000
030
005
−1
100
010
001
=
−100
0 2 0
0 0 4
27
The system of equations becomes:
−100
0 2 0
0 0 4
x
y
z
=
0
0
0
Solving this system, we get x=y= 0 and zcan be any non-zero scalar.
Therefore, an eigenvector corresponding to λ= 1 is any vector of the form
0
0
1
.
Step 3: Lastly, repeat the process for λ= 3:
A−λI =
000
030
005
−3
100
010
001
=
−300
0 0 0
0 0 2
The system of equations becomes:
−300
0 0 0
0 0 2
x
y
z
=
0
0
0
Solving this system, we get y=z= 0 and xcan be any non-zero scalar.
Therefore, an eigenvector corresponding to λ= 3 is any vector of the form
1
0
0
.
Question 32
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 3. Given
that corresponding eigenvectors are
v1=
1
0
1
,v2=
1
1
0
,and v3=
1
0
−1
,
write the matrix A.
Solution
To write the matrix A, we can use the formula for diagonalizing a matrix:
A=P DP −1where Dis a diagonal matrix with eigenvalues of Aon the diagonal
sorted according to their respective eigenvectors in P.
Step 1: Compute the matrix Pusing the given eigenvectors.
P=v1v2v3=
1 1 1
0 1 0
1 0 −1
28
Step 2: Compute the inverse of matrix Pto get P−1.To find P−1,
we need to solve the equation P P −1=I, where Iis the identity matrix.
1 1 1
0 1 0
1 0 −1
a d g
b e h
c f i
=
100
010
001
Solving this system of equations gives P−1.
P−1=
1−1 1
010
−1 0 −1
Step 3: Compute the diagonal matrix Dwith eigenvalues on the
diagonal.
D=
200
0−1 0
003
Step 4: Write the matrix Ausing the formula A=P DP −1.Substi-
tuting the values of P,D, and P−1into the formula:
A=P DP −1=
1 1 1
0 1 0
1 0 −1
200
0−1 0
003
1−1 1
010
−1 0 −1
Calculating the matrix product gives us the matrix A.
Question 33
Question
Let A be a 3x3 matrix with eigenvalues 2, 3, and 4. If B = A2−3A+ 5I, find
the eigenvalues of B.
Solution
To find the eigenvalues of B, we will first find the matrix A and then use the
properties of eigenvalues to find the eigenvalues of B.
Step 1: Finding A From the given eigenvalues of A, we know that the
characteristic equation of A is given by:
det(A−λI) = 0
where λrepresents an eigenvalue.
Substitute the given eigenvalues into the characteristic equation to find the
matrix A.
29
Step 2: Finding B Now, we can find B using the equation stated in the
question: B = A2−3A+ 5I.
Substitute the matrix A into the expression for B to calculate the matrix B.
Step 3: Finding the eigenvalues of B The eigenvalues of B are equal to
the eigenvalues of the matrix B.
Calculate the eigenvalues of B by finding the roots of the characteristic
equation of B, given by:
det(B−λI)=0
where λrepresents an eigenvalue.
Solve the characteristic equation to find the eigenvalues of B.
Therefore, the eigenvalues of matrix B are the roots of the characteristic
equation.
Question 34
Question
Let A=3 1
3 1. Determine the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0.
det 3−λ1
3 1 −λ= 0.
Step 2: Expanding the determinant, we get:
(3 −λ)(1 −λ)−3=0.
Step 3: Simplifying the equation gives:
λ2−4λ= 0.
Step 4: Factoring out λ, we have:
λ(λ−4) = 0.
Step 5: This equation has two solutions: λ= 0 and λ= 4. Therefore, the
eigenvalues of Aare 0 and 4.
Step 6: To find the eigenvector corresponding to λ= 0, we substitute λ= 0
back into the equation (A−λI)v= 0.
(A−0I)v=3 1
3 1v=0
0.
30
Step 7: Solving the system of equations, we get v=t−1
3for some scalar
t= 0.
Step 8: Therefore, the eigenvector corresponding to λ= 0 is v1=−1
3.
Step 9: To find the eigenvector corresponding to λ= 4, we substitute λ= 4
back into the equation (A−λI)v= 0.
(A−4I)v=−1 1
3−3v=0
0.
Step 10: Solving the system of equations, we get v=t1
1for some scalar
t= 0.
Step 11: Therefore, the eigenvector corresponding to λ= 4 is v2=1
1.
Question 35
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 5. Find the
eigenvalues of the matrix A2−4A+ 3I, where Iis the 3 ×3 identity matrix.
Solution
Given that Ahas eigenvalues λ1= 2, λ2= 3, and λ3= 5, we want to find the
eigenvalues of the matrix A2−4A+ 3I.
Step 1: Find the eigenvalues of A2The eigenvalues of A2are simply the
squares of the eigenvalues of A. Therefore, the eigenvalues of A2are λ2
1= 22= 4,
λ2
2= 32= 9, and λ2
3= 52= 25.
Step 2: Find the eigenvalues of −4AThe eigenvalues of −4Aare simply
−4 times the eigenvalues of A. Therefore, the eigenvalues of −4Aare −4·2 = −8,
−4·3 = −12, and −4·5 = −20.
Step 3: Find the eigenvalues of A2−4A+ 3ITo find the eigenvalues of
A2−4A+ 3I, we add the corresponding eigenvalues of A2,−4A, and 3I. Thus,
the eigenvalues of A2−4A+ 3Iare:
4−8 + 3 = −1,9−12 + 3 = 0,and 25 −20 + 3 = 8.
Therefore, the eigenvalues of the matrix A2−4A+ 3Iare −1, 0, and 8.
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