MATH 334 - DIFFERENTIAL
EQUATIONS - Matrix methods and
eigenvalue problems
Question Bank - Set 1
Liberty University
Question 1
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 3,1,−2. Find the eigenvalues of
the matrix B= 2A2−A+I, where Iis the 3 ×3 identity matrix.
Solution
Step 1: First, calculate A2.
A2=A·A
Step 2: Next, calculate 2A2and −A. Then, add 2A2−A+I. Step 3: Using
the eigenvalues of A, find the eigenvalues of 2A2−A+I.
Question 2
Question
Let Abe a 3×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Determine
the eigenvalues of the matrix 2A2−3A+I, where Iis the 3 ×3 identity matrix.
Solution
Let λbe an eigenvalue of the matrix 2A2−3A+I, and let vbe the corresponding
eigenvector. Then we have
(2A2−3A+I)v=λv.
Expanding the left side gives
2A2v−3Av +Iv =λv.
Since Av =λv for some eigenvalue λ, we can rewrite the equation as
2λ2v−3λv +v=λv.
Dividing through by v(which is nonzero since it is an eigenvector) gives us
the characteristic equation for the eigenvalues λ:
2λ2−3λ+ 1 = 0.
Now, we solve this quadratic equation to find the eigenvalues. Factoring or
using the quadratic formula, we get
λ=3±√32−4·2·1
2·2=3±√1
4=3±1
4.
Therefore, the eigenvalues of 2A2−3A+Iare λ=4
4= 1 and λ=2
4=1
2.
Question 3
Question
Let Abe a 3 ×3 matrix given by
A=
210
121
012
.
Find the eigenvalues and eigenvectors of A.
Solution
Step 1: Find the characteristic equation. The characteristic equation of
a matrix Ais given by det(A−λI) = 0, where λis an eigenvalue and Iis the
identity matrix. For a 3 ×3 matrix A, this characteristic equation will be a
cubic equation in λ.
First, we find A−λI:
A−λI =
210
121
012
−λ
100
010
001
=
2−λ1 0
1 2 −λ1
0 1 2 −λ
.
Next, we find the determinant of A−λI and set it equal to zero to find the
characteristic equation. This gives:
det(A−λI) = (2 −λ)((2 −λ)2−1) −1(1 −λ) = λ3−6λ2+ 11λ−6 = 0.
2
Step 2: Solve the characteristic equation. To find the eigenvalues, we
solve the characteristic equation λ3−6λ2+ 11λ−6 = 0. By inspection, we find
that one root is λ= 1.
To find the other roots, we can use synthetic division or other root finding
methods. The characteristic equation factors as (λ−1)(λ−2)(λ−3) = 0, so
the eigenvalues are λ1= 1, λ2= 2,and λ3= 3.
Step 3: Find the eigenvectors. For each eigenvalue, we solve the system
(A−λI)v=0to find the eigenvector associated with that eigenvalue. We start
by finding the eigenvector corresponding to λ= 1.
For λ= 1, we solve the system (A−I)v=0:
(A−I)v=
110
111
011
v=
0
0
0
.
Putting this system into row-echelon form, we get:
110
111
011
→
110
001
011
→
1 0 −1
0 1 1
0 0 1
.
So, the eigenvector corresponding to λ= 1 is v1=
−1
1
1
.
We can proceed similarly to find the eigenvectors corresponding to the eigen-
values λ= 2 and λ= 3.
Question 4
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−3. If the
corresponding eigenvectors are v1=
1
−1
1
,v2=
2
0
−1
, and v3=
1
0
1
, find
A.
Solution
Step 1: Recall that if λis an eigenvalue of Awith eigenvector v, we have
Av=λv. This means that for each eigenvalue, we have the following equations:
For λ1= 2 and v1=
1
−1
1
:
Av1= 2v1
3
A
1
−1
1
= 2
1
−1
1
A
1
−1
1
=
2
−2
2
Step 2: Similarly, for λ2= 1 and v2=
2
0
−1
:
Av2= 1v2
A
2
0
−1
= 1
2
0
−1
A
2
0
−1
=
2
0
−1
Step 3: Lastly, for λ3=−3 and v3=
1
0
1
:
Av3=−3v3
A
1
0
1
=−3
1
0
1
A
1
0
1
=
−3
0
−3
Step 4: Combining the results from Steps 1, 2, and 3:
A=
2 2 −3
−2 0 0
2−1−3
Question 5
Question
Let A=
401
041
116
. Find the characteristic polynomial of Aand hence deter-
mine the eigenvalues of A.
4
Solution
Step 1: To find the characteristic polynomial of A, we need to calculate det(A−
λI), where Iis the identity matrix and λis an eigenvalue.
A−λI =
4−λ0 1
0 4 −λ1
1 1 6 −λ
Step 2: Next, compute the determinant of A−λI.
det(A−λI) = (4 −λ)[(4 −λ)(6 −λ)−1] −0 + 1[(0)(6 −λ)−1(1)]
det(A−λI) = (4 −λ)[24 −10λ+λ2−1] −1
det(A−λI) = (4 −λ)(23 −10λ+λ2)+1
det(A−λI) = 92 −40λ+ 4λ2−23λ+ 10λ2−1
det(A−λI)=4λ2−63λ+ 91
Step 3: The characteristic polynomial is given by det(A−λI) = 0. So, we
set 4λ2−63λ+ 91 = 0. Solving this quadratic equation gives us the eigenvalues.
Step 4: Using the quadratic formula, we have:
λ=63 ±p632−4(4)(91)
2(4) =63 ±√3969 −1456
8
λ=63 ±√2513
8
Therefore, the eigenvalues of matrix Aare λ1=63+√2513
8and λ2=63−√2513
8.
Question 6
Question
Let Abe a 4 ×4 matrix with eigenvalues λ1= 2, λ2=−1, λ3= 3, and λ4= 4.
If Bis the inverse of A, determine the eigenvalues of B.
Solution
Let Abe a 4 ×4 matrix with eigenvalues λ1= 2, λ2=−1, λ3= 3, and λ4= 4.
We are asked to find the eigenvalues of the inverse matrix B.
Since Ais invertible, none of the eigenvalues are zero. Therefore, the eigen-
values of A−1are the reciprocal of the eigenvalues of A. In other words, if λis
an eigenvalue of A, then 1
λis an eigenvalue of A−1.
Therefore, the eigenvalues of Bare 1
2,−1, 1
3, and 1
4.
5
Question 7
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1=−2, λ2= 1, and λ3= 3. Find
the corresponding eigenvectors for each eigenvalue.
Solution
Step 1: For λ1=−2, We solve (A−λ1I)x=0:
511
151
115
x1
x2
x3
=
0
0
0
Performing row operations, we get the RREF matrix:
1 0 −1
0 1 1
0 0 0
So, the eigenvector corresponding to λ1=−2 is
1
−1
1
.
Step 2: For λ2= 1, We solve (A−λ2I)x=0:
−3 1 1
1−3 1
1 1 −3
x1
x2
x3
=
0
0
0
Performing row operations, we get the RREF matrix:
1−1−1
0 0 0
0 0 0
So, the eigenvector corresponding to λ2= 1 is
1
1
0
.
Step 3: For λ3= 3, We solve (A−λ3I)x=0:
−5 1 1
1−5 1
1 1 −5
x1
x2
x3
=
0
0
0
Performing row operations, we get the RREF matrix:
1−1
5−1
5
0 0 0
0 0 0
6
So, the eigenvector corresponding to λ3= 3 is
1
5
5
.
Question 8
Question
Let Abe a square matrix such that A2−5A+ 6I=O, where Iis the identity
matrix. Find the eigenvalues of A.
Solution
Step 1: Given that A2−5A+ 6I=O, we can rewrite this as A2−5A+ 6I= 0,
where 0 is the zero matrix. This suggests that the characteristic polynomial of
Ais p(λ) = λ2−5λ+ 6.
Step 2: To find the eigenvalues of A, we need to solve the characteristic
equation p(λ) = 0. Therefore, we solve λ2−5λ+ 6 = 0.
Step 3: Factoring the quadratic equation, we get (λ−2)(λ−3) = 0. There-
fore, the eigenvalues are λ1= 2 and λ2= 3.
Step 4: Hence, the eigenvalues of matrix Aare 2 and 3.
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−3.
Determine the eigenvectors corresponding to each eigenvalue.
Solution
We know that the eigenvectors corresponding to each eigenvalue λiare the
solutions to the system (A−λiI)v=0, where Iis the identity matrix and vis
the eigenvector corresponding to λi.
Step 1: Find the eigenvector corresponding to λ1= 2 Let v1=
x
y
z
be the eigenvector corresponding to λ1. Substitute λ1= 2 into the equation
(A−λ1I)v1=0and solve for v1.
−1 1 1
0 1 −1
1−1 3
x
y
z
=
0
0
0
7
Step 2: Solve for v1Setting up the system of equations:
−x+y+z= 0
y−z= 0
x−y+ 3z= 0
Solving this system, we find v1=
1
1
−1
.
Step 3: Find the eigenvector corresponding to λ2= 1 Let v2=
x
y
z
be the eigenvector corresponding to λ2. Substitute λ2= 1 into the equation
(A−λ2I)v2=0and solve for v2.
1 1 1
0 0 −1
1−1 4
x
y
z
=
0
0
0
Step 4: Solve for v2Setting up the system of equations:
x+y+z= 0
−z= 0
x−y+ 4z= 0
Solving this system, we find v2=
1
−1
1
.
Step 5: Find the eigenvector corresponding to λ3=−3 Let v3=
x
y
z
be the eigenvector corresponding to λ3. Substitute λ3=−3 into the equation
(A−λ3I)v3=0and solve for v3.
5 1 1
0 4 −1
1−1 6
x
y
z
=
0
0
0
Step 6: Solve for v3Setting up the system of equations:
5x+y+z= 0
4y−z= 0
x−y+ 6z= 0
Solving this system, we find v3=
1
1
1
.
8
Question 10
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If
the eigenvectors corresponding to λ1and λ2are v1=
1
2
1
and v2=
2
0
1
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: The eigenvector corresponding to λ3can be found by solving the equa-
tion (A−3I)v3= 0, where v3is the eigenvector corresponding to λ3.
Step 2: First, we find the matrix A−3I:
A−3I=
100
020
003
−
300
030
003
=
−200
0−1 0
0 0 0
Step 3: Next, we solve the equation (−2)v31 = 0, (−1)v32 = 0, and 0v33 = 0
where v3=
v31
v32
v33
.
Step 4: From the first equation −2v31 = 0, we get v31 = 0.
Step 5: From the second equation −v32 = 0, we get v32 = 0.
Step 6: From the third equation 0 = 0, we see that v33 can be any value.
Step 7: Therefore, the eigenvector v3corresponding to λ3= 3 is
0
0
1
.
Question 11
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2=−2, and λ3= 3. If
B=A2−3A+ 2I, find the eigenvalues of B.
Solution
Step 1: First, we need to find the matrix A2. Given that the eigenvalues of ma-
trix Aare λ1= 1, λ2=−2, and λ3= 3, weknowthatAissimilartoadiagonalmatrixD
=
100
0−2 0
003
.
9
Step 2: Next, we calculate A2by squaring the diagonal matrix D.A2=
D2=
100
0−2 0
003
2
=
100
040
009
.
Step 3: Now, let’s find Bby substituting A2into B=A2−3A+ 2I.
B=
100
040
009
−3
100
0−2 0
003
+ 2
100
010
001
. Simplifying, we get B=
0 0 0
0 10 0
0 0 16
.
Step 4: Finally, we find the eigenvalues of Bby observing that Bis diagonal
with entries 0, 10, and 16. Therefore, the eigenvalues of Bare 0, 10, and 16.
Question 12
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 5. Given
that the eigenvectors corresponding to λ1and λ2are v1=
1
1
1
and v2=
−1
1
2
,
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: Recall that the eigenvectors of a matrix Aare the non-zero vectors
vwhich satisfy the eigenvalue equation Av=λv, where λis the eigenvalue
corresponding to v.
Step 2: Since we are given the eigenvalues and eigenvectors of A, we can
construct the matrix Aas follows:
A=P DP −1,
where Pis the matrix whose columns are the eigenvectors of Aand Dis the
diagonal matrix with the eigenvalues of Aon the diagonal.
Step 3: Recall that the matrix Pis given by
P=v1v2v3.
Step 4: From the given information, we have P=
1−1 ?
11?
12?
. To find the
third eigenvector v3, we solve the system of equations Av3=λ3v3.
Step 5: Substituting λ3= 5, we have Av3= 5v3. Therefore, we need to
solve the system of equations (A−5I)v3= 0, where Iis the identity matrix.
10
Step 6: Substituting A, we have
(A−5I)v3=
−3−1 ?
1−6 ?
1 2 −2
x
y
z
=
0
0
0
.
Step 7: Solving the system of linear equations using Gaussian elimination,
we find the solution v3=
−1
1
1
.
Therefore, the eigenvector corresponding to the eigenvalue λ3= 5 is v3=
−1
1
1
.
Question 13
Question
Let Abe a 3 ×3 matrix with eigenvalues 1, 2, and 3. Determine the eigenvalues
of the matrix B=A3−2A2+A.
Solution
Step 1: First, let’s find the eigenvalues of matrix B.
Step 2: We know that if λis an eigenvalue of matrix A, then λis an eigenvalue
of any polynomial f(A). Specifically, if λis an eigenvalue of A, then λ3,λ2, and
λare eigenvalues of A3,A2, and A, respectively.
Step 3: Therefore, the eigenvalues of matrix Bare 13−2(1)2+ 1 = 0,
23−2(2)2+ 2 = 2, and 33−2(3)2+ 3 = 12.
Step 4: So the eigenvalues of matrix Bare 0, 2, and 12.
Question 14
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3=−3. Deter-
mine the characteristic polynomial of Aand find the eigenvectors corresponding
to each eigenvalue.
Solution
Step 1: The characteristic polynomial of a matrix Ais given by |A−λI|= 0,
where Iis the identity matrix.
11
The characteristic polynomial is then:
det
A11 −λ A12 A13
A21 A22 −λ A23
A31 A32 A33 −λ
= 0
For our 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3=−3, the
characteristic polynomial will be:
det
A11 −2A12 A13
A21 A22 + 1 A23
A31 A32 A33 + 3
= 0
Step 2: Next, find the eigenvector corresponding to each eigenvalue by
solving the equation (A−λI)v= 0 for each eigenvalue.
For λ1= 2:
A11 −2A12 A13
A21 A22 −2A23
A31 A32 A33 −2
x
y
z
=
0
0
0
Step 3: Solve the system of equations for λ1= 2 to find the eigenvector v1.
Similarly, repeat Steps 2 and 3 for eigenvalues λ2=−1 and λ3=−3 to find
the eigenvectors v2and v3respectively.
Question 15
Question
Let Abe a 3×3 matrix with characteristic polynomial p(λ) = λ3−7λ2+14λ−8.
If one of the eigenvalues of Ais 2, find the other two eigenvalues of A.
Solution
Step 1: Since 2 is an eigenvalue of A, we know that p(2) = 0. So, we have:
23−7(2)2+ 14(2) −8 = 0
8−28 + 28 −8=0
0=0
This confirms that λ= 2 is indeed an eigenvalue of A.
Step 2: By the Factor Theorem, we know that p(λ) can be factored as
(λ−2)(λ2−5λ+ 4). Therefore, the remaining eigenvalues are the roots of the
quadratic λ2−5λ+ 4.
Step 3: To find the remaining eigenvalues, we solve λ2−5λ+ 4 = 0:
λ=5±√52−4·4
2
12
λ=5±√25 −16
2
λ=5±√9
2
λ=5±3
2
So, the other two eigenvalues are λ= 4 and λ= 1.
Question 16
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1=−2, λ2= 1, and λ3= 4. Given
that the corresponding eigenvectors are v1=
1
−1
2
,v2=
0
1
−1
, and v3=
1
1
1
,
respectively, find the matrix exponential eAt.
Solution
Step 1: We can express Ain terms of its eigenvectors and eigenvalues as A=
P DP −1, where P= [v1,v2,v3] and D= diag(λ1, λ2, λ3). Therefore, we have:
P=
1 0 1
−111
2−1 1
and D=
−200
0 1 0
0 0 4
Step 2: Next, we need to compute eAt . Since A=P DP −1, we can express
eAt =P eDt P−1.
Step 3: To compute eDt, we replace λiby eλitalong the diagonal of D.
Therefore, we get:
eDt =
e−2t0 0
0et0
0 0 e4t
Step 4: Therefore, to find eAt, we have:
eAt =P eDt P−1
=
1 0 1
−111
2−1 1
e−2t0 0
0et0
0 0 e4t
1 0 1
−111
2−1 1
−1
Step 5: After performing the matrix multiplication and inversion, we arrive
at the final expression for eAt.
13
Question 17
Question
Let A be a 3x3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3=−1. If the
eigenvectors corresponding to λ1and λ2are
v1=
1
0
1
and v2=
1
1
1
respectively, find a basis for the eigenspace corresponding to λ3.
Solution
Step 1: The eigenspace corresponding to λ3=−1 is the null space of the matrix
A−λ3I.
A−λ3I=A+I=
1 0 1
0 2 1
0 0 −2
Step 2: To find a basis for the eigenspace corresponding to λ3, we need to
solve the equation (A+I)v=0, where v=
x
y
z
.
1 0 1
0 2 1
0 0 −2
x
y
z
=
0
0
0
Step 3: Solving the system of equations, we get:
x+z= 0
2y+z= 0
−2z= 0
Step 4: From the third equation, z= 0. Substituting z= 0 into the first
two equations, we get:
x= 0
2y= 0 =⇒y= 0
Therefore, a basis for the eigenspace corresponding to λ3is the vector v3=
0
0
0
.
14
Question 18
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3=−3. If
B=A2−2A+I, where Iis the identity matrix, find the eigenvalues of B.
Solution
Step 1: First, let’s find the matrix B. Given that B=A2−2A+I, we can
substitute the eigenvalues of Ainto the expression to find the corresponding
eigenvectors. Let v1, v2,and v3be the eigenvectors corresponding to eigenvalues
1,2,and −3 respectively.
Step 2: For λ= 1, we have: (A2−2A+I)v1= 1v1
Expanding this equation and substituting Av1=λv1gives: A2v1−2Av1+
v1=v1
Step 3: For λ= 2, we have: (A2−2A+I)v2= 2v2
Similarly, we get: A2v2−2Av2+v2= 2v2
Step 4: For λ=−3, we have: A2v3−2Av3+v3=−3v3
This simplifies to: A2v3−2Av3+v3=−3v3
Step 5: Now, we can find the eigenvalues of Bby evaluating Bvifor i= 1,2,3
using the properties of matrices and the eigenvectors of A.
Bv1= (A2−2A+I)v1=v1
Bv2= (A2−2A+I)v2= 2v2
Bv3= (A2−2A+I)v3= (−3)v3
Step 6: Therefore, the eigenvalues of Bare: λ′
1= 1, λ′
2= 2, and λ′
3=−3.
Question 19
Question
Let Abe a 3 ×3 matrix with eigenvalues −2, 1, and 4. If v1,v2, and v3are the
corresponding eigenvectors, find the matrix A.
Solution
Step 1: Recall that if λis an eigenvalue of a matrix A, and vis the corresponding
eigenvector, then Av =λv.
Step 2: Let’s denote the matrix Aas:
A=
a b c
d e f
g h i
Step 3: Since −2, 1, and 4 are eigenvalues of A, we have:
Av1=−2v1, Av2=v2, Av3= 4v3
15
Step 4: Writing out these equations explicitly, we get:
a b c
d e f
g h i
x1
y1
z1
=
−2x1
−2y1
−2z1
a b c
d e f
g h i
x2
y2
z2
=
x2
y2
z2
a b c
d e f
g h i
x3
y3
z3
=
4x3
4y3
4z3
Step 5: Solving these equations, we find that Ais:
A=
−200
0 1 0
0 0 4
Question 20
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 3. Suppose
that the corresponding eigenvectors are v1=
1
0
1
,v2=
1
1
1
, and v3=
1
2
1
,
normalized to have a length of 1. Find A.
Solution
To find the matrix A, we can use the fact that the matrix Acan be written as
A=P DP −1, where Pis the matrix whose columns are the eigenvectors of A
and Dis the diagonal matrix consisting of the eigenvalues.
Step 1: Finding the matrix P
The matrix Pis formed by the eigenvectors of A.
P=
111
012
111
Step 2: Finding the diagonal matrix D
The diagonal matrix Dis formed by the eigenvalues of A.
D=
200
0−1 0
003
16
Step 3: Finding matrix A
To find A, we use the formula A=P DP −1.
A=P DP −1
We first calculate P−1using the inverse matrix formula:
P−1=1
det(P)adj(P)
det(P) = 1(1 −2) −1(1 −2) + 1(1 −2) = 0
Since the determinant is 0, Pis not invertible. Hence, matrix Acannot be
uniquely determined in this case.
Question 21
Question
Let Abe a 3×3 matrix with characteristic polynomial p(λ) = λ3−5λ2+ 8λ−4.
If the eigenvalues of Aare λ1, λ2,and λ3, find the eigenvalues of A2−4A+ 3I,
where Iis the 3 ×3 identity matrix.
Solution
Step 1: First, we need to find the eigenvalues of A. Given that the characteristic
polynomial of Ais p(λ) = λ3−5λ2+ 8λ−4, the eigenvalues of Aare the roots
of this polynomial. So, the eigenvalues of Aare λ1= 1, λ2= 2,and λ3= 2.
Step 2: Now, we need to find the eigenvalues of A2−4A+ 3I. Let B=A2−
4A+3I. We know that if λis an eigenvalue of A, then λ2is an eigenvalue of A2,
−4λis an eigenvalue of −4A, and 3 is an eigenvalue of 3I. Thus, the eigenvalues
of Bare λ2
1−4λ1+ 3 = 0, λ2
2−4λ2+ 3 = 0,and λ2
3−4λ3+ 3 = 0. Solving these
equations gives the eigenvalues of B: For λ1= 1: 12−4(1) + 3 = 0 =⇒0 = 0
For λ2= 2: 22−4(2) + 3 = 0 =⇒1 = 0, which is not possible. For λ3= 2:
22−4(2) + 3 = 0 =⇒1 = 0, which is not possible. Therefore, the eigenvalues
of A2−4A+ 3Iare 0 and 1.
Question 22
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 5. If
B= 2A2−5A+ 4I, where Iis the 3 ×3 identity matrix, find the eigenvalues of
B.
17
Solution
Step 1: We first calculate B.
B= 2A2−5A+ 4I
= 2A2−5A+ 4
100
010
001
Step 2: Since Ais a square matrix, we can use the spectral decomposition
to write Ain terms of its eigenvectors and eigenvalues:
A=P DP −1
where Dis a diagonal matrix with the eigenvalues of Aon the diagonal.
Step 3: Substituting this into the expression for B, we have
B= 2(P DP −1)2−5(P DP −1)+4I
Step 4: Simplifying, we get
B= 2P D2P−1−5P DP −1+ 4I
Step 5: Since the eigenvectors of Aform a basis, Pis invertible. We can
compute BP −1to find the eigenvalues of B.
BP −1= 2P D2−5D+ 4I
Step 6: The eigenvalues of Bare the diagonal entries of 2P D2−5D+ 4I,
which are obtained by multiplying each eigenvalue of Aby the corresponding
coefficient in 2P D2−5D+ 4I. Therefore, the eigenvalues of Bare λ′
1= 42·2−
5·2 + 4 = 27, λ′
2= 42·3−5·3 + 4 = 44, and λ′
3= 42·5−5·5 + 4 = 84.
Question 23
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 a basis for the eigenspace
corresponding to λ2.
Solution
Step 1: Recall that the eigenspace corresponding to an eigenvalue λis the set
of all vectors vsuch that (A−λI)v=0.
18
Step 2: Let’s find the eigenvectors corresponding to λ2= 1. We need to
solve the equation (A−λ2I)v=0, where Iis the 3 ×3 identity matrix.
Step 3: Substituting λ2= 1 and expanding the equation, we get:
(A−I)v=
101
001
214
x
y
z
=
0
0
0
Step 4: Row reducing the augmented matrix [A—0], we get:
1010
0010
2140
→
1010
0100
0120
→
1010
0100
0020
Step 5: From the row-reduced form, we have the solution z= 0, and x=
−z= 0 then y= 0. So the basis for the eigenspace corresponding to λ2= 1 is
v2=
0
0
0
.
Question 24
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3=−2. If
the corresponding eigenvectors are v1=
1
0
1
,v2=
1
1
0
, and v3=
1
1
1
, find
a matrix Psuch that A=P DP −1, where Dis the diagonal matrix with the
eigenvalues on the diagonal.
Solution
Step 1: We construct the matrix Pby using the eigenvectors as columns.
P=
111
011
101
Step 2: We form the diagonal matrix Dwith the eigenvalues on the diagonal.
D=
2 0 0
0−1 0
0 0 −2
Step 3: We find the inverse of matrix P, denoted as P−1. We use the formula:
P−1=1
det(P)adj(P), where adj(P) is the adjugate of P. First, calculate the
19
determinant of P: det(P) = (1)(1)(1) + (1)(1)(1) + (1)(0)(0) −(1)(0)(1) −
(1)(1)(1) −(1)(0)(1) = 1.
Step 4: Next, find the adjugate of P.
adj(P) =
1−1 1
−110
1 0 1
Step 5: Calculate the inverse of Pusing the formula:
P−1=1
det(P)·adj(P) =
1−1 1
−110
1 0 1
Step 6: Finally, we can express Ain terms of P,D, and P−1as A=P DP −1.
Therefore,
A=
111
011
101
2 0 0
0−1 0
0 0 −2
1−1 1
−110
1 0 1
Question 25
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If
B=A2−2A+I, where Iis the 3 ×3 identity matrix, determine the eigenvalues
of B.
Solution
Step 1: First, we find the eigenvalues of matrix Bby substituting Ainto the
expression B=A2−2A+I.
Step 2: We have B=A2−2A+I. Substituting the given eigenvalues of A,
we get
B=
100
020
003
2
−2
100
020
003
+
100
010
001
B=
100
040
009
−
200
040
006
+
100
010
001
B=
000
010
004
Step 3: The eigenvalues of matrix Bare the values of λthat satisfy the
equation |B−λI|= 0.
20
Step 4: Solving for the eigenvalues of B, we have
−λ0 0
0 1 −λ0
0 0 4 −λ
= 0
−λ(1 −λ)(4 −λ)=0
λ= 0,1,4
Therefore, the eigenvalues of matrix Bare λ1= 0, λ2= 1, and λ3= 4.
Question 26
Question
Let Abe a 3 ×3 matrix with eigenvalues λ=−1,2,4. Find the eigenvectors
corresponding to each eigenvalue.
Solution
1. For eigenvalue λ=−1:
(a) To find the eigenvector corresponding to λ=−1, we solve the system
of equations (A−λI)v=0, where vis the eigenvector.
(b) Given Ahas eigenvalue λ=−1, we have A+I=
011
031
025
.
(c) Subtracting λI from Agives us A−λI =
111
041
026
.
(d) Setting A−λI equal to 0, we get the system of equations:
v1+v2+v3= 0
4v2+v3= 0
2v2+ 6v3= 0
.
(e) Solving this system, we find that v1=−2, v2= 1, and v3=−1.
Therefore, the eigenvector corresponding to λ=−1 is v1=
−2
1
−1
.
2. For eigenvalue λ= 2:
(a) Following the same process as above, we find that the eigenvector
corresponding to λ= 2 is v2=
1
0
0
.
3. For eigenvalue λ= 4:
21
(a) Following the same process as above, we find that the eigenvector
corresponding to λ= 4 is v3=
1
−1
1
.
Question 27
Question
Let Abe a 3×3 matrix with eigenvalues −2, 1, and 5 corresponding to eigenvec-
tors v1,v2, and v3, respectively. Determine a matrix Bsuch that AB =BA.
Solution
To find a matrix Bsuch that AB =BA, we can use the fact that Ais diago-
nalizable. Since Ahas 3 linearly independent eigenvectors, we can write:
A=P DP −1
where Dis the diagonal matrix with the eigenvalues of Aalong the diagonal
and Pis the matrix whose columns are the corresponding eigenvectors of A.
Thus,
A=P DP −1=v1v2v3
−200
0 1 0
0 0 5
v1v2v3−1
Step 1: Calculate P−1Since the columns of Pare the eigenvectors of A,
we have:
P=v1v2v3
To find P−1, we can use the formula: P−1=1
det(P)adj(P), where adj(P) is the
adjugate of P.
Step 2: Calculate D D is the diagonal matrix with the eigenvalues of A
along the diagonal, given by:
D=
−200
0 1 0
0 0 5
Step 3: Compute AB and BA Now we can form the matrix Bsuch that
AB =BA by inspection:
B=P DP −1
Thus, Bis the matrix that commutes with A.
22
Question 28
Question
Let A=2−1
−1 2 . 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 an eigenvalue and Iis the identity matrix.
A−λI=2−1
−1 2 −λ1 0
0 1=2−λ−1
−1 2 −λ
|A−λI|= (2 −λ)(2 −λ)−(−1)(−1) = λ2−4λ+ 3 = 0
Step 2: Solve the characteristic equation to find the eigenvalues.
λ2−4λ+ 3 = 0
(λ−3)(λ−1) = 0
λ1= 3 and λ2= 1
Step 3: To find the eigenvectors, we substitute each eigenvalue back into the
equation (A−λI)v=0, where vis the eigenvector. For λ1= 3:
−1−1
−1−1x
y=0
0
x=−y
So, eigenvectors corresponding to λ1= 3 are of the form −1
1.
For λ2= 1: 1−1
−1 1 x
y=0
0
x=y
So, eigenvectors corresponding to λ2= 1 are of the form 1
1.
Therefore, the eigenvalues of Aare λ1= 3 and λ2= 1, with corresponding
eigenvectors −1
1and 1
1respectively.
23
Question 29
Question
Let Abe a 3 ×3 matrix given by
A=
121
111
002
.
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 λis the eigenvalue and Iis the identity matrix.
A−λI =
1−λ2 1
1 1 −λ1
0 0 2 −λ
.
|A−λI|= (1 −λ)((1 −λ)(2 −λ)−1) −2((1)(2 −λ)) + 0 = 0.
Step 2: Simplify the characteristic equation to find the eigenvalues.
(1 −λ)((1 −λ)(2 −λ)−1) −2((1)(2 −λ)) = 0.
(1 −λ)((2 −λ)−(1 + λ)) −4+2λ= 0.
(1 −λ)(1 −λ)−4+2λ= 0.
(1 −λ)2−4+2λ= 0.
λ2−2λ−3 = 0.
Step 3: Solve the quadratic equation to find the eigenvalues.
λ2−2λ−3=0.
(λ−3)(λ+ 1) = 0.
λ= 3,−1.
Step 4: Next, we find the eigenvectors associated with each eigenvalue.
For λ= 3: Substitute λ= 3 into (A−λI)v= 0 to find the eigenvector v.
(A−3I)v=
−2 2 1
1−2 1
0 0 −1
v=
0
0
0
.
Step 5: Solving the system of equations, we get v=
1
1
0
.
24
For λ=−1: Substitute λ=−1 into (A−λI)v= 0 to find the eigenvector
v.
(A+I)v=
021
121
001
v=
0
0
0
.
Step 6: Solving the system of equations, we get v=
−1
1
2
.
Therefore, the eigenvalues of matrix Aare λ1= 3 and λ2=−1, with
corresponding eigenvectors v1=
1
1
0
and v2=
−1
1
2
.
Question 30
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3,and λ3=−4. If the
eigenvector corresponding to λ1is
1
2
−1
, find the eigenvectors corresponding
to λ2and λ3.
Solution
Step 1: The eigenvector corresponding to λ2can be found by solving the system
of equations (A−λI)v= 0, where λ= 3 and vis the eigenvector.
A−λI =
2 0 0
0 3 0
0 0 −4
−3
100
010
001
=
−1 0 0
000
0 0 −7
Setting this matrix multiplied by the eigenvector v=
x
y
z
equal to the zero
vector gives the equations −x= 0, −7z= 0. Therefore, the eigenvector corre-
sponding to λ2is
0
1
0
.
Step 2: Similarly, the eigenvector corresponding to λ3can be found by solv-
ing the system of equations (A−λI)v= 0, where λ=−4.
A−λI =
2 0 0
0 3 0
0 0 −4
+ 4
100
010
001
=
−200
0−1 0
0 0 0
25
Setting this matrix multiplied by the eigenvector v=
x
y
z
equal to the zero
vector gives the equations −2x= 0, −y= 0. Therefore, the eigenvector corre-
sponding to λ3is
1
0
0
.
Question 31
Question
Let Abe a 3 ×3 matrix given by
A=
2−3 1
1 1 0
0 1 −2
.
Find the eigenvalues and corresponding eigenvectors of A.
Solution
Step 1: To find the eigenvalues of the matrix A, we solve the characteristic
equation |A−λI|= 0, where Iis the identity matrix.
|A−λI|=
2−λ−3 1
1 1 −λ0
0 1 −2−λ
= 0.
Step 2: Expanding the determinant, we get
(2 −λ)
1−λ0
1−2−λ
+ 3
1 0
0−2−λ
−
1−3
0 1 −λ
= 0.
Step 3: Simplifying the determinants, we get
(2 −λ)[(1 −λ)(−2−λ)] + 3(−2−λ)−(1 −λ)=0.
Step 4: Expanding and simplifying further, we obtain the characteristic
equation
λ3−λ2−6λ−7=0.
Step 5: Using synthetic division or other methods, we find that one of the
roots is λ= 1. We can then factor out (λ−1) from the characteristic equation
to get
(λ−1)(λ2−7λ−7) = 0.
Step 6: Solving λ2−7λ−7 = 0, we find the other two roots to be λ=7±√57
2.
26
Step 7: Next, we find the eigenvectors corresponding to each eigenvalue by
solving (A−λI)v= 0 for each eigenvalue. Let’s start with λ= 1:
A−I=
1−3 1
1 0 0
0 1 −3
→
1−3 1
0 3 −1
0 1 −3
→
1 0 −8
0 1 −3
0 0 0
.
Step 8: Solving the system A−I= 0, we get the eigenvector corresponding
to λ= 1 as
8
3
1
.
Step 9: For the other two eigenvalues, λ=7±√57
2, repeat the process to find
the corresponding eigenvectors.
Question 32
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3= 0. Given
that one eigenvector corresponding to λ2is
1
1
2
, find the matrix A.
Solution
Step 1: Since Ais a 3 ×3 matrix, it has 3 eigenvalues. We are given that the
eigenvalues of Aare λ1= 2, λ2= 1, and λ3= 0.
Step 2: We are also given that one eigenvector corresponding to λ2= 1 is
1
1
2
. Let’s denote this eigenvector as v2=
1
1
2
.
Step 3: Recall that for a square matrix Aand an eigenvector vcorresponding
to eigenvalue λ, the following equation holds: Av=λv.
Step 4: Applying this to our given eigenvector v2with eigenvalue λ2= 1,
we get:
Av2=λ2v2
A
1
1
2
= 1 ·
1
1
2
a b c
d e f
g h i
1
1
2
=
1
1
2
Step 5: Solving the matrix equation above, we get the following system of
equations:
a+b+ 2c= 1
27
d+e+ 2f= 1
g+h+ 2i= 2
Step 6: From the given eigenvalues, we know that λ1= 2 corresponds to an
eigenvector v1, and λ3= 0 corresponds to an eigenvector v3. We can use the
eigenpairs to find the other columns of matrix A.
Step 7: After solving the system of equations and substituting, we find the
matrix Ato be:
A=
4−3 7
3−2 5
−2 1 −2
Question 33
Question
Let Abe a 3×3 matrix with eigenvalues λ1= 2, λ2= 1, λ3=−3. If det(A) = 4,
find the eigenvalues of the matrix B= 3A2−2A+I, where Iis the identity
matrix.
Solution
Step 1: First, let’s find the eigenvalues of matrix B= 3A2−2A+I.
B= 3A2−2A+I
= 3A2−2A+I
= 3A2−2A+I·A
= 3A2−2A+A
= 3A2−A.
Step 2: Now, let’s find the eigenvalues of matrix 3A2−A.
3A2−A= 3(AA)−A
= 3Aλ1−λ1I
= 3(2A)−2I
= 6A−2I
= 2(3A−I).
Step 3: The eigenvalues are the scalars αfor which the equation (3A2−A)v=
αv has a nontrivial solution v= 0. Therefore, the eigenvalues of Bare the
eigenvalues of 3A2−A, which are the eigenvalues of 3A−I, which are related
to the eigenvalues of Aby λ= 3λi−1.
Hence, the eigenvalues of matrix Bare 3(2) −1 = 5, 3(1) −1 = 2, and
3(−3) −1 = −10.
28
Question 34
Question
Let Abe a 3×3 matrix with eigenvalues −1, 2, and 3. Given that the eigenvector
corresponding to the eigenvalue −1 is
1
1
1
and the eigenvector corresponding to
the eigenvalue 2 is
−2
1
1
, find the eigenvector corresponding to the eigenvalue
3.
Solution
To find the eigenvector corresponding to the eigenvalue 3, we can use the fact
that the sum of the eigenvalues of a matrix is equal to its trace. The trace of
a matrix is the sum of the elements on its main diagonal. Since Ais a 3 ×3
matrix, its trace is equal to the sum of its eigenvalues:
Trace(A) = −1 + 2 + 3 = 4.
The trace of Ais also equal to the sum of its diagonal elements:
Trace(A) = a11 +a22 +a33.
Since we know the eigenvalues of A, we can use these to find diagonal elements
of the matrix:
a11 =−1, a22 = 2, a33 = 3.
Therefore, the matrix Ais
A=
−100
0 2 0
0 0 3
.
To find the eigenvector corresponding to the eigenvalue 3, we need to solve
the system of equations (A−3I)v=0, where vis the eigenvector we are looking
for and Iis the identity matrix.
A−3I=
−1−3 0 0
0 2 −3 0
0 0 3 −3
=
−400
0−1 0
0 0 0
.
Solving the system (A−3I)v=0, we get:
−4v1= 0 =⇒v1= 0
−v2= 0 =⇒v2= 0.
So, the eigenvector corresponding to the eigenvalue 3 is
0
0
1
.
29
Question 35
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 5. Find
the eigenvectors associated with each eigenvalue.
Solution
To find the eigenvectors associated with each eigenvalue, we solve the equation
(A−λI)v=0, where Ais the given matrix, λis the eigenvalue, vis the
eigenvector, and Iis the identity matrix.
Step 1: Eigenvector for λ1= 2
Substitute λ= 2 into the equation (A−λI)v=0:
(A−2I)v=0
011
0−3 0
003
x
y
z
=
0
0
0
Solving this system of equations, we get y=z. So, the eigenvector for λ1= 2
is any vector of the form
1
1
1
.
Step 2: Eigenvector for λ2=−1
Substitute λ=−1 into the equation (A−λI)v=0:
(A+I)v=0
211
020
006
x
y
z
=
0
0
0
Solving this system of equations, we get x=−y. So, the eigenvector for
λ2=−1 is any vector of the form
−1
1
k
.
Step 3: Eigenvector for λ3= 5
Substitute λ= 5 into the equation (A−λI)v=0:
(A−5I)v=0
−3 1 1
0−6 0
0 0 −2
x
y
z
=
0
0
0
Solving this system of equations, we get x=1
3yand x=−1
3z. So, the
eigenvector for λ3= 5 is any vector of the form
1
3
−3
.
30
Question 7
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1=−2, λ2= 1, and λ3= 3. Find
the corresponding eigenvectors for each eigenvalue.
Solution
Step 1: For λ1=−2, We solve (A−λ1I)x=0:
511
151
115
x1
x2
x3
=
0
0
0
Performing row operations, we get the RREF matrix:
1 0 −1
0 1 1
0 0 0
So, the eigenvector corresponding to λ1=−2 is
1
−1
1
.
Step 2: For λ2= 1, We solve (A−λ2I)x=0:
−3 1 1
1−3 1
1 1 −3
x1
x2
x3
=
0
0
0
Performing row operations, we get the RREF matrix:
1−1−1
0 0 0
0 0 0
So, the eigenvector corresponding to λ2= 1 is
1
1
0
.
Step 3: For λ3= 3, We solve (A−λ3I)x=0:
−5 1 1
1−5 1
1 1 −5
x1
x2
x3
=
0
0
0
Performing row operations, we get the RREF matrix:
1−1
5−1
5
0 0 0
0 0 0
6
So, the eigenvector corresponding to λ3= 3 is
1
5
5
.
Question 8
Question
Let Abe a square matrix such that A2−5A+ 6I=O, where Iis the identity
matrix. Find the eigenvalues of A.
Solution
Step 1: Given that A2−5A+ 6I=O, we can rewrite this as A2−5A+ 6I= 0,
where 0 is the zero matrix. This suggests that the characteristic polynomial of
Ais p(λ) = λ2−5λ+ 6.
Step 2: To find the eigenvalues of A, we need to solve the characteristic
equation p(λ) = 0. Therefore, we solve λ2−5λ+ 6 = 0.
Step 3: Factoring the quadratic equation, we get (λ−2)(λ−3) = 0. There-
fore, the eigenvalues are λ1= 2 and λ2= 3.
Step 4: Hence, the eigenvalues of matrix Aare 2 and 3.
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−3.
Determine the eigenvectors corresponding to each eigenvalue.
Solution
We know that the eigenvectors corresponding to each eigenvalue λiare the
solutions to the system (A−λiI)v=0, where Iis the identity matrix and vis
the eigenvector corresponding to λi.
Step 1: Find the eigenvector corresponding to λ1= 2 Let v1=
x
y
z
be the eigenvector corresponding to λ1. Substitute λ1= 2 into the equation
(A−λ1I)v1=0and solve for v1.
−1 1 1
0 1 −1
1−1 3
x
y
z
=
0
0
0
7
Step 2: Solve for v1Setting up the system of equations:
−x+y+z= 0
y−z= 0
x−y+ 3z= 0
Solving this system, we find v1=
1
1
−1
.
Step 3: Find the eigenvector corresponding to λ2= 1 Let v2=
x
y
z
be the eigenvector corresponding to λ2. Substitute λ2= 1 into the equation
(A−λ2I)v2=0and solve for v2.
1 1 1
0 0 −1
1−1 4
x
y
z
=
0
0
0
Step 4: Solve for v2Setting up the system of equations:
x+y+z= 0
−z= 0
x−y+ 4z= 0
Solving this system, we find v2=
1
−1
1
.
Step 5: Find the eigenvector corresponding to λ3=−3 Let v3=
x
y
z
be the eigenvector corresponding to λ3. Substitute λ3=−3 into the equation
(A−λ3I)v3=0and solve for v3.
5 1 1
0 4 −1
1−1 6
x
y
z
=
0
0
0
Step 6: Solve for v3Setting up the system of equations:
5x+y+z= 0
4y−z= 0
x−y+ 6z= 0
Solving this system, we find v3=
1
1
1
.
8
Question 10
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If
the eigenvectors corresponding to λ1and λ2are v1=
1
2
1
and v2=
2
0
1
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: The eigenvector corresponding to λ3can be found by solving the equa-
tion (A−3I)v3= 0, where v3is the eigenvector corresponding to λ3.
Step 2: First, we find the matrix A−3I:
A−3I=
100
020
003
−
300
030
003
=
−200
0−1 0
0 0 0
Step 3: Next, we solve the equation (−2)v31 = 0, (−1)v32 = 0, and 0v33 = 0
where v3=
v31
v32
v33
.
Step 4: From the first equation −2v31 = 0, we get v31 = 0.
Step 5: From the second equation −v32 = 0, we get v32 = 0.
Step 6: From the third equation 0 = 0, we see that v33 can be any value.
Step 7: Therefore, the eigenvector v3corresponding to λ3= 3 is
0
0
1
.
Question 11
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2=−2, and λ3= 3. If
B=A2−3A+ 2I, find the eigenvalues of B.
Solution
Step 1: First, we need to find the matrix A2. Given that the eigenvalues of ma-
trix Aare λ1= 1, λ2=−2, and λ3= 3, weknowthatAissimilartoadiagonalmatrixD
=
100
0−2 0
003
.
9
Step 2: Next, we calculate A2by squaring the diagonal matrix D.A2=
D2=
100
0−2 0
003
2
=
100
040
009
.
Step 3: Now, let’s find Bby substituting A2into B=A2−3A+ 2I.
B=
100
040
009
−3
100
0−2 0
003
+ 2
100
010
001
. Simplifying, we get B=
0 0 0
0 10 0
0 0 16
.
Step 4: Finally, we find the eigenvalues of Bby observing that Bis diagonal
with entries 0, 10, and 16. Therefore, the eigenvalues of Bare 0, 10, and 16.
Question 12
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 5. Given
that the eigenvectors corresponding to λ1and λ2are v1=
1
1
1
and v2=
−1
1
2
,
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: Recall that the eigenvectors of a matrix Aare the non-zero vectors
vwhich satisfy the eigenvalue equation Av=λv, where λis the eigenvalue
corresponding to v.
Step 2: Since we are given the eigenvalues and eigenvectors of A, we can
construct the matrix Aas follows:
A=P DP −1,
where Pis the matrix whose columns are the eigenvectors of Aand Dis the
diagonal matrix with the eigenvalues of Aon the diagonal.
Step 3: Recall that the matrix Pis given by
P=v1v2v3.
Step 4: From the given information, we have P=
1−1 ?
11?
12?
. To find the
third eigenvector v3, we solve the system of equations Av3=λ3v3.
Step 5: Substituting λ3= 5, we have Av3= 5v3. Therefore, we need to
solve the system of equations (A−5I)v3= 0, where Iis the identity matrix.
10
Step 6: Substituting A, we have
(A−5I)v3=
−3−1 ?
1−6 ?
1 2 −2
x
y
z
=
0
0
0
.
Step 7: Solving the system of linear equations using Gaussian elimination,
we find the solution v3=
−1
1
1
.
Therefore, the eigenvector corresponding to the eigenvalue λ3= 5 is v3=
−1
1
1
.
Question 13
Question
Let Abe a 3 ×3 matrix with eigenvalues 1, 2, and 3. Determine the eigenvalues
of the matrix B=A3−2A2+A.
Solution
Step 1: First, let’s find the eigenvalues of matrix B.
Step 2: We know that if λis an eigenvalue of matrix A, then λis an eigenvalue
of any polynomial f(A). Specifically, if λis an eigenvalue of A, then λ3,λ2, and
λare eigenvalues of A3,A2, and A, respectively.
Step 3: Therefore, the eigenvalues of matrix Bare 13−2(1)2+ 1 = 0,
23−2(2)2+ 2 = 2, and 33−2(3)2+ 3 = 12.
Step 4: So the eigenvalues of matrix Bare 0, 2, and 12.
Question 14
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3=−3. Deter-
mine the characteristic polynomial of Aand find the eigenvectors corresponding
to each eigenvalue.
Solution
Step 1: The characteristic polynomial of a matrix Ais given by |A−λI|= 0,
where Iis the identity matrix.
11
The characteristic polynomial is then:
det
A11 −λ A12 A13
A21 A22 −λ A23
A31 A32 A33 −λ
= 0
For our 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3=−3, the
characteristic polynomial will be:
det
A11 −2A12 A13
A21 A22 + 1 A23
A31 A32 A33 + 3
= 0
Step 2: Next, find the eigenvector corresponding to each eigenvalue by
solving the equation (A−λI)v= 0 for each eigenvalue.
For λ1= 2:
A11 −2A12 A13
A21 A22 −2A23
A31 A32 A33 −2
x
y
z
=
0
0
0
Step 3: Solve the system of equations for λ1= 2 to find the eigenvector v1.
Similarly, repeat Steps 2 and 3 for eigenvalues λ2=−1 and λ3=−3 to find
the eigenvectors v2and v3respectively.
Question 15
Question
Let Abe a 3×3 matrix with characteristic polynomial p(λ) = λ3−7λ2+14λ−8.
If one of the eigenvalues of Ais 2, find the other two eigenvalues of A.
Solution
Step 1: Since 2 is an eigenvalue of A, we know that p(2) = 0. So, we have:
23−7(2)2+ 14(2) −8 = 0
8−28 + 28 −8=0
0=0
This confirms that λ= 2 is indeed an eigenvalue of A.
Step 2: By the Factor Theorem, we know that p(λ) can be factored as
(λ−2)(λ2−5λ+ 4). Therefore, the remaining eigenvalues are the roots of the
quadratic λ2−5λ+ 4.
Step 3: To find the remaining eigenvalues, we solve λ2−5λ+ 4 = 0:
λ=5±√52−4·4
2
12
λ=5±√25 −16
2
λ=5±√9
2
λ=5±3
2
So, the other two eigenvalues are λ= 4 and λ= 1.
Question 16
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1=−2, λ2= 1, and λ3= 4. Given
that the corresponding eigenvectors are v1=
1
−1
2
,v2=
0
1
−1
, and v3=
1
1
1
,
respectively, find the matrix exponential eAt.
Solution
Step 1: We can express Ain terms of its eigenvectors and eigenvalues as A=
P DP −1, where P= [v1,v2,v3] and D= diag(λ1, λ2, λ3). Therefore, we have:
P=
1 0 1
−111
2−1 1
and D=
−200
0 1 0
0 0 4
Step 2: Next, we need to compute eAt . Since A=P DP −1, we can express
eAt =P eDt P−1.
Step 3: To compute eDt, we replace λiby eλitalong the diagonal of D.
Therefore, we get:
eDt =
e−2t0 0
0et0
0 0 e4t
Step 4: Therefore, to find eAt, we have:
eAt =P eDt P−1
=
1 0 1
−111
2−1 1
e−2t0 0
0et0
0 0 e4t
1 0 1
−111
2−1 1
−1
Step 5: After performing the matrix multiplication and inversion, we arrive
at the final expression for eAt.
13
Question 17
Question
Let A be a 3x3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3=−1. If the
eigenvectors corresponding to λ1and λ2are
v1=
1
0
1
and v2=
1
1
1
respectively, find a basis for the eigenspace corresponding to λ3.
Solution
Step 1: The eigenspace corresponding to λ3=−1 is the null space of the matrix
A−λ3I.
A−λ3I=A+I=
1 0 1
0 2 1
0 0 −2
Step 2: To find a basis for the eigenspace corresponding to λ3, we need to
solve the equation (A+I)v=0, where v=
x
y
z
.
1 0 1
0 2 1
0 0 −2
x
y
z
=
0
0
0
Step 3: Solving the system of equations, we get:
x+z= 0
2y+z= 0
−2z= 0
Step 4: From the third equation, z= 0. Substituting z= 0 into the first
two equations, we get:
x= 0
2y= 0 =⇒y= 0
Therefore, a basis for the eigenspace corresponding to λ3is the vector v3=
0
0
0
.
14
Question 18
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3=−3. If
B=A2−2A+I, where Iis the identity matrix, find the eigenvalues of B.
Solution
Step 1: First, let’s find the matrix B. Given that B=A2−2A+I, we can
substitute the eigenvalues of Ainto the expression to find the corresponding
eigenvectors. Let v1, v2,and v3be the eigenvectors corresponding to eigenvalues
1,2,and −3 respectively.
Step 2: For λ= 1, we have: (A2−2A+I)v1= 1v1
Expanding this equation and substituting Av1=λv1gives: A2v1−2Av1+
v1=v1
Step 3: For λ= 2, we have: (A2−2A+I)v2= 2v2
Similarly, we get: A2v2−2Av2+v2= 2v2
Step 4: For λ=−3, we have: A2v3−2Av3+v3=−3v3
This simplifies to: A2v3−2Av3+v3=−3v3
Step 5: Now, we can find the eigenvalues of Bby evaluating Bvifor i= 1,2,3
using the properties of matrices and the eigenvectors of A.
Bv1= (A2−2A+I)v1=v1
Bv2= (A2−2A+I)v2= 2v2
Bv3= (A2−2A+I)v3= (−3)v3
Step 6: Therefore, the eigenvalues of Bare: λ′
1= 1, λ′
2= 2, and λ′
3=−3.
Question 19
Question
Let Abe a 3 ×3 matrix with eigenvalues −2, 1, and 4. If v1,v2, and v3are the
corresponding eigenvectors, find the matrix A.
Solution
Step 1: Recall that if λis an eigenvalue of a matrix A, and vis the corresponding
eigenvector, then Av =λv.
Step 2: Let’s denote the matrix Aas:
A=
a b c
d e f
g h i
Step 3: Since −2, 1, and 4 are eigenvalues of A, we have:
Av1=−2v1, Av2=v2, Av3= 4v3
15
Step 4: Writing out these equations explicitly, we get:
a b c
d e f
g h i
x1
y1
z1
=
−2x1
−2y1
−2z1
a b c
d e f
g h i
x2
y2
z2
=
x2
y2
z2
a b c
d e f
g h i
x3
y3
z3
=
4x3
4y3
4z3
Step 5: Solving these equations, we find that Ais:
A=
−200
0 1 0
0 0 4
Question 20
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 3. Suppose
that the corresponding eigenvectors are v1=
1
0
1
,v2=
1
1
1
, and v3=
1
2
1
,
normalized to have a length of 1. Find A.
Solution
To find the matrix A, we can use the fact that the matrix Acan be written as
A=P DP −1, where Pis the matrix whose columns are the eigenvectors of A
and Dis the diagonal matrix consisting of the eigenvalues.
Step 1: Finding the matrix P
The matrix Pis formed by the eigenvectors of A.
P=
111
012
111
Step 2: Finding the diagonal matrix D
The diagonal matrix Dis formed by the eigenvalues of A.
D=
200
0−1 0
003
16
Step 3: Finding matrix A
To find A, we use the formula A=P DP −1.
A=P DP −1
We first calculate P−1using the inverse matrix formula:
P−1=1
det(P)adj(P)
det(P) = 1(1 −2) −1(1 −2) + 1(1 −2) = 0
Since the determinant is 0, Pis not invertible. Hence, matrix Acannot be
uniquely determined in this case.
Question 21
Question
Let Abe a 3×3 matrix with characteristic polynomial p(λ) = λ3−5λ2+ 8λ−4.
If the eigenvalues of Aare λ1, λ2,and λ3, find the eigenvalues of A2−4A+ 3I,
where Iis the 3 ×3 identity matrix.
Solution
Step 1: First, we need to find the eigenvalues of A. Given that the characteristic
polynomial of Ais p(λ) = λ3−5λ2+ 8λ−4, the eigenvalues of Aare the roots
of this polynomial. So, the eigenvalues of Aare λ1= 1, λ2= 2,and λ3= 2.
Step 2: Now, we need to find the eigenvalues of A2−4A+ 3I. Let B=A2−
4A+3I. We know that if λis an eigenvalue of A, then λ2is an eigenvalue of A2,
−4λis an eigenvalue of −4A, and 3 is an eigenvalue of 3I. Thus, the eigenvalues
of Bare λ2
1−4λ1+ 3 = 0, λ2
2−4λ2+ 3 = 0,and λ2
3−4λ3+ 3 = 0. Solving these
equations gives the eigenvalues of B: For λ1= 1: 12−4(1) + 3 = 0 =⇒0 = 0
For λ2= 2: 22−4(2) + 3 = 0 =⇒1 = 0, which is not possible. For λ3= 2:
22−4(2) + 3 = 0 =⇒1 = 0, which is not possible. Therefore, the eigenvalues
of A2−4A+ 3Iare 0 and 1.
Question 22
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 5. If
B= 2A2−5A+ 4I, where Iis the 3 ×3 identity matrix, find the eigenvalues of
B.
17
Solution
Step 1: We first calculate B.
B= 2A2−5A+ 4I
= 2A2−5A+ 4
100
010
001
Step 2: Since Ais a square matrix, we can use the spectral decomposition
to write Ain terms of its eigenvectors and eigenvalues:
A=P DP −1
where Dis a diagonal matrix with the eigenvalues of Aon the diagonal.
Step 3: Substituting this into the expression for B, we have
B= 2(P DP −1)2−5(P DP −1)+4I
Step 4: Simplifying, we get
B= 2P D2P−1−5P DP −1+ 4I
Step 5: Since the eigenvectors of Aform a basis, Pis invertible. We can
compute BP −1to find the eigenvalues of B.
BP −1= 2P D2−5D+ 4I
Step 6: The eigenvalues of Bare the diagonal entries of 2P D2−5D+ 4I,
which are obtained by multiplying each eigenvalue of Aby the corresponding
coefficient in 2P D2−5D+ 4I. Therefore, the eigenvalues of Bare λ′
1= 42·2−
5·2 + 4 = 27, λ′
2= 42·3−5·3 + 4 = 44, and λ′
3= 42·5−5·5 + 4 = 84.
Question 23
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 a basis for the eigenspace
corresponding to λ2.
Solution
Step 1: Recall that the eigenspace corresponding to an eigenvalue λis the set
of all vectors vsuch that (A−λI)v=0.
18
Step 2: Let’s find the eigenvectors corresponding to λ2= 1. We need to
solve the equation (A−λ2I)v=0, where Iis the 3 ×3 identity matrix.
Step 3: Substituting λ2= 1 and expanding the equation, we get:
(A−I)v=
101
001
214
x
y
z
=
0
0
0
Step 4: Row reducing the augmented matrix [A—0], we get:
1010
0010
2140
→
1010
0100
0120
→
1010
0100
0020
Step 5: From the row-reduced form, we have the solution z= 0, and x=
−z= 0 then y= 0. So the basis for the eigenspace corresponding to λ2= 1 is
v2=
0
0
0
.
Question 24
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3=−2. If
the corresponding eigenvectors are v1=
1
0
1
,v2=
1
1
0
, and v3=
1
1
1
, find
a matrix Psuch that A=P DP −1, where Dis the diagonal matrix with the
eigenvalues on the diagonal.
Solution
Step 1: We construct the matrix Pby using the eigenvectors as columns.
P=
111
011
101
Step 2: We form the diagonal matrix Dwith the eigenvalues on the diagonal.
D=
2 0 0
0−1 0
0 0 −2
Step 3: We find the inverse of matrix P, denoted as P−1. We use the formula:
P−1=1
det(P)adj(P), where adj(P) is the adjugate of P. First, calculate the
19
determinant of P: det(P) = (1)(1)(1) + (1)(1)(1) + (1)(0)(0) −(1)(0)(1) −
(1)(1)(1) −(1)(0)(1) = 1.
Step 4: Next, find the adjugate of P.
adj(P) =
1−1 1
−110
1 0 1
Step 5: Calculate the inverse of Pusing the formula:
P−1=1
det(P)·adj(P) =
1−1 1
−110
1 0 1
Step 6: Finally, we can express Ain terms of P,D, and P−1as A=P DP −1.
Therefore,
A=
111
011
101
2 0 0
0−1 0
0 0 −2
1−1 1
−110
1 0 1
Question 25
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If
B=A2−2A+I, where Iis the 3 ×3 identity matrix, determine the eigenvalues
of B.
Solution
Step 1: First, we find the eigenvalues of matrix Bby substituting Ainto the
expression B=A2−2A+I.
Step 2: We have B=A2−2A+I. Substituting the given eigenvalues of A,
we get
B=
100
020
003
2
−2
100
020
003
+
100
010
001
B=
100
040
009
−
200
040
006
+
100
010
001
B=
000
010
004
Step 3: The eigenvalues of matrix Bare the values of λthat satisfy the
equation |B−λI|= 0.
20
Step 4: Solving for the eigenvalues of B, we have
−λ0 0
0 1 −λ0
0 0 4 −λ
= 0
−λ(1 −λ)(4 −λ)=0
λ= 0,1,4
Therefore, the eigenvalues of matrix Bare λ1= 0, λ2= 1, and λ3= 4.
Question 26
Question
Let Abe a 3 ×3 matrix with eigenvalues λ=−1,2,4. Find the eigenvectors
corresponding to each eigenvalue.
Solution
1. For eigenvalue λ=−1:
(a) To find the eigenvector corresponding to λ=−1, we solve the system
of equations (A−λI)v=0, where vis the eigenvector.
(b) Given Ahas eigenvalue λ=−1, we have A+I=
011
031
025
.
(c) Subtracting λI from Agives us A−λI =
111
041
026
.
(d) Setting A−λI equal to 0, we get the system of equations:
v1+v2+v3= 0
4v2+v3= 0
2v2+ 6v3= 0
.
(e) Solving this system, we find that v1=−2, v2= 1, and v3=−1.
Therefore, the eigenvector corresponding to λ=−1 is v1=
−2
1
−1
.
2. For eigenvalue λ= 2:
(a) Following the same process as above, we find that the eigenvector
corresponding to λ= 2 is v2=
1
0
0
.
3. For eigenvalue λ= 4:
21
(a) Following the same process as above, we find that the eigenvector
corresponding to λ= 4 is v3=
1
−1
1
.
Question 27
Question
Let Abe a 3×3 matrix with eigenvalues −2, 1, and 5 corresponding to eigenvec-
tors v1,v2, and v3, respectively. Determine a matrix Bsuch that AB =BA.
Solution
To find a matrix Bsuch that AB =BA, we can use the fact that Ais diago-
nalizable. Since Ahas 3 linearly independent eigenvectors, we can write:
A=P DP −1
where Dis the diagonal matrix with the eigenvalues of Aalong the diagonal
and Pis the matrix whose columns are the corresponding eigenvectors of A.
Thus,
A=P DP −1=v1v2v3
−200
0 1 0
0 0 5
v1v2v3−1
Step 1: Calculate P−1Since the columns of Pare the eigenvectors of A,
we have:
P=v1v2v3
To find P−1, we can use the formula: P−1=1
det(P)adj(P), where adj(P) is the
adjugate of P.
Step 2: Calculate D D is the diagonal matrix with the eigenvalues of A
along the diagonal, given by:
D=
−200
0 1 0
0 0 5
Step 3: Compute AB and BA Now we can form the matrix Bsuch that
AB =BA by inspection:
B=P DP −1
Thus, Bis the matrix that commutes with A.
22
Question 28
Question
Let A=2−1
−1 2 . 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 an eigenvalue and Iis the identity matrix.
A−λI=2−1
−1 2 −λ1 0
0 1=2−λ−1
−1 2 −λ
|A−λI|= (2 −λ)(2 −λ)−(−1)(−1) = λ2−4λ+ 3 = 0
Step 2: Solve the characteristic equation to find the eigenvalues.
λ2−4λ+ 3 = 0
(λ−3)(λ−1) = 0
λ1= 3 and λ2= 1
Step 3: To find the eigenvectors, we substitute each eigenvalue back into the
equation (A−λI)v=0, where vis the eigenvector. For λ1= 3:
−1−1
−1−1x
y=0
0
x=−y
So, eigenvectors corresponding to λ1= 3 are of the form −1
1.
For λ2= 1: 1−1
−1 1 x
y=0
0
x=y
So, eigenvectors corresponding to λ2= 1 are of the form 1
1.
Therefore, the eigenvalues of Aare λ1= 3 and λ2= 1, with corresponding
eigenvectors −1
1and 1
1respectively.
23
Question 29
Question
Let Abe a 3 ×3 matrix given by
A=
121
111
002
.
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 λis the eigenvalue and Iis the identity matrix.
A−λI =
1−λ2 1
1 1 −λ1
0 0 2 −λ
.
|A−λI|= (1 −λ)((1 −λ)(2 −λ)−1) −2((1)(2 −λ)) + 0 = 0.
Step 2: Simplify the characteristic equation to find the eigenvalues.
(1 −λ)((1 −λ)(2 −λ)−1) −2((1)(2 −λ)) = 0.
(1 −λ)((2 −λ)−(1 + λ)) −4+2λ= 0.
(1 −λ)(1 −λ)−4+2λ= 0.
(1 −λ)2−4+2λ= 0.
λ2−2λ−3 = 0.
Step 3: Solve the quadratic equation to find the eigenvalues.
λ2−2λ−3=0.
(λ−3)(λ+ 1) = 0.
λ= 3,−1.
Step 4: Next, we find the eigenvectors associated with each eigenvalue.
For λ= 3: Substitute λ= 3 into (A−λI)v= 0 to find the eigenvector v.
(A−3I)v=
−2 2 1
1−2 1
0 0 −1
v=
0
0
0
.
Step 5: Solving the system of equations, we get v=
1
1
0
.
24
For λ=−1: Substitute λ=−1 into (A−λI)v= 0 to find the eigenvector
v.
(A+I)v=
021
121
001
v=
0
0
0
.
Step 6: Solving the system of equations, we get v=
−1
1
2
.
Therefore, the eigenvalues of matrix Aare λ1= 3 and λ2=−1, with
corresponding eigenvectors v1=
1
1
0
and v2=
−1
1
2
.
Question 30
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3,and λ3=−4. If the
eigenvector corresponding to λ1is
1
2
−1
, find the eigenvectors corresponding
to λ2and λ3.
Solution
Step 1: The eigenvector corresponding to λ2can be found by solving the system
of equations (A−λI)v= 0, where λ= 3 and vis the eigenvector.
A−λI =
2 0 0
0 3 0
0 0 −4
−3
100
010
001
=
−1 0 0
000
0 0 −7
Setting this matrix multiplied by the eigenvector v=
x
y
z
equal to the zero
vector gives the equations −x= 0, −7z= 0. Therefore, the eigenvector corre-
sponding to λ2is
0
1
0
.
Step 2: Similarly, the eigenvector corresponding to λ3can be found by solv-
ing the system of equations (A−λI)v= 0, where λ=−4.
A−λI =
2 0 0
0 3 0
0 0 −4
+ 4
100
010
001
=
−200
0−1 0
0 0 0
25
Setting this matrix multiplied by the eigenvector v=
x
y
z
equal to the zero
vector gives the equations −2x= 0, −y= 0. Therefore, the eigenvector corre-
sponding to λ3is
1
0
0
.
Question 31
Question
Let Abe a 3 ×3 matrix given by
A=
2−3 1
1 1 0
0 1 −2
.
Find the eigenvalues and corresponding eigenvectors of A.
Solution
Step 1: To find the eigenvalues of the matrix A, we solve the characteristic
equation |A−λI|= 0, where Iis the identity matrix.
|A−λI|=
2−λ−3 1
1 1 −λ0
0 1 −2−λ
= 0.
Step 2: Expanding the determinant, we get
(2 −λ)
1−λ0
1−2−λ
+ 3
1 0
0−2−λ
−
1−3
0 1 −λ
= 0.
Step 3: Simplifying the determinants, we get
(2 −λ)[(1 −λ)(−2−λ)] + 3(−2−λ)−(1 −λ)=0.
Step 4: Expanding and simplifying further, we obtain the characteristic
equation
λ3−λ2−6λ−7=0.
Step 5: Using synthetic division or other methods, we find that one of the
roots is λ= 1. We can then factor out (λ−1) from the characteristic equation
to get
(λ−1)(λ2−7λ−7) = 0.
Step 6: Solving λ2−7λ−7 = 0, we find the other two roots to be λ=7±√57
2.
26
Step 7: Next, we find the eigenvectors corresponding to each eigenvalue by
solving (A−λI)v= 0 for each eigenvalue. Let’s start with λ= 1:
A−I=
1−3 1
1 0 0
0 1 −3
→
1−3 1
0 3 −1
0 1 −3
→
1 0 −8
0 1 −3
0 0 0
.
Step 8: Solving the system A−I= 0, we get the eigenvector corresponding
to λ= 1 as
8
3
1
.
Step 9: For the other two eigenvalues, λ=7±√57
2, repeat the process to find
the corresponding eigenvectors.
Question 32
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3= 0. Given
that one eigenvector corresponding to λ2is
1
1
2
, find the matrix A.
Solution
Step 1: Since Ais a 3 ×3 matrix, it has 3 eigenvalues. We are given that the
eigenvalues of Aare λ1= 2, λ2= 1, and λ3= 0.
Step 2: We are also given that one eigenvector corresponding to λ2= 1 is
1
1
2
. Let’s denote this eigenvector as v2=
1
1
2
.
Step 3: Recall that for a square matrix Aand an eigenvector vcorresponding
to eigenvalue λ, the following equation holds: Av=λv.
Step 4: Applying this to our given eigenvector v2with eigenvalue λ2= 1,
we get:
Av2=λ2v2
A
1
1
2
= 1 ·
1
1
2
a b c
d e f
g h i
1
1
2
=
1
1
2
Step 5: Solving the matrix equation above, we get the following system of
equations:
a+b+ 2c= 1
27
d+e+ 2f= 1
g+h+ 2i= 2
Step 6: From the given eigenvalues, we know that λ1= 2 corresponds to an
eigenvector v1, and λ3= 0 corresponds to an eigenvector v3. We can use the
eigenpairs to find the other columns of matrix A.
Step 7: After solving the system of equations and substituting, we find the
matrix Ato be:
A=
4−3 7
3−2 5
−2 1 −2
Question 33
Question
Let Abe a 3×3 matrix with eigenvalues λ1= 2, λ2= 1, λ3=−3. If det(A) = 4,
find the eigenvalues of the matrix B= 3A2−2A+I, where Iis the identity
matrix.
Solution
Step 1: First, let’s find the eigenvalues of matrix B= 3A2−2A+I.
B= 3A2−2A+I
= 3A2−2A+I
= 3A2−2A+I·A
= 3A2−2A+A
= 3A2−A.
Step 2: Now, let’s find the eigenvalues of matrix 3A2−A.
3A2−A= 3(AA)−A
= 3Aλ1−λ1I
= 3(2A)−2I
= 6A−2I
= 2(3A−I).
Step 3: The eigenvalues are the scalars αfor which the equation (3A2−A)v=
αv has a nontrivial solution v= 0. Therefore, the eigenvalues of Bare the
eigenvalues of 3A2−A, which are the eigenvalues of 3A−I, which are related
to the eigenvalues of Aby λ= 3λi−1.
Hence, the eigenvalues of matrix Bare 3(2) −1 = 5, 3(1) −1 = 2, and
3(−3) −1 = −10.
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Question 34
Question
Let Abe a 3×3 matrix with eigenvalues −1, 2, and 3. Given that the eigenvector
corresponding to the eigenvalue −1 is
1
1
1
and the eigenvector corresponding to
the eigenvalue 2 is
−2
1
1
, find the eigenvector corresponding to the eigenvalue
3.
Solution
To find the eigenvector corresponding to the eigenvalue 3, we can use the fact
that the sum of the eigenvalues of a matrix is equal to its trace. The trace of
a matrix is the sum of the elements on its main diagonal. Since Ais a 3 ×3
matrix, its trace is equal to the sum of its eigenvalues:
Trace(A) = −1 + 2 + 3 = 4.
The trace of Ais also equal to the sum of its diagonal elements:
Trace(A) = a11 +a22 +a33.
Since we know the eigenvalues of A, we can use these to find diagonal elements
of the matrix:
a11 =−1, a22 = 2, a33 = 3.
Therefore, the matrix Ais
A=
−100
0 2 0
0 0 3
.
To find the eigenvector corresponding to the eigenvalue 3, we need to solve
the system of equations (A−3I)v=0, where vis the eigenvector we are looking
for and Iis the identity matrix.
A−3I=
−1−3 0 0
0 2 −3 0
0 0 3 −3
=
−400
0−1 0
0 0 0
.
Solving the system (A−3I)v=0, we get:
−4v1= 0 =⇒v1= 0
−v2= 0 =⇒v2= 0.
So, the eigenvector corresponding to the eigenvalue 3 is
0
0
1
.
29
Question 35
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 5. Find
the eigenvectors associated with each eigenvalue.
Solution
To find the eigenvectors associated with each eigenvalue, we solve the equation
(A−λI)v=0, where Ais the given matrix, λis the eigenvalue, vis the
eigenvector, and Iis the identity matrix.
Step 1: Eigenvector for λ1= 2
Substitute λ= 2 into the equation (A−λI)v=0:
(A−2I)v=0
011
0−3 0
003
x
y
z
=
0
0
0
Solving this system of equations, we get y=z. So, the eigenvector for λ1= 2
is any vector of the form
1
1
1
.
Step 2: Eigenvector for λ2=−1
Substitute λ=−1 into the equation (A−λI)v=0:
(A+I)v=0
211
020
006
x
y
z
=
0
0
0
Solving this system of equations, we get x=−y. So, the eigenvector for
λ2=−1 is any vector of the form
−1
1
k
.
Step 3: Eigenvector for λ3= 5
Substitute λ= 5 into the equation (A−λI)v=0:
(A−5I)v=0
−3 1 1
0−6 0
0 0 −2
x
y
z
=
0
0
0
Solving this system of equations, we get x=1
3yand x=−1
3z. So, the
eigenvector for λ3= 5 is any vector of the form
1
3
−3
.
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