MATH 334 - DIFFERENTIAL
EQUATIONS - Matrix methods and
eigenvalue problems
Question Bank - Set 4
Liberty University
Question 1
Question
Let Abe a square matrix such that A2−4A+ 3I= 0, where Iis the identity
matrix. Find all possible eigenvalues of A.
Solution
Step 1: We are given that A2−4A+ 3I= 0. Let’s rewrite this equation as
A2−3A−A+ 3I= 0.
Step 2: We can factorize the above equation as A(A−3I)−(A−3I) = 0.
Step 3: Factoring out (A−3I), we get (A−3I)(A−I) = 0.
Step 4: For the product of two matrices to be zero, at least one of the
matrices must be singular. So, either A−3Iis singular or A−Iis singular.
Step 5: For A−3Ito be singular, we must have det(A−3I) = 0.
Step 6: Thus, the possible eigenvalues of Aare λ= 3 or λ= 1.
Question 2
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 4, and λ3=−3. If v1,
v2, and v3are the corresponding eigenvectors, find the matrix A.
Solution
Given that the matrix Ahas eigenvalues λ1= 2, λ2= 4, and λ3=−3 with
corresponding eigenvectors v1,v2, and v3, we have:
A=SΛS−1
where S= [v1v2v3] and Λ is a diagonal matrix with the eigenvalues on the
diagonal. We can express Susing the eigenvectors:
S= [v1v2v3] =
v11v21v31
v12v22v32
v13v23v33
Since v1,v2, and v3are eigenvectors of A, we have:
Av1= 2v1, Av2= 4v2, Av3=−3v3
Multiplying Aby v1,v2, and v3respectively, we get:
Av1=
2
0
0
= 2
1
0
0
Av2=
0
4
0
= 4
0
1
0
Av3=
0
0
−3
=−3
0
0
1
Therefore,
S=
100
010
001
Finally, we can find the matrix Aas:
A=SΛS−1=
100
010
001
2 0 0
0 4 0
0 0 −3
100
010
001
Question 3
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 4. If
the eigenvectors corresponding to λ1and λ3are v1=
1
2
1
and v3=
1
1
2
,
respectively, determine the eigenvector corresponding to λ2.
2
Solution
Step 1: Recall that the eigenvectors of a matrix Aare the non-zero vectors v
such that Av=λv, where λis an eigenvalue of A.
Step 2: Since λ1= 2 is an eigenvalue of Awith corresponding eigenvector
v1=
1
2
1
, we have Av1= 2v1.
Step 3: Similarly, since λ3= 4 is an eigenvalue of Awith corresponding
eigenvector v3=
1
1
2
, we have Av3= 4v3.
Step 4: We can express the matrix Ausing its eigenvalues and eigenvectors
as A=P DP −1, where P= [v1v2v3] is the matrix whose columns are the
eigenvectors, and D= diag(λ1, λ2, λ3) is the diagonal matrix of the eigenvalues.
Step 5: To find the eigenvector corresponding to λ2= 3, we first calculate
the matrix D:
D=
200
030
004
Step 6: Next, we calculate the matrix Pusing the given eigenvectors:
P=
1⋆1
2⋆1
1⋆2
Step 7: To solve for the missing column vector in P, substitute the known
eigenvectors into A=P DP −1and solve for the missing column vector.
Step 8: After obtaining the full matrix P, the eigenvector corresponding to
λ2= 3 will be the second column of P.
Question 4
Question
Let Abe a 2 ×2 matrix such that A2−3A+ 2I=O, where Iis the 2 ×2
identity matrix. If λ1and λ2are the eigenvalues of A, find the values of λ1and
λ2.
Solution
Step 1: First, we will use the given relationship A2−3A+ 2I=Oto find the
possible values for the matrix A.
A2−3A+ 2I=O
3
A2−3A+ 2 1 0
0 1=0 0
0 0
A2−3A+2 0
0 2=0 0
0 0
A2−3A=−2 0
0−2
Step 2: Next, we will find the characteristic equation of matrix Ato deter-
mine its eigenvalues λ1and λ2.
det(A−λI) = det a b
c d−λ1 0
0 1= 0
det a−λ b
c d −λ= (a−λ)(d−λ)−bc = 0
Step 3: Solving the characteristic equation, we have:
(λ−1)(λ−1) = 0
λ2−2λ+ 1 = 0
(λ−1)2= 0
Step 4: Since the characteristic equation has a repeated root, the eigenvalues
λ1and λ2are both equal to 1. Therefore, the values of λ1and λ2are both λ= 1.
Question 5
Question
Let Abe a 3 ×3 matrix with eigenvalues 1,2,and 3. Find the eigenvectors
corresponding to each eigenvalue.
Solution
Step 1: Recall that the eigenvectors of a matrix Acorresponding to an eigenvalue
λare the nonzero solutions to the equation (A−λI)v=0, where Iis the identity
matrix.
Step 2: For λ= 1, we solve the equation (A−I)v=0:
0 1 1
−101
−1−1 2
x
y
z
=
0
0
0
Step 3: Row-reducing the augmented matrix gives
0 1 1 |0
−101|0
−1−1 2 |0
→
1 0 −1|0
0 1 1 |0
0 0 0 |0
4
Step 4: We see that the solutions to the system are x=−t,y=−t, and
z=t, where tis a scalar. Therefore, the eigenvector corresponding to λ= 1 is
−1
−1
1
.
Step 5: Repeat the above steps for λ= 2 and λ= 3 to find the eigenvectors
corresponding to those eigenvalues.
Step 6: The eigenvector corresponding to λ= 2 is
1
−1
1
, and the eigenvector
corresponding to λ= 3 is
1
1
1
.
Question 6
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1,and λ3= 4. Find
the matrix A.
Solution
Step 1: Recall that the eigenvalues of a matrix are the roots of the characteristic
equation |A−λI|= 0, where Iis the identity matrix.
Step 2: The characteristic equation for a 3 ×3 matrix Ais given by:
|A−λI|=
a11 −λ a12 a13
a21 a22 −λ a23
a31 a32 a33 −λ
Step 3: Substitute the eigenvalues λ1= 2, λ2=−1,and λ3= 4 into the
characteristic equation |A−λI|= 0.
Step 4: For λ1= 2:
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
= 0
Step 5: For λ2=−1:
a11 + 1 a12 a13
a21 a22 + 1 a23
a31 a32 a33 + 1
= 0
Step 6: For λ3= 4:
a11 −4a12 a13
a21 a22 −4a23
a31 a32 a33 −4
= 0
5
Step 7: Solve the three characteristic equations simultaneously to find the
matrix A.
Question 7
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1=−2, λ2= 1, and λ3= 4. If v1,
v2, and v3are the corresponding eigenvectors, find the matrix A.
Solution
Step 1: Recall that if λis an eigenvalue of matrix Aand vis the corresponding
eigenvector, then Av=λv.
Step 2: We can express the matrix Ain terms of its eigenvectors and eigen-
values. Let Pbe the matrix whose columns are the eigenvectors of A:
P= [v1v2v3]
Step 3: The matrix Acan be diagonalized using matrix P:
A=P DP −1
where Dis the diagonal matrix with eigenvalues of matrix Aon the diagonal:
D=
−200
0 1 0
0 0 4
Step 4: We can find the inverse of matrix Pby using the formula:
P−1=1
det(P)adj(P)
where adj(P) is the adjugate of matrix P.
Step 5: Since the columns of Pare linearly independent, we have:
det(P)=1
Step 6: Therefore, the matrix Acan be expressed as:
A=P DP −1
Question 8
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 4. If
the eigenvectors corresponding to λ1and λ2are v1=
1
1
1
and v2=
1
−1
0
respectively, find the eigenvector corresponding to λ3.
6
Solution
Given the eigenvalues and eigenvectors of matrix A, we have the equation Av=
λv, where vis the eigenvector corresponding to λ.
Step 1: Construct the matrix equation Let v3=
x
y
z
be the eigenvec-
tor corresponding to λ3= 4. The matrix equation for this eigenvector is:
Av3=λ3v3
a b c
d e f
g h i
x
y
z
= 4
x
y
z
Step 2: Write out the matrix equation Expanding the matrix equation
gives us the system of equations:
ax +by +cz = 4x
dx +ey +fz = 4y
gx +hy +iz = 4z
Step 3: Substitute the values from the given eigenvectors Now we
substitute the values from the given eigenvectors v1and v2into the system of
equations: For v1=
1
1
1
:
a+b+c= 4
d+e+f= 4
g+h+i= 4
For v2=
1
−1
0
:
a−b= 4
d−e=−4
g= 0
Step 4: Solve the system of equations From the equations derived in
Step 3, we can solve for the values of a,b,c,d,e,f,g,h, and i. Ultimately, we
find the eigenvector corresponding to λ3by solving the system of equations.
7
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 4, λ2= 2, and λ3=−1. If
the eigenvectors corresponding to λ1and λ2are v1=
1
1
1
and v2=
1
−1
0
,
respectively, find a basis for the eigenspace corresponding to λ3.
Solution
Step 1: Recall that the eigenspace corresponding to an eigenvalue λis the set
of all vectors vsuch that Av =λv.
Step 2: Let v=
x
y
z
be an eigenvector corresponding to λ3. Then we have:
Av =λ3v
A
x
y
z
=−1
x
y
z
Step 3: This can be rewritten as:
4 1 2
2 1 −1
1 2 1
x
y
z
=
−x
−y
−z
Step 4: By solving the system of equations represented by this matrix equa-
tion, we can find the eigenvectors corresponding to λ3. This will give us the
basis for the eigenspace corresponding to λ3.
Question 10
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 2,−1,3. If det(A) = 10, find the
inverse of A.
Solution
Step 1: The determinant of a matrix is the product of its eigenvalues. Given
that det(A) = 10 and the eigenvalues of Aare λ= 2,−1,3, we have:
2×(−1) ×3 = 10
−6 = 10
8
This is a contradiction, so there must be an error in the given information. So,
let’s check the information provided.
Step 2: Let’s verify the determinant of matrix Ausing the product of its
eigenvalues formula. The determinant of a matrix Acan be expressed as:
det(A) = λ1λ2λ3
det(A) = 2 ×(−1) ×3 = −6
Step 3: Find the adjugate of Ausing its cofactor matrix. The cofactor
matrix Ccan be found by finding the determinant of each of the 3x3 matrices
obtained by removing one row and one column from A. The adjugate A∗is then
obtained by taking the transpose of C.
A=
a b c
d e f
g h i
Calculating the cofactor matrix:
C=
−3 6 −3
5−10 5
−2 4 −2
Calculating the adjugate:
A∗=
−3 5 −2
6−10 4
−3 5 −2
Step 4: Find the inverse of Ausing the formula: A−1=1
det(A)A∗.
A−1=1
−6
−3 5 −2
6−10 4
−3 5 −2
A−1=
1
2−5
3
1
3
−15
3−2
3
1
2−5
3
1
3
Therefore, the inverse of matrix Ais
1
2−5
3
1
3
−15
3−2
3
1
2−5
3
1
3
.
Question 11
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1=−2, λ2= 1, and λ3= 4. Find a
basis for each eigenspace corresponding to these eigenvalues.
9
Solution
Step 1: To find the eigenspace corresponding to each eigenvalue, we need to
solve the system of equations (A−λI)v=0, where λis the eigenvalue and vis
the eigenvector.
Step 2: For λ=−2, we solve (A+ 2I)v=0:
−2 + 2 1 0
2 1 + 2 −1
−1−2 2 + 2
v1
v2
v3
=
0
0
0
Solving this system gives the eigenspace for λ=−2.
Step 3: For λ= 1, we solve (A−I)v=0:
−1 + 1 1 0
2 1 −1−1
−1−2 2 −1
v1
v2
v3
=
0
0
0
Solving this system gives the eigenspace for λ= 1.
Step 4: For λ= 4, we solve (A−4I)v=0:
−4 + 4 1 0
2 1 −4−1
−1−2 2 −4
v1
v2
v3
=
0
0
0
Solving this system gives the eigenspace for λ= 4.
Question 12
Question
Let Abe a 3 ×3 matrix given by
A=
2−1 0
−1 2 −1
0−1 2
Find the eigenvalues and eigenvectors of A.
Solution
Step 1: Find the eigenvalues of A.To find the eigenvalues of A, we need to
solve the characteristic equation |A−λI|= 0, where λrepresents the eigenvalue
and Iis the identity matrix. The characteristic equation is given by:
|A−λI|=
2−λ−1 0
−1 2 −λ−1
0−1 2 −λ
= 0
10
Expanding the determinant, we get
(2 −λ)((2 −λ)2−1) + 1(2 −λ)=0
Solving this equation will give us the eigenvalues.
Step 2: Solve for the eigenvalues. Expanding the equation (2 −λ)((2 −
λ)2−1) + (2 −λ) = 0, we get
(2 −λ)(λ2−4λ+ 3) + (2 −λ)=0
Simplifying further, we have
(λ−1)(λ2−4λ+ 3) = 0
Now, solving the quadratic equation λ2−4λ+ 3 = 0, we find the eigenvalues to
be λ= 1,3,and 1.
Step 3: Find the eigenvectors corresponding to each eigenvalue.
For λ= 1:To find the eigenvectors corresponding to λ= 1, we substitute
λ= 1 into the equation Av=λv, where vis the eigenvector. This gives the
equation
(A−I)v=0
Where Iis the identity matrix. By solving the system of equations (A−I)v=0,
we can find the eigenvector associated with λ= 1.
For λ= 3:Similarly, we solve the equation (A−3I)v=0to find the
eigenvector associated with λ= 3.
After solving these systems of equations, we can find the corresponding eigen-
vectors for each eigenvalue.
Question 13
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 2,3,and −1.Determine the
eigenvectors corresponding to each eigenvalue.
Solution
Given that Ais a 3×3 matrix with eigenvalues λ= 2,3,and −1,we can find the
eigenvectors corresponding to each eigenvalue by solving the system of equations
(A−λI)v=0,where vis the eigenvector and Iis the identity matrix.
For λ= 2:Step 1: Set λ= 2 in the equation (A−λI)v=0to get
(A−2I)v=0.Step 2: Write the system of equations as
−123
1 0 3
2 4 0
x
y
z
=
0
0
0
.
11
Step 3: Solve the system of equations to obtain the eigenvector corresponding
to λ= 2.
For λ= 3:Step 1: Set λ= 3 in the equation (A−λI)v=0to get
(A−3I)v=0.Step 2: Write the system of equations as
−1 2 3
1−1 3
2 4 −3
x
y
z
=
0
0
0
.
Step 3: Solve the system of equations to obtain the eigenvector corresponding
to λ= 3.
For λ=−1:Step 1: Set λ=−1 in the equation (A−λI)v=0to get
(A+I)v=0.Step 2: Write the system of equations as
−323
1−2 3
2 4 1
x
y
z
=
0
0
0
.
Step 3: Solve the system of equations to obtain the eigenvector corresponding
to λ=−1.
Question 14
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 5, λ2=−2, and λ3=−1. If
the eigenvectors corresponding to λ1and λ2are v1=
1
0
1
and v2=
0
1
−1
respectively, find a basis for the eigenspace corresponding to λ3.
Solution
Step 1: Recall that the eigenspace corresponding to an eigenvalue λis the set
of all vectors vsuch that (A−λI)v=0.
Step 2: We are given that Ais a 3 ×3 matrix with eigenvalues λ1= 5,
λ2=−2, and λ3=−1.
Step 3: To find the eigenvectors corresponding to λ3=−1, we solve the
system (A+I)v=0, where Iis the 3 ×3 identity matrix.
Step 4: Substituting λ=−1 and expanding the matrix equation, we get:
611
0−1−1
−1−1−2
x
y
z
=
0
0
0
Step 5: Solving the system of equations, we get:
6x+y+z= 0
−y−z= 0
−x−y−2z= 0
12
Step 6: From the second equation, we have y=−z. Substituting y=−z
into the first and third equation, we have:
6x−z= 0
−x−z= 0
Step 7: Solving the system of equations, we find x=z.
Step 8: Therefore, the eigenvectors corresponding to λ3=−1 are of the
form
z
−z
z
, where zis a real number.
Step 9: One basis for the eigenspace corresponding to λ3=−1 is the vector
v3=
1
−1
1
.
Question 15
Question
Let Abe a 3 ×3 matrix given by
A=
211
132
122
.
Find the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0, where λis the eigenvalue and Iis the identity matrix.
Step 2: The characteristic equation is given by
2−λ1 1
1 3 −λ2
1 2 2 −λ
= 0.
Step 3: Expanding the determinant, we get
(2 −λ)
3−λ2
2 2 −λ−
1 2
1 2 −λ
+
1 3 −λ
1 2
= 0.
Step 4: Simplifying further, we have
(2 −λ)[(3 −λ)(2 −λ)−4] −[(2 −λ)−2] + [2 −3 + λ] = 0.
Step 5: Solving this equation gives us the eigenvalues λ= 1,2,4.
13
Step 6: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute back into the equation (A−λI)v=0and solve for v.
Step 7: For λ= 1, solving (A−I)v=0gives us the eigenvector
v1=
1
1
−1
.
Step 8: For λ= 2, solving (A−2I)v=0gives us the eigenvector
v2=
−1
1
1
.
Step 9: For λ= 4, solving (A−4I)v=0gives us the eigenvector
v3=
1
1
1
.
Step 10: Therefore, the eigenvalues of Aare λ= 1,2,4 with eigenvectors
v1=
1
1
−1
,v2=
−1
1
1
, and v3=
1
1
1
respectively.
Question 16
Question
Let Abe a 3 ×3 matrix given by
A=
4 1 −1
−1 3 −1
2−2 3
Find the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: Find the characteristic polynomial The characteristic polynomial
of matrix Ais given by det(A−λI) = 0, where λis the eigenvalue we are trying
to find.
A−λI =
4−λ1−1
−1 3 −λ−1
2−2 3 −λ
Next, we find the determinant of the matrix A−λI to form the characteristic
polynomial.
14
Step 2: Calculate the determinant
det(A−λI) = det
4−λ1−1
−1 3 −λ−1
2−2 3 −λ
Expanding along the first row, we get:
= (4 −λ)det 3−λ−1
−2 3 −λ−1·det −1−1
2 3 −λ−(−1) ·det −1 3 −λ
2−2
Solving these determinants will give us the characteristic polynomial.
Step 3: Solve for the characteristic polynomial Solving the determi-
nants, we get:
= (4−λ)((3−λ)(3−λ)−(−1)(−2))−((−1)((3−λ)(−1)−(−1)(2))+(−1)((−1)(−2)−(3−λ)(2))
= 0
This simplifies to the characteristic polynomial:
λ3−10λ2+ 29λ−24 = 0
Step 4: Find the eigenvalues Now, we need to find the roots of the
characteristic polynomial λ3−10λ2+ 29λ−24 = 0. By inspection, we can see
that λ= 4 is a root of the polynomial.
Step 5: Find the remaining eigenvalues Using polynomial long division
or synthetic division, we can divide the characteristic polynomial by (λ−4) to
find the remaining roots.
Step 6: Find the eigenvectors For each eigenvalue, we can find the
eigenvector by substituting back into the equation (A−λI)v=0and solving
for v.
For each eigenvalue, solve the system of equations to find the eigenvector.
Question 17
Question
Let Abe a 3×3 matrix with eigenvalues λ1= 2, λ2= 1+i, and λ3= 1−i. Find
the characteristic polynomial of Aand determine whether Ais diagonalizable.
Solution
Step 1: The characteristic polynomial of a matrix Ais given by
det(A−λI)=0,
where Iis the identity matrix. For a 3 ×3 matrix A, this will be a cubic
polynomial.
15
Step 2: For eigenvalue λ1= 2,
A−2I=
a−2b c
d e −2f
g h i −2
.
Taking the determinant and setting it equal to zero gives the coefficient of the
characteristic polynomial for λ1.
Step 3: For eigenvalue λ2= 1 + i,
A−(1 + i)I=
a−(1 + i)b c
d e −(1 + i)f
g h i −(1 + i)
.
Taking the determinant and setting it equal to zero gives the coefficient of the
characteristic polynomial for λ2.
Step 4: For eigenvalue λ3= 1 −i,
A−(1 −i)I=
a−(1 −i)b c
d e −(1 −i)f
g h i −(1 −i)
.
Taking the determinant and setting it equal to zero gives the coefficient of the
characteristic polynomial for λ3.
Step 5: The characteristic polynomial of Awill be a cubic polynomial with
roots λ1= 2, λ2= 1 + i, and λ3= 1 −i.
Step 6: To determine whether Ais diagonalizable, we need to check if the
algebraic multiplicity of each eigenvalue is equal to its geometric multiplicity. If
they are all equal, then Ais diagonalizable.
Question 18
Question
Let Abe a 3x3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−3. Find the
eigenvectors associated with each eigenvalue.
Solution
Step 1: Find the eigenvector associated with λ1= 2. To find the eigenvec-
tor, we need to solve the equation (A−λI)v= 0, where vis the eigenvector
corresponding to λ1.
Substitute λ1= 2 into the equation to get:
(A−2I)v= 0
16
Then, substitute Aas the given 3x3 matrix with eigenvalues and Ias the
identity matrix:
2−1 0
1 1 0
−1−1 5
v= 0
Perform row operations to row-reduce the augmented matrix to row-echelon
form and solve for v.
Step 2: Find the eigenvector associated with λ2= 1. Follow the same steps
as in Step 1, but this time use λ2= 1:
(A−I)v= 0
Substitute the values and solve for v.
Step 3: Find the eigenvector associated with λ3=−3. Again, follow the
same steps as in Step 1, but use λ3=−3:
(A+ 3I)v= 0
Substitute the values and solve for v.
Question 19
Question
Let Abe a 3 ×3 matrix given by
A=
3−1 0
1 2 −1
−1−1 3
Determine the eigenvalues and eigenvectors of the matrix A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation given
by det(A−λI) = 0, where Iis the identity matrix.
A−λI =
3−λ−1 0
1 2 −λ−1
−1−1 3 −λ
The characteristic equation becomes:
det(A−λI) = det
3−λ−1 0
1 2 −λ−1
−1−1 3 −λ
= 0
17
Step 2: Expanding the determinant along the first row, we get:
(3−λ) [(2 −λ)(3 −λ)−(−1)(−1)]+1 [−1(3 −λ)−(−1)(−1)]+(−1) [−1(−1) −1(2 −λ)] = 0
Solving this equation gives us the eigenvalues of the matrix.
Step 3: Once we have the eigenvalues, we can find the eigenvectors corre-
sponding to each eigenvalue by solving (A−λI)v=0, where vis the eigenvector
corresponding to the eigenvalue λ.
Step 4: Repeat step 3 for each eigenvalue to find the corresponding eigen-
vectors. The eigenvectors form a basis for each eigenspace of the matrix.
Question 20
Question
Let Abe a square matrix given by
A=
2−1 0
1 3 −1
0−1 2
.
Find the eigenvalues and corresponding 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 matrix A:
A−λI =
2−λ−1 0
1 3 −λ−1
0−1 2 −λ
.
Calculating the determinant and setting it to zero gives:
det(A−λI) = (2 −λ)((3 −λ)(2 −λ)−1) + (−1)(−1) = 0.
Solving this equation will give the eigenvalues.
Step 2: Find the eigenvalues Expanding the determinant equation, we
get:
(2 −λ)((3 −λ)(2 −λ)−1) + 1 = 0
(2 −λ)(6 −5λ+λ2−1) + 1 = 0
(2 −λ)(5λ−λ2+ 5) + 1 = 0
−λ3+ 8λ2−15λ+ 11 = 0
Hence, the characteristic equation is −λ3+ 8λ2−15λ+ 11 = 0.
18
Step 3: Find the eigenvectors Once we have found the eigenvalues,
we can substitute them back into (A−λI)x=0to find the corresponding
eigenvectors.
For each eigenvalue, solve (A−λI)x=0to find the eigenvector.
This process will give us the eigenvalues and corresponding eigenvectors for
matrix A.
Question 21
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 1,2,3. Find the eigenvalues of
the matrix B=A2+ 2A+I, where Iis the identity matrix.
Solution
Given that the eigenvalues of matrix Aare λ= 1,2,3, we want to find the
eigenvalues of the matrix B=A2+ 2A+I.
Step 1: First, we find the matrix B=A2+ 2A+I.
B=A2+ 2A+I
Step 2: Since we know the eigenvalues of matrix A, we can use the eigen-
value equation Av =λv to find the eigenvalues of matrix B. Let vbe an
eigenvector of Acorresponding to eigenvalue λ.
Step 3: We have:
Bv = (A2+ 2A+I)v
=A2v+ 2Av +Iv
=A(Av)+2Av +v
=A(λv) + 2(λv) + v
=λAv + 2λv +v
=λ2v+ 2λv +v
= (λ2+ 2λ+ 1)v
= (λ+ 1)2v
= (1 + 1)2v(when λ= 1)
= 4v(when λ= 1)
Step 4: Therefore, the eigenvalue of Bwhen λ= 1 is λ= 4.
Step 5: Similarly, we can repeat the above calculations for the other eigen-
values of Ato find the eigenvalues of B.
Bv = (A2+ 2A+I)v= (2 + 1)2v= 9v(when λ= 2)
Bv = (A2+ 2A+I)v= (3 + 1)2v= 16v(when λ= 3)
Step 6: Therefore, the eigenvalues of matrix B=A2+2A+Iare λ= 4,9,16.
19
Question 22
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If tr(A)
denotes the trace of A(sum of the diagonal entries), find the possible values of
tr(A).
Solution
Step 1: The trace of a matrix is the sum of its diagonal entries. Therefore, we
have:
tr(A) = λ1+λ2+λ3
Step 2: Substituting the given eigenvalues into the formula, we get:
tr(A)=1+2+3
tr(A) = 6
Therefore, the only possible value for tr(A) is 6 .
Question 23
Question
Let Abe a 3 ×3 real matrix with eigenvalues λ= 1,3,5 and corresponding
eigenvectors
v1=
1
0
1
,v2=
1
1
1
,v3=
0
1
1
.
Find A.
Solution
To find the matrix A, we can use the formula A=P DP −1, where Pis the
matrix whose columns are the eigenvectors of A, and Dis the diagonal matrix
with the corresponding eigenvalues on the diagonal.
Step 1: Construct matrix P
The matrix Pis formed by the eigenvectors, so
P=
110
011
111
.
Step 2: Construct matrix D
20
The matrix Dis a diagonal matrix with the eigenvalues on the diagonal, so
D=
100
030
005
.
Step 3: Calculate P−1
Next, we need to find the inverse of matrix Pin order to compute A=
P DP −1. The inverse of Pcan be found using standard techniques. We have
P−1=1
det(P)
A−B C
−D E −F
G−H I
,
where det(P) is the determinant of P.
Finally, combining P,D, and P−1, we can calculate A:
A=P DP −1.
Therefore, the matrix Acan be determined by the above calculations.
Question 24
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2=−2, and λ3= 4. If v1=
1
1
1
is an eigenvector corresponding to λ1, find an eigenvector v3corresponding
to λ3.
Solution
Step 1: Recall that eigenvectors v1, v2, . . . , vnassociated with distinct eigenval-
ues λ1, λ2, . . . , λnof a matrix Aare linearly independent. Therefore, we can
find the third eigenvector by solving the system of equations (A−λI)v= 0
where Ais the given matrix, λ=λ3= 4, and v=
x
y
z
.
Step 2: Subtract λI from Ato get:
A−4I=
−300
0 6 0
0 0 0
Step 3: Setting (A−4I)v= 0 gives the following system of equations:
−3x= 0
6y= 0
0z= 0
21
Step 4: The solution to the system of equations is x= 0 and y= 0. There-
fore, there are infinitely many solutions, which corresponds to the eigenvectors
of the matrix A. One such eigenvector is v3=
0
0
1
.
Thus, an eigenvector corresponding to λ3= 4 is v3=
0
0
1
.
Question 25
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and a repeated
eigenvalue λ3= 3. If the corresponding eigenvectors are v1=
1
0
1
,v2=
1
1
0
,
and v3=
1
1
a
, find the value of a.
Solution
Step 1: Since Ais a 3 ×3 matrix, it has 3 eigenvalues (which may repeat) and 3
corresponding eigenvectors. Therefore, the eigenvectors v1,v2, and v3provided
are indeed the eigenvectors corresponding to the eigenvalues λ1,λ2, and λ3.
Step 2: For each eigenvector, we have Avi=λivi, where i∈ {1,2,3}.
Step 3: Using the given information, we have: Av1= 1
1
0
1
=
1
0
1
Av2=
2
1
1
0
=
2
2
0
Av3= 3
1
1
a
=
3
3
3a
Step 4: Comparing Av3with λ3v3, we get:
3
3
3a
=
3
3
3a
This implies
3a= 3, so a= 1.
Therefore, the value of ais 1 .
22
Question 26
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Given
that corresponding eigenvectors are v1=
1
1
1
,v2=
1
2
3
, and v3=
0
1
−1
, find
the matrix A.
Solution
Step 1: Recall that if λis an eigenvalue of A, then there exists a non-zero vector
vsuch that Av=λv. In other words, vis an eigenvector corresponding to
eigenvalue λ.
Step 2: Using the given eigenvectors and eigenvalues, we can write Av1=
1·v1,Av2= 2 ·v2, and Av3= 3 ·v3.
Step 3: Writing the matrix Ain terms of the eigenvectors and eigenvalues:
A= [v1v2v3]
λ10 0
0λ20
0 0 λ3
[v1v2v3]−1
Step 4: First, find [v1v2v3]:
[v1v2v3] =
1 1 0
1 2 1
1 3 −1
Step 5: Next, find [v1v2v3]−1using a calculator or matrix methods.
Step 6: Substitute the matrices [v1v2v3] and [v1v2v3]−1into the formula
for Aand simplify to find the matrix A.
Question 27
Question
Let Abe a 3×3 matrix with eigenvalues λ= 1,3,5. Given that Ais symmetric,
find the eigenvectors corresponding to each eigenvalue.
Solution
Step 1: Recall that eigenvectors are the vectors that satisfy the equation (A−
λI)x=0, where λis the eigenvalue.
Step 2: For λ= 1, we have:
A−λI =
000
020
004
−
100
010
001
=
−100
0 1 0
0 0 3
23
Step 3: By solving (A−λI)x=0for λ= 1, we get the eigenvector v1=
1
0
0
.
Step 4: For λ= 3, we have:
A−λI =
−200
0 0 0
0 0 2
Step 5: By solving (A−λI)x=0for λ= 3, we get the eigenvector v2=
0
1
0
.
Step 6: For λ= 5, we have:
A−λI =
−400
0−2 0
0 0 0
Step 7: By solving (A−λI)x=0for λ= 5, we get the eigenvector v3=
0
0
1
.
Therefore, the eigenvectors corresponding to the eigenvalues λ= 1,3,5 are
v1=
1
0
0
,v2=
0
1
0
, and v3=
0
0
1
.
Question 28
Question
Let Abe a 3x3 matrix with eigenvalues λ1= 2, λ2= 3,and λ3= 4. Given
that the eigenvectors corresponding to λ1and λ2are v1=
1
0
1
and v2=
1
1
0
,
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: Recall that for an n×nmatrix A, the eigenvectors corresponding to
distinct eigenvalues are linearly independent.
Step 2: Since we know that v1and v2are eigenvectors of Acorresponding
to λ1= 2 and λ2= 3 respectively, we can form a basis for the eigenspace
corresponding to λ3= 4.
Step 3: Let v3=
x
y
z
be the eigenvector corresponding to λ3. We know
that Av3=λ3v3, so we have:
24
A
x
y
z
= 4
x
y
z
Step 4: This gives us the system of equations:
3x+y= 4x
x+ 3y= 4y
z= 4z
Step 5: Simplifying the first two equations:
(−x+y= 0
x−y= 0
Step 6: Solving this system of equations, we find x=y. This implies that
the eigenvector corresponding to λ3= 4 is v3=
1
1
z
.
Step 7: Therefore, the eigenvector corresponding to λ3= 4 is v3=
1
1
0
.
Question 29
Question
Let Abe a 3 ×3 matrix given by
A=
211
121
112
Determine the eigenvalues and corresponding eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of A, we need to solve the characteristic equation
|A−λI|= 0, where λis the eigenvalue and Iis the identity matrix.
A−λI =
2−λ1 1
1 2 −λ1
1 1 2 −λ
Step 2: Setting the determinant of A−λI to zero, we have
2−λ1 1
1 2 −λ1
1 1 2 −λ
= 0
25
Step 3: Expanding the determinant using the cofactor expansion along the
first row, we get the characteristic polynomial
(2 −λ)[(2 −λ)(2 −λ)−1] −[(2 −λ)−1 + 1] + 1 = 0
Step 4: Simplifying, we have
(2 −λ)[(2 −λ)2−1] −2(2 −λ)=0
(2 −λ)(4 −4λ+λ2−1) −2(2 −λ) = 0
(2 −λ)(3 −4λ+λ2) = 2(2 −λ)
Step 5: This leads to the characteristic equation −λ3+ 6λ2−9λ= 0.
Step 6: Factoring out λ, we get λ(λ−3)(λ−3) = 0.
Step 7: Therefore, the eigenvalues are λ= 0 with a multiplicity of 1 and
λ= 3 with a multiplicity of 2.
Step 8: Next, we find the eigenvectors corresponding to each eigenvalue. For
λ= 0, we solve (A−0I)X= 0 to find the eigenvector.
Step 9: Row-reducing the augmented matrix (A−0I|0), we get
211|0
121|0
112|0
Step 10: Performing row operations, we find the eigenvector corresponding
to λ= 0 as X=
−1
1
1
.
Step 11: For λ= 3, we solve (A−3I)X= 0 to find the eigenvector.
Step 12: Row-reducing the augmented matrix (A−3I|0), we get
−1 1 1 |0
1−1 1 |0
1 1 −1|0
Step 13: Performing row operations, we find the eigenvector corresponding
to λ= 3 as X=
1
1
1
.
Therefore, the eigenvalues of matrix Aare λ= 0 with eigenvector
−1
1
1
and λ= 3 with eigenvector
1
1
1
.
26
Question 30
Question
Let Abe a 3 ×3 matrix with eigenvalues 1, 2, and 3. If B= 2A2−A+I, where
Iis the 3 ×3 identity matrix, find the eigenvalues of B.
Solution
Step 1: First, let’s find the eigenvalues of A. The characteristic equation of a
3×3 matrix Ais given by det(A−λI) = 0. This gives us:
det
a11 −λ a12 a13
a21 a22 −λ a23
a31 a32 a33 −λ
= 0
Substitute the given eigenvalues 1, 2, and 3 into the characteristic equation to
determine aij , as follows:
det
a11 −1a12 a13
a21 a22 −1a23
a31 a32 a33 −1
= 0
det
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
= 0
det
a11 −3a12 a13
a21 a22 −3a23
a31 a32 a33 −3
= 0
This will give us the values of the elements of A.
Step 2: Calculate the matrix Busing the formula B= 2A2−A+I.
Step 3: Let’s find the eigenvalues of matrix B. Since Bis defined as a
function of matrix A, we can relate the eigenvalues of Bto the eigenvalues of A
using the following property: If λis an eigenvalue of matrix A, then f(λ) is an
eigenvalue of f(A) for any polynomial function f(x).
Therefore, the eigenvalues of Bare {2(1)2−1+1,2(2)2−2+1,2(3)2−3+1}.
Calculating these values gives us the final answer for the eigenvalues of B.
Question 31
Question
Let Abe a 3 ×3 matrix given by
A=
110
121
012
.
Find the eigenvalues and eigenvectors of matrix A.
27
Solution
Step 1: To find the eigenvalues of matrix A, we solve the characteristic equation
|A−λI|= 0, where Iis the identity matrix. We have:
A−λI =
110
121
012
−λ
100
010
001
=
1−λ1 0
1 2 −λ1
0 1 2 −λ
.
The characteristic equation becomes |A−λI|= (1 −λ)((2 −λ)(2 −λ)−1) −
(1 ·(2 −λ)) = 0. This simplifies to λ3−5λ2+ 6λ−2 = 0.
Step 2: To find the roots of the characteristic equation, we can use numerical
methods or simply observe that λ= 1 is a root. So we divide the characteristic
equation by (λ−1) to get a quadratic equation which can be factored to find
the remaining roots. Dividing by (λ−1), we get:
(λ3−5λ2+ 6λ−2)∇ · (λ−1) = λ2−4λ+ 2 = 0.
This quadratic equation does not factor nicely, so we compute the remaining
eigenvalues as:
λ=4±√42−4∗1∗2
2=4±√8
2= 2 ±√2.
Step 3: Now that we have found the eigenvalues λ= 1,2±√2, we can find
the corresponding eigenvectors by solving the system (A−λI)v= 0 for each
eigenvalue.
For λ= 1, we have:
(A−I)v=
010
111
011
v= 0.
From this system of equations, we can find linearly independent eigenvectors
with components v1= 1, v2=−1, v3= 1.
For λ= 2 + √2, we have:
(A−(2 + √2)I)v=
−1−√2 1 0
1−√2 1
0 1 −√2
v= 0.
Solving this system of equations, we find an eigenvector v=
1
2(1 + √2)
1
.
For λ= 2 −√2, we have:
(A−(2 −√2)I)v=
−1 + √2 1 0
1√2 1
0 1 √2
v= 0.
28
Solving this system of equations, we find an eigenvector v=
1
2(1 −√2)
1
.
Therefore, the eigenvalues of matrix Aare λ= 1,2±√2, with corresponding
eigenvectors v
Question 32
Question
Let Abe a square matrix. If Ahas eigenvalues 2 and −1 with corresponding
eigenvectors v1=1
1and v2=1
−1, find a matrix Bsuch that B2=A.
Solution
Step 1: To find matrix B, we first express Ain terms of its eigenvectors and
eigenvalues. Since v1and v2are eigenvectors corresponding to eigenvalues 2
and −1 respectively, we can write:
A=P DP −1
where
P=v1v2=1 1
1−1
and
D=2 0
0−1
Step 2: Recall that for a diagonal matrix D,Dnis obtained by raising each
diagonal element to the power of n. Therefore, A2= (P DP −1)2=P D2P−1.
Step 3: Calculate D2:
D2=2 0
0−12
=4 0
0 1
Step 4: Now, compute A2:
A2=P D2P−1=1 1
1−14 0
0 1−1
2
1
2
1
2
1
2
Step 5: Simplify the expression to find A2and identify a suitable matrix B
such that B2=A.
A2=4 0
4 0−1
2
1
2
1
2
1
2=−2 2
−2 2
Therefore, a suitable matrix Bsuch that B2=Ais:
B=0 1
1 1
29
Question 33
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 4, λ2= 1,and λ3=−2. Find a
basis for each eigenspace corresponding to these eigenvalues.
Solution
Step 1: To find the eigenvectors corresponding to λ= 4, we need to solve the
system (A−4I)v=0, where Iis the identity matrix.
Step 2: Setting up the equation (A−4I)v=0, we have
0 2 −1
3 0 1
1 2 0
x
y
z
=
0
0
0
Step 3: Solving the system of equations, we get x= 0, y = 1, z = 3. There-
fore, an eigenvector corresponding to λ= 4 is v1=
0
1
3
.
Step 4: Following the same process, we can find eigenvectors corresponding
to λ= 1 and λ=−2. The eigenvectors corresponding to λ= 1 and λ=−2 are
v2=
1
1
1
and v3=
−1
1
−1
, respectively.
Therefore, bases for the eigenspaces are: - Eigenspace corresponding to
λ= 4: {
0
1
3
}- Eigenspace corresponding to λ= 1: {
1
1
1
}- Eigenspace corre-
sponding to λ=−2: {
−1
1
−1
}
Question 34
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−1. If the
corresponding eigenvectors are given by
v1=
1
0
−1
,v2=
1
1
0
,and v3=
1
2
−2
,
find the matrix A.
30
Solution
Step 1: Recall that a matrix can be constructed using its eigenvectors and
eigenvalues. Since Ais a 3 ×3 matrix, we can write it as
A=P DP −1,
where
P=v1v2v3and D=
λ10 0
0λ20
0 0 λ3
.
Step 2: Substituting the given values into Pand D, we have
P=
111
012
−1 0 −2
and D=
2 0 0
0 1 0
0 0 −1
.
Step 3: Calculate P−1. To find the inverse of P, we solve the equation
P−1P=I, where Iis the identity matrix. Therefore,
P−1=
−2 2 −1
1 0 −1
2−1 1
.
Step 4: Compute A=P DP −1. We have
A=
111
012
−1 0 −2
2 0 0
0 1 0
0 0 −1
−2 2 −1
1 0 −1
2−1 1
.
Step 5: Multiplying the matrices, we get
A=
111
012
−1 0 −2
2 0 0
0 1 0
0 0 −1
−2 2 −1
1 0 −1
2−1 1
=
2 1 −1
2 1 −2
6 2 −3
.
Therefore, the matrix Ais
2 1 −1
2 1 −2
6 2 −3
.
Question 35
Question
Let Abe a 3 ×3 matrix given by
A=
311
131
113
.
Find the eigenvalues of matrix Aand determine the corresponding eigenvec-
tors.
31
Solution
Step 1: Find the characteristic polynomial of matrix A.The character-
istic polynomial of a matrix Ais given by det(A−λI), where λis the eigenvalue
and Iis the identity matrix of the same size as A.
For our matrix A, we have:
A−λI =
3−λ1 1
1 3 −λ1
1 1 3 −λ
.
The determinant of A−λI is:
det(A−λI) = (3 −λ)((3 −λ)2−1) −(3 −λ−1) + 1 −(3 −λ)
det(A−λI) = (3 −λ)(9 −6λ+λ2−1) −2+1−3 + λ
det(A−λI) = (3 −λ)(λ2−6λ+ 8) −4 + λ
det(A−λI) = 3λ2−18λ+ 24 −λ3+ 6λ2−8λ−4 + λ
det(A−λI) = −λ3+ 9λ2−21λ+ 20.
So, the characteristic polynomial is −λ3+ 9λ2−21λ+ 20.
Step 2: Find the eigenvalues of matrix A.To find the eigenvalues, we
need to solve the equation −λ3+ 9λ2−21λ+ 20 = 0. By inspection, we can see
that one of the eigenvalues is λ= 1.
Let’s factor the polynomial to find the other eigenvalues:
−λ3+ 9λ2−21λ+ 20 = −(λ−1)(λ2−8λ+ 20).
Using the quadratic formula, we can find the remaining eigenvalues to be
λ= 4 and λ= 5.
Step 3: Find the eigenvectors corresponding to each eigenvalue.
Now, we will find the eigenvectors corresponding to each eigenvalue.
For λ= 1, we need to solve the system of equations (A−I)v= 0:
(A−I)v=
211
121
112
v=
0
0
0
.
Solving this system, we get v=
1
−1
0
.
Similarly, for λ= 4 and λ= 5, we can find the eigenvectors to be v=
1
1
−2
and v=
1
0
−1
, respectively.
Therefore, the eigenvalues of matrix Aare 1, 4, and 5, with corresponding
eigenvectors
1
−1
0
,
1
1
−2
, and
1
0
−1
.
32
Solution
Given that the matrix Ahas eigenvalues λ1= 2, λ2= 4, and λ3=−3 with
corresponding eigenvectors v1,v2, and v3, we have:
A=SΛS−1
where S= [v1v2v3] and Λ is a diagonal matrix with the eigenvalues on the
diagonal. We can express Susing the eigenvectors:
S= [v1v2v3] =
v11v21v31
v12v22v32
v13v23v33
Since v1,v2, and v3are eigenvectors of A, we have:
Av1= 2v1, Av2= 4v2, Av3=−3v3
Multiplying Aby v1,v2, and v3respectively, we get:
Av1=
2
0
0
= 2
1
0
0
Av2=
0
4
0
= 4
0
1
0
Av3=
0
0
−3
=−3
0
0
1
Therefore,
S=
100
010
001
Finally, we can find the matrix Aas:
A=SΛS−1=
100
010
001
2 0 0
0 4 0
0 0 −3
100
010
001
Question 3
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 4. If
the eigenvectors corresponding to λ1and λ3are v1=
1
2
1
and v3=
1
1
2
,
respectively, determine the eigenvector corresponding to λ2.
2
Solution
Step 1: Recall that the eigenvectors of a matrix Aare the non-zero vectors v
such that Av=λv, where λis an eigenvalue of A.
Step 2: Since λ1= 2 is an eigenvalue of Awith corresponding eigenvector
v1=
1
2
1
, we have Av1= 2v1.
Step 3: Similarly, since λ3= 4 is an eigenvalue of Awith corresponding
eigenvector v3=
1
1
2
, we have Av3= 4v3.
Step 4: We can express the matrix Ausing its eigenvalues and eigenvectors
as A=P DP −1, where P= [v1v2v3] is the matrix whose columns are the
eigenvectors, and D= diag(λ1, λ2, λ3) is the diagonal matrix of the eigenvalues.
Step 5: To find the eigenvector corresponding to λ2= 3, we first calculate
the matrix D:
D=
200
030
004
Step 6: Next, we calculate the matrix Pusing the given eigenvectors:
P=
1⋆1
2⋆1
1⋆2
Step 7: To solve for the missing column vector in P, substitute the known
eigenvectors into A=P DP −1and solve for the missing column vector.
Step 8: After obtaining the full matrix P, the eigenvector corresponding to
λ2= 3 will be the second column of P.
Question 4
Question
Let Abe a 2 ×2 matrix such that A2−3A+ 2I=O, where Iis the 2 ×2
identity matrix. If λ1and λ2are the eigenvalues of A, find the values of λ1and
λ2.
Solution
Step 1: First, we will use the given relationship A2−3A+ 2I=Oto find the
possible values for the matrix A.
A2−3A+ 2I=O
3
A2−3A+ 2 1 0
0 1=0 0
0 0
A2−3A+2 0
0 2=0 0
0 0
A2−3A=−2 0
0−2
Step 2: Next, we will find the characteristic equation of matrix Ato deter-
mine its eigenvalues λ1and λ2.
det(A−λI) = det a b
c d−λ1 0
0 1= 0
det a−λ b
c d −λ= (a−λ)(d−λ)−bc = 0
Step 3: Solving the characteristic equation, we have:
(λ−1)(λ−1) = 0
λ2−2λ+ 1 = 0
(λ−1)2= 0
Step 4: Since the characteristic equation has a repeated root, the eigenvalues
λ1and λ2are both equal to 1. Therefore, the values of λ1and λ2are both λ= 1.
Question 5
Question
Let Abe a 3 ×3 matrix with eigenvalues 1,2,and 3. Find the eigenvectors
corresponding to each eigenvalue.
Solution
Step 1: Recall that the eigenvectors of a matrix Acorresponding to an eigenvalue
λare the nonzero solutions to the equation (A−λI)v=0, where Iis the identity
matrix.
Step 2: For λ= 1, we solve the equation (A−I)v=0:
0 1 1
−101
−1−1 2
x
y
z
=
0
0
0
Step 3: Row-reducing the augmented matrix gives
0 1 1 |0
−101|0
−1−1 2 |0
→
1 0 −1|0
0 1 1 |0
0 0 0 |0
4
Step 4: We see that the solutions to the system are x=−t,y=−t, and
z=t, where tis a scalar. Therefore, the eigenvector corresponding to λ= 1 is
−1
−1
1
.
Step 5: Repeat the above steps for λ= 2 and λ= 3 to find the eigenvectors
corresponding to those eigenvalues.
Step 6: The eigenvector corresponding to λ= 2 is
1
−1
1
, and the eigenvector
corresponding to λ= 3 is
1
1
1
.
Question 6
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1,and λ3= 4. Find
the matrix A.
Solution
Step 1: Recall that the eigenvalues of a matrix are the roots of the characteristic
equation |A−λI|= 0, where Iis the identity matrix.
Step 2: The characteristic equation for a 3 ×3 matrix Ais given by:
|A−λI|=
a11 −λ a12 a13
a21 a22 −λ a23
a31 a32 a33 −λ
Step 3: Substitute the eigenvalues λ1= 2, λ2=−1,and λ3= 4 into the
characteristic equation |A−λI|= 0.
Step 4: For λ1= 2:
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
= 0
Step 5: For λ2=−1:
a11 + 1 a12 a13
a21 a22 + 1 a23
a31 a32 a33 + 1
= 0
Step 6: For λ3= 4:
a11 −4a12 a13
a21 a22 −4a23
a31 a32 a33 −4
= 0
5
Step 7: Solve the three characteristic equations simultaneously to find the
matrix A.
Question 7
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1=−2, λ2= 1, and λ3= 4. If v1,
v2, and v3are the corresponding eigenvectors, find the matrix A.
Solution
Step 1: Recall that if λis an eigenvalue of matrix Aand vis the corresponding
eigenvector, then Av=λv.
Step 2: We can express the matrix Ain terms of its eigenvectors and eigen-
values. Let Pbe the matrix whose columns are the eigenvectors of A:
P= [v1v2v3]
Step 3: The matrix Acan be diagonalized using matrix P:
A=P DP −1
where Dis the diagonal matrix with eigenvalues of matrix Aon the diagonal:
D=
−200
0 1 0
0 0 4
Step 4: We can find the inverse of matrix Pby using the formula:
P−1=1
det(P)adj(P)
where adj(P) is the adjugate of matrix P.
Step 5: Since the columns of Pare linearly independent, we have:
det(P)=1
Step 6: Therefore, the matrix Acan be expressed as:
A=P DP −1
Question 8
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 4. If
the eigenvectors corresponding to λ1and λ2are v1=
1
1
1
and v2=
1
−1
0
respectively, find the eigenvector corresponding to λ3.
6
Solution
Given the eigenvalues and eigenvectors of matrix A, we have the equation Av=
λv, where vis the eigenvector corresponding to λ.
Step 1: Construct the matrix equation Let v3=
x
y
z
be the eigenvec-
tor corresponding to λ3= 4. The matrix equation for this eigenvector is:
Av3=λ3v3
a b c
d e f
g h i
x
y
z
= 4
x
y
z
Step 2: Write out the matrix equation Expanding the matrix equation
gives us the system of equations:
ax +by +cz = 4x
dx +ey +fz = 4y
gx +hy +iz = 4z
Step 3: Substitute the values from the given eigenvectors Now we
substitute the values from the given eigenvectors v1and v2into the system of
equations: For v1=
1
1
1
:
a+b+c= 4
d+e+f= 4
g+h+i= 4
For v2=
1
−1
0
:
a−b= 4
d−e=−4
g= 0
Step 4: Solve the system of equations From the equations derived in
Step 3, we can solve for the values of a,b,c,d,e,f,g,h, and i. Ultimately, we
find the eigenvector corresponding to λ3by solving the system of equations.
7
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 4, λ2= 2, and λ3=−1. If
the eigenvectors corresponding to λ1and λ2are v1=
1
1
1
and v2=
1
−1
0
,
respectively, find a basis for the eigenspace corresponding to λ3.
Solution
Step 1: Recall that the eigenspace corresponding to an eigenvalue λis the set
of all vectors vsuch that Av =λv.
Step 2: Let v=
x
y
z
be an eigenvector corresponding to λ3. Then we have:
Av =λ3v
A
x
y
z
=−1
x
y
z
Step 3: This can be rewritten as:
4 1 2
2 1 −1
1 2 1
x
y
z
=
−x
−y
−z
Step 4: By solving the system of equations represented by this matrix equa-
tion, we can find the eigenvectors corresponding to λ3. This will give us the
basis for the eigenspace corresponding to λ3.
Question 10
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 2,−1,3. If det(A) = 10, find the
inverse of A.
Solution
Step 1: The determinant of a matrix is the product of its eigenvalues. Given
that det(A) = 10 and the eigenvalues of Aare λ= 2,−1,3, we have:
2×(−1) ×3 = 10
−6 = 10
8
This is a contradiction, so there must be an error in the given information. So,
let’s check the information provided.
Step 2: Let’s verify the determinant of matrix Ausing the product of its
eigenvalues formula. The determinant of a matrix Acan be expressed as:
det(A) = λ1λ2λ3
det(A) = 2 ×(−1) ×3 = −6
Step 3: Find the adjugate of Ausing its cofactor matrix. The cofactor
matrix Ccan be found by finding the determinant of each of the 3x3 matrices
obtained by removing one row and one column from A. The adjugate A∗is then
obtained by taking the transpose of C.
A=
a b c
d e f
g h i
Calculating the cofactor matrix:
C=
−3 6 −3
5−10 5
−2 4 −2
Calculating the adjugate:
A∗=
−3 5 −2
6−10 4
−3 5 −2
Step 4: Find the inverse of Ausing the formula: A−1=1
det(A)A∗.
A−1=1
−6
−3 5 −2
6−10 4
−3 5 −2
A−1=
1
2−5
3
1
3
−15
3−2
3
1
2−5
3
1
3
Therefore, the inverse of matrix Ais
1
2−5
3
1
3
−15
3−2
3
1
2−5
3
1
3
.
Question 11
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1=−2, λ2= 1, and λ3= 4. Find a
basis for each eigenspace corresponding to these eigenvalues.
9
Solution
Step 1: To find the eigenspace corresponding to each eigenvalue, we need to
solve the system of equations (A−λI)v=0, where λis the eigenvalue and vis
the eigenvector.
Step 2: For λ=−2, we solve (A+ 2I)v=0:
−2 + 2 1 0
2 1 + 2 −1
−1−2 2 + 2
v1
v2
v3
=
0
0
0
Solving this system gives the eigenspace for λ=−2.
Step 3: For λ= 1, we solve (A−I)v=0:
−1 + 1 1 0
2 1 −1−1
−1−2 2 −1
v1
v2
v3
=
0
0
0
Solving this system gives the eigenspace for λ= 1.
Step 4: For λ= 4, we solve (A−4I)v=0:
−4 + 4 1 0
2 1 −4−1
−1−2 2 −4
v1
v2
v3
=
0
0
0
Solving this system gives the eigenspace for λ= 4.
Question 12
Question
Let Abe a 3 ×3 matrix given by
A=
2−1 0
−1 2 −1
0−1 2
Find the eigenvalues and eigenvectors of A.
Solution
Step 1: Find the eigenvalues of A.To find the eigenvalues of A, we need to
solve the characteristic equation |A−λI|= 0, where λrepresents the eigenvalue
and Iis the identity matrix. The characteristic equation is given by:
|A−λI|=
2−λ−1 0
−1 2 −λ−1
0−1 2 −λ
= 0
10
Expanding the determinant, we get
(2 −λ)((2 −λ)2−1) + 1(2 −λ)=0
Solving this equation will give us the eigenvalues.
Step 2: Solve for the eigenvalues. Expanding the equation (2 −λ)((2 −
λ)2−1) + (2 −λ) = 0, we get
(2 −λ)(λ2−4λ+ 3) + (2 −λ)=0
Simplifying further, we have
(λ−1)(λ2−4λ+ 3) = 0
Now, solving the quadratic equation λ2−4λ+ 3 = 0, we find the eigenvalues to
be λ= 1,3,and 1.
Step 3: Find the eigenvectors corresponding to each eigenvalue.
For λ= 1:To find the eigenvectors corresponding to λ= 1, we substitute
λ= 1 into the equation Av=λv, where vis the eigenvector. This gives the
equation
(A−I)v=0
Where Iis the identity matrix. By solving the system of equations (A−I)v=0,
we can find the eigenvector associated with λ= 1.
For λ= 3:Similarly, we solve the equation (A−3I)v=0to find the
eigenvector associated with λ= 3.
After solving these systems of equations, we can find the corresponding eigen-
vectors for each eigenvalue.
Question 13
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 2,3,and −1.Determine the
eigenvectors corresponding to each eigenvalue.
Solution
Given that Ais a 3×3 matrix with eigenvalues λ= 2,3,and −1,we can find the
eigenvectors corresponding to each eigenvalue by solving the system of equations
(A−λI)v=0,where vis the eigenvector and Iis the identity matrix.
For λ= 2:Step 1: Set λ= 2 in the equation (A−λI)v=0to get
(A−2I)v=0.Step 2: Write the system of equations as
−123
1 0 3
2 4 0
x
y
z
=
0
0
0
.
11
Step 3: Solve the system of equations to obtain the eigenvector corresponding
to λ= 2.
For λ= 3:Step 1: Set λ= 3 in the equation (A−λI)v=0to get
(A−3I)v=0.Step 2: Write the system of equations as
−1 2 3
1−1 3
2 4 −3
x
y
z
=
0
0
0
.
Step 3: Solve the system of equations to obtain the eigenvector corresponding
to λ= 3.
For λ=−1:Step 1: Set λ=−1 in the equation (A−λI)v=0to get
(A+I)v=0.Step 2: Write the system of equations as
−323
1−2 3
2 4 1
x
y
z
=
0
0
0
.
Step 3: Solve the system of equations to obtain the eigenvector corresponding
to λ=−1.
Question 14
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 5, λ2=−2, and λ3=−1. If
the eigenvectors corresponding to λ1and λ2are v1=
1
0
1
and v2=
0
1
−1
respectively, find a basis for the eigenspace corresponding to λ3.
Solution
Step 1: Recall that the eigenspace corresponding to an eigenvalue λis the set
of all vectors vsuch that (A−λI)v=0.
Step 2: We are given that Ais a 3 ×3 matrix with eigenvalues λ1= 5,
λ2=−2, and λ3=−1.
Step 3: To find the eigenvectors corresponding to λ3=−1, we solve the
system (A+I)v=0, where Iis the 3 ×3 identity matrix.
Step 4: Substituting λ=−1 and expanding the matrix equation, we get:
611
0−1−1
−1−1−2
x
y
z
=
0
0
0
Step 5: Solving the system of equations, we get:
6x+y+z= 0
−y−z= 0
−x−y−2z= 0
12
Step 6: From the second equation, we have y=−z. Substituting y=−z
into the first and third equation, we have:
6x−z= 0
−x−z= 0
Step 7: Solving the system of equations, we find x=z.
Step 8: Therefore, the eigenvectors corresponding to λ3=−1 are of the
form
z
−z
z
, where zis a real number.
Step 9: One basis for the eigenspace corresponding to λ3=−1 is the vector
v3=
1
−1
1
.
Question 15
Question
Let Abe a 3 ×3 matrix given by
A=
211
132
122
.
Find the eigenvalues and eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0, where λis the eigenvalue and Iis the identity matrix.
Step 2: The characteristic equation is given by
2−λ1 1
1 3 −λ2
1 2 2 −λ
= 0.
Step 3: Expanding the determinant, we get
(2 −λ)
3−λ2
2 2 −λ−
1 2
1 2 −λ
+
1 3 −λ
1 2
= 0.
Step 4: Simplifying further, we have
(2 −λ)[(3 −λ)(2 −λ)−4] −[(2 −λ)−2] + [2 −3 + λ] = 0.
Step 5: Solving this equation gives us the eigenvalues λ= 1,2,4.
13
Step 6: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute back into the equation (A−λI)v=0and solve for v.
Step 7: For λ= 1, solving (A−I)v=0gives us the eigenvector
v1=
1
1
−1
.
Step 8: For λ= 2, solving (A−2I)v=0gives us the eigenvector
v2=
−1
1
1
.
Step 9: For λ= 4, solving (A−4I)v=0gives us the eigenvector
v3=
1
1
1
.
Step 10: Therefore, the eigenvalues of Aare λ= 1,2,4 with eigenvectors
v1=
1
1
−1
,v2=
−1
1
1
, and v3=
1
1
1
respectively.
Question 16
Question
Let Abe a 3 ×3 matrix given by
A=
4 1 −1
−1 3 −1
2−2 3
Find the eigenvalues and eigenvectors of matrix A.
Solution
Step 1: Find the characteristic polynomial The characteristic polynomial
of matrix Ais given by det(A−λI) = 0, where λis the eigenvalue we are trying
to find.
A−λI =
4−λ1−1
−1 3 −λ−1
2−2 3 −λ
Next, we find the determinant of the matrix A−λI to form the characteristic
polynomial.
14
Step 2: Calculate the determinant
det(A−λI) = det
4−λ1−1
−1 3 −λ−1
2−2 3 −λ
Expanding along the first row, we get:
= (4 −λ)det 3−λ−1
−2 3 −λ−1·det −1−1
2 3 −λ−(−1) ·det −1 3 −λ
2−2
Solving these determinants will give us the characteristic polynomial.
Step 3: Solve for the characteristic polynomial Solving the determi-
nants, we get:
= (4−λ)((3−λ)(3−λ)−(−1)(−2))−((−1)((3−λ)(−1)−(−1)(2))+(−1)((−1)(−2)−(3−λ)(2))
= 0
This simplifies to the characteristic polynomial:
λ3−10λ2+ 29λ−24 = 0
Step 4: Find the eigenvalues Now, we need to find the roots of the
characteristic polynomial λ3−10λ2+ 29λ−24 = 0. By inspection, we can see
that λ= 4 is a root of the polynomial.
Step 5: Find the remaining eigenvalues Using polynomial long division
or synthetic division, we can divide the characteristic polynomial by (λ−4) to
find the remaining roots.
Step 6: Find the eigenvectors For each eigenvalue, we can find the
eigenvector by substituting back into the equation (A−λI)v=0and solving
for v.
For each eigenvalue, solve the system of equations to find the eigenvector.
Question 17
Question
Let Abe a 3×3 matrix with eigenvalues λ1= 2, λ2= 1+i, and λ3= 1−i. Find
the characteristic polynomial of Aand determine whether Ais diagonalizable.
Solution
Step 1: The characteristic polynomial of a matrix Ais given by
det(A−λI)=0,
where Iis the identity matrix. For a 3 ×3 matrix A, this will be a cubic
polynomial.
15
Step 2: For eigenvalue λ1= 2,
A−2I=
a−2b c
d e −2f
g h i −2
.
Taking the determinant and setting it equal to zero gives the coefficient of the
characteristic polynomial for λ1.
Step 3: For eigenvalue λ2= 1 + i,
A−(1 + i)I=
a−(1 + i)b c
d e −(1 + i)f
g h i −(1 + i)
.
Taking the determinant and setting it equal to zero gives the coefficient of the
characteristic polynomial for λ2.
Step 4: For eigenvalue λ3= 1 −i,
A−(1 −i)I=
a−(1 −i)b c
d e −(1 −i)f
g h i −(1 −i)
.
Taking the determinant and setting it equal to zero gives the coefficient of the
characteristic polynomial for λ3.
Step 5: The characteristic polynomial of Awill be a cubic polynomial with
roots λ1= 2, λ2= 1 + i, and λ3= 1 −i.
Step 6: To determine whether Ais diagonalizable, we need to check if the
algebraic multiplicity of each eigenvalue is equal to its geometric multiplicity. If
they are all equal, then Ais diagonalizable.
Question 18
Question
Let Abe a 3x3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−3. Find the
eigenvectors associated with each eigenvalue.
Solution
Step 1: Find the eigenvector associated with λ1= 2. To find the eigenvec-
tor, we need to solve the equation (A−λI)v= 0, where vis the eigenvector
corresponding to λ1.
Substitute λ1= 2 into the equation to get:
(A−2I)v= 0
16
Then, substitute Aas the given 3x3 matrix with eigenvalues and Ias the
identity matrix:
2−1 0
1 1 0
−1−1 5
v= 0
Perform row operations to row-reduce the augmented matrix to row-echelon
form and solve for v.
Step 2: Find the eigenvector associated with λ2= 1. Follow the same steps
as in Step 1, but this time use λ2= 1:
(A−I)v= 0
Substitute the values and solve for v.
Step 3: Find the eigenvector associated with λ3=−3. Again, follow the
same steps as in Step 1, but use λ3=−3:
(A+ 3I)v= 0
Substitute the values and solve for v.
Question 19
Question
Let Abe a 3 ×3 matrix given by
A=
3−1 0
1 2 −1
−1−1 3
Determine the eigenvalues and eigenvectors of the matrix A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation given
by det(A−λI) = 0, where Iis the identity matrix.
A−λI =
3−λ−1 0
1 2 −λ−1
−1−1 3 −λ
The characteristic equation becomes:
det(A−λI) = det
3−λ−1 0
1 2 −λ−1
−1−1 3 −λ
= 0
17
Step 2: Expanding the determinant along the first row, we get:
(3−λ) [(2 −λ)(3 −λ)−(−1)(−1)]+1 [−1(3 −λ)−(−1)(−1)]+(−1) [−1(−1) −1(2 −λ)] = 0
Solving this equation gives us the eigenvalues of the matrix.
Step 3: Once we have the eigenvalues, we can find the eigenvectors corre-
sponding to each eigenvalue by solving (A−λI)v=0, where vis the eigenvector
corresponding to the eigenvalue λ.
Step 4: Repeat step 3 for each eigenvalue to find the corresponding eigen-
vectors. The eigenvectors form a basis for each eigenspace of the matrix.
Question 20
Question
Let Abe a square matrix given by
A=
2−1 0
1 3 −1
0−1 2
.
Find the eigenvalues and corresponding 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 matrix A:
A−λI =
2−λ−1 0
1 3 −λ−1
0−1 2 −λ
.
Calculating the determinant and setting it to zero gives:
det(A−λI) = (2 −λ)((3 −λ)(2 −λ)−1) + (−1)(−1) = 0.
Solving this equation will give the eigenvalues.
Step 2: Find the eigenvalues Expanding the determinant equation, we
get:
(2 −λ)((3 −λ)(2 −λ)−1) + 1 = 0
(2 −λ)(6 −5λ+λ2−1) + 1 = 0
(2 −λ)(5λ−λ2+ 5) + 1 = 0
−λ3+ 8λ2−15λ+ 11 = 0
Hence, the characteristic equation is −λ3+ 8λ2−15λ+ 11 = 0.
18
Step 3: Find the eigenvectors Once we have found the eigenvalues,
we can substitute them back into (A−λI)x=0to find the corresponding
eigenvectors.
For each eigenvalue, solve (A−λI)x=0to find the eigenvector.
This process will give us the eigenvalues and corresponding eigenvectors for
matrix A.
Question 21
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 1,2,3. Find the eigenvalues of
the matrix B=A2+ 2A+I, where Iis the identity matrix.
Solution
Given that the eigenvalues of matrix Aare λ= 1,2,3, we want to find the
eigenvalues of the matrix B=A2+ 2A+I.
Step 1: First, we find the matrix B=A2+ 2A+I.
B=A2+ 2A+I
Step 2: Since we know the eigenvalues of matrix A, we can use the eigen-
value equation Av =λv to find the eigenvalues of matrix B. Let vbe an
eigenvector of Acorresponding to eigenvalue λ.
Step 3: We have:
Bv = (A2+ 2A+I)v
=A2v+ 2Av +Iv
=A(Av)+2Av +v
=A(λv) + 2(λv) + v
=λAv + 2λv +v
=λ2v+ 2λv +v
= (λ2+ 2λ+ 1)v
= (λ+ 1)2v
= (1 + 1)2v(when λ= 1)
= 4v(when λ= 1)
Step 4: Therefore, the eigenvalue of Bwhen λ= 1 is λ= 4.
Step 5: Similarly, we can repeat the above calculations for the other eigen-
values of Ato find the eigenvalues of B.
Bv = (A2+ 2A+I)v= (2 + 1)2v= 9v(when λ= 2)
Bv = (A2+ 2A+I)v= (3 + 1)2v= 16v(when λ= 3)
Step 6: Therefore, the eigenvalues of matrix B=A2+2A+Iare λ= 4,9,16.
19
Question 22
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If tr(A)
denotes the trace of A(sum of the diagonal entries), find the possible values of
tr(A).
Solution
Step 1: The trace of a matrix is the sum of its diagonal entries. Therefore, we
have:
tr(A) = λ1+λ2+λ3
Step 2: Substituting the given eigenvalues into the formula, we get:
tr(A)=1+2+3
tr(A) = 6
Therefore, the only possible value for tr(A) is 6 .
Question 23
Question
Let Abe a 3 ×3 real matrix with eigenvalues λ= 1,3,5 and corresponding
eigenvectors
v1=
1
0
1
,v2=
1
1
1
,v3=
0
1
1
.
Find A.
Solution
To find the matrix A, we can use the formula A=P DP −1, where Pis the
matrix whose columns are the eigenvectors of A, and Dis the diagonal matrix
with the corresponding eigenvalues on the diagonal.
Step 1: Construct matrix P
The matrix Pis formed by the eigenvectors, so
P=
110
011
111
.
Step 2: Construct matrix D
20
The matrix Dis a diagonal matrix with the eigenvalues on the diagonal, so
D=
100
030
005
.
Step 3: Calculate P−1
Next, we need to find the inverse of matrix Pin order to compute A=
P DP −1. The inverse of Pcan be found using standard techniques. We have
P−1=1
det(P)
A−B C
−D E −F
G−H I
,
where det(P) is the determinant of P.
Finally, combining P,D, and P−1, we can calculate A:
A=P DP −1.
Therefore, the matrix Acan be determined by the above calculations.
Question 24
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2=−2, and λ3= 4. If v1=
1
1
1
is an eigenvector corresponding to λ1, find an eigenvector v3corresponding
to λ3.
Solution
Step 1: Recall that eigenvectors v1, v2, . . . , vnassociated with distinct eigenval-
ues λ1, λ2, . . . , λnof a matrix Aare linearly independent. Therefore, we can
find the third eigenvector by solving the system of equations (A−λI)v= 0
where Ais the given matrix, λ=λ3= 4, and v=
x
y
z
.
Step 2: Subtract λI from Ato get:
A−4I=
−300
0 6 0
0 0 0
Step 3: Setting (A−4I)v= 0 gives the following system of equations:
−3x= 0
6y= 0
0z= 0
21
Step 4: The solution to the system of equations is x= 0 and y= 0. There-
fore, there are infinitely many solutions, which corresponds to the eigenvectors
of the matrix A. One such eigenvector is v3=
0
0
1
.
Thus, an eigenvector corresponding to λ3= 4 is v3=
0
0
1
.
Question 25
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and a repeated
eigenvalue λ3= 3. If the corresponding eigenvectors are v1=
1
0
1
,v2=
1
1
0
,
and v3=
1
1
a
, find the value of a.
Solution
Step 1: Since Ais a 3 ×3 matrix, it has 3 eigenvalues (which may repeat) and 3
corresponding eigenvectors. Therefore, the eigenvectors v1,v2, and v3provided
are indeed the eigenvectors corresponding to the eigenvalues λ1,λ2, and λ3.
Step 2: For each eigenvector, we have Avi=λivi, where i∈ {1,2,3}.
Step 3: Using the given information, we have: Av1= 1
1
0
1
=
1
0
1
Av2=
2
1
1
0
=
2
2
0
Av3= 3
1
1
a
=
3
3
3a
Step 4: Comparing Av3with λ3v3, we get:
3
3
3a
=
3
3
3a
This implies
3a= 3, so a= 1.
Therefore, the value of ais 1 .
22
Question 26
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Given
that corresponding eigenvectors are v1=
1
1
1
,v2=
1
2
3
, and v3=
0
1
−1
, find
the matrix A.
Solution
Step 1: Recall that if λis an eigenvalue of A, then there exists a non-zero vector
vsuch that Av=λv. In other words, vis an eigenvector corresponding to
eigenvalue λ.
Step 2: Using the given eigenvectors and eigenvalues, we can write Av1=
1·v1,Av2= 2 ·v2, and Av3= 3 ·v3.
Step 3: Writing the matrix Ain terms of the eigenvectors and eigenvalues:
A= [v1v2v3]
λ10 0
0λ20
0 0 λ3
[v1v2v3]−1
Step 4: First, find [v1v2v3]:
[v1v2v3] =
1 1 0
1 2 1
1 3 −1
Step 5: Next, find [v1v2v3]−1using a calculator or matrix methods.
Step 6: Substitute the matrices [v1v2v3] and [v1v2v3]−1into the formula
for Aand simplify to find the matrix A.
Question 27
Question
Let Abe a 3×3 matrix with eigenvalues λ= 1,3,5. Given that Ais symmetric,
find the eigenvectors corresponding to each eigenvalue.
Solution
Step 1: Recall that eigenvectors are the vectors that satisfy the equation (A−
λI)x=0, where λis the eigenvalue.
Step 2: For λ= 1, we have:
A−λI =
000
020
004
−
100
010
001
=
−100
0 1 0
0 0 3
23
Step 3: By solving (A−λI)x=0for λ= 1, we get the eigenvector v1=
1
0
0
.
Step 4: For λ= 3, we have:
A−λI =
−200
0 0 0
0 0 2
Step 5: By solving (A−λI)x=0for λ= 3, we get the eigenvector v2=
0
1
0
.
Step 6: For λ= 5, we have:
A−λI =
−400
0−2 0
0 0 0
Step 7: By solving (A−λI)x=0for λ= 5, we get the eigenvector v3=
0
0
1
.
Therefore, the eigenvectors corresponding to the eigenvalues λ= 1,3,5 are
v1=
1
0
0
,v2=
0
1
0
, and v3=
0
0
1
.
Question 28
Question
Let Abe a 3x3 matrix with eigenvalues λ1= 2, λ2= 3,and λ3= 4. Given
that the eigenvectors corresponding to λ1and λ2are v1=
1
0
1
and v2=
1
1
0
,
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: Recall that for an n×nmatrix A, the eigenvectors corresponding to
distinct eigenvalues are linearly independent.
Step 2: Since we know that v1and v2are eigenvectors of Acorresponding
to λ1= 2 and λ2= 3 respectively, we can form a basis for the eigenspace
corresponding to λ3= 4.
Step 3: Let v3=
x
y
z
be the eigenvector corresponding to λ3. We know
that Av3=λ3v3, so we have:
24
A
x
y
z
= 4
x
y
z
Step 4: This gives us the system of equations:
3x+y= 4x
x+ 3y= 4y
z= 4z
Step 5: Simplifying the first two equations:
(−x+y= 0
x−y= 0
Step 6: Solving this system of equations, we find x=y. This implies that
the eigenvector corresponding to λ3= 4 is v3=
1
1
z
.
Step 7: Therefore, the eigenvector corresponding to λ3= 4 is v3=
1
1
0
.
Question 29
Question
Let Abe a 3 ×3 matrix given by
A=
211
121
112
Determine the eigenvalues and corresponding eigenvectors of matrix A.
Solution
Step 1: To find the eigenvalues of A, we need to solve the characteristic equation
|A−λI|= 0, where λis the eigenvalue and Iis the identity matrix.
A−λI =
2−λ1 1
1 2 −λ1
1 1 2 −λ
Step 2: Setting the determinant of A−λI to zero, we have
2−λ1 1
1 2 −λ1
1 1 2 −λ
= 0
25
Step 3: Expanding the determinant using the cofactor expansion along the
first row, we get the characteristic polynomial
(2 −λ)[(2 −λ)(2 −λ)−1] −[(2 −λ)−1 + 1] + 1 = 0
Step 4: Simplifying, we have
(2 −λ)[(2 −λ)2−1] −2(2 −λ)=0
(2 −λ)(4 −4λ+λ2−1) −2(2 −λ) = 0
(2 −λ)(3 −4λ+λ2) = 2(2 −λ)
Step 5: This leads to the characteristic equation −λ3+ 6λ2−9λ= 0.
Step 6: Factoring out λ, we get λ(λ−3)(λ−3) = 0.
Step 7: Therefore, the eigenvalues are λ= 0 with a multiplicity of 1 and
λ= 3 with a multiplicity of 2.
Step 8: Next, we find the eigenvectors corresponding to each eigenvalue. For
λ= 0, we solve (A−0I)X= 0 to find the eigenvector.
Step 9: Row-reducing the augmented matrix (A−0I|0), we get
211|0
121|0
112|0
Step 10: Performing row operations, we find the eigenvector corresponding
to λ= 0 as X=
−1
1
1
.
Step 11: For λ= 3, we solve (A−3I)X= 0 to find the eigenvector.
Step 12: Row-reducing the augmented matrix (A−3I|0), we get
−1 1 1 |0
1−1 1 |0
1 1 −1|0
Step 13: Performing row operations, we find the eigenvector corresponding
to λ= 3 as X=
1
1
1
.
Therefore, the eigenvalues of matrix Aare λ= 0 with eigenvector
−1
1
1
and λ= 3 with eigenvector
1
1
1
.
26
Question 30
Question
Let Abe a 3 ×3 matrix with eigenvalues 1, 2, and 3. If B= 2A2−A+I, where
Iis the 3 ×3 identity matrix, find the eigenvalues of B.
Solution
Step 1: First, let’s find the eigenvalues of A. The characteristic equation of a
3×3 matrix Ais given by det(A−λI) = 0. This gives us:
det
a11 −λ a12 a13
a21 a22 −λ a23
a31 a32 a33 −λ
= 0
Substitute the given eigenvalues 1, 2, and 3 into the characteristic equation to
determine aij , as follows:
det
a11 −1a12 a13
a21 a22 −1a23
a31 a32 a33 −1
= 0
det
a11 −2a12 a13
a21 a22 −2a23
a31 a32 a33 −2
= 0
det
a11 −3a12 a13
a21 a22 −3a23
a31 a32 a33 −3
= 0
This will give us the values of the elements of A.
Step 2: Calculate the matrix Busing the formula B= 2A2−A+I.
Step 3: Let’s find the eigenvalues of matrix B. Since Bis defined as a
function of matrix A, we can relate the eigenvalues of Bto the eigenvalues of A
using the following property: If λis an eigenvalue of matrix A, then f(λ) is an
eigenvalue of f(A) for any polynomial function f(x).
Therefore, the eigenvalues of Bare {2(1)2−1+1,2(2)2−2+1,2(3)2−3+1}.
Calculating these values gives us the final answer for the eigenvalues of B.
Question 31
Question
Let Abe a 3 ×3 matrix given by
A=
110
121
012
.
Find the eigenvalues and eigenvectors of matrix A.
27
Solution
Step 1: To find the eigenvalues of matrix A, we solve the characteristic equation
|A−λI|= 0, where Iis the identity matrix. We have:
A−λI =
110
121
012
−λ
100
010
001
=
1−λ1 0
1 2 −λ1
0 1 2 −λ
.
The characteristic equation becomes |A−λI|= (1 −λ)((2 −λ)(2 −λ)−1) −
(1 ·(2 −λ)) = 0. This simplifies to λ3−5λ2+ 6λ−2 = 0.
Step 2: To find the roots of the characteristic equation, we can use numerical
methods or simply observe that λ= 1 is a root. So we divide the characteristic
equation by (λ−1) to get a quadratic equation which can be factored to find
the remaining roots. Dividing by (λ−1), we get:
(λ3−5λ2+ 6λ−2)∇ · (λ−1) = λ2−4λ+ 2 = 0.
This quadratic equation does not factor nicely, so we compute the remaining
eigenvalues as:
λ=4±√42−4∗1∗2
2=4±√8
2= 2 ±√2.
Step 3: Now that we have found the eigenvalues λ= 1,2±√2, we can find
the corresponding eigenvectors by solving the system (A−λI)v= 0 for each
eigenvalue.
For λ= 1, we have:
(A−I)v=
010
111
011
v= 0.
From this system of equations, we can find linearly independent eigenvectors
with components v1= 1, v2=−1, v3= 1.
For λ= 2 + √2, we have:
(A−(2 + √2)I)v=
−1−√2 1 0
1−√2 1
0 1 −√2
v= 0.
Solving this system of equations, we find an eigenvector v=
1
2(1 + √2)
1
.
For λ= 2 −√2, we have:
(A−(2 −√2)I)v=
−1 + √2 1 0
1√2 1
0 1 √2
v= 0.
28
Solving this system of equations, we find an eigenvector v=
1
2(1 −√2)
1
.
Therefore, the eigenvalues of matrix Aare λ= 1,2±√2, with corresponding
eigenvectors v
Question 32
Question
Let Abe a square matrix. If Ahas eigenvalues 2 and −1 with corresponding
eigenvectors v1=1
1and v2=1
−1, find a matrix Bsuch that B2=A.
Solution
Step 1: To find matrix B, we first express Ain terms of its eigenvectors and
eigenvalues. Since v1and v2are eigenvectors corresponding to eigenvalues 2
and −1 respectively, we can write:
A=P DP −1
where
P=v1v2=1 1
1−1
and
D=2 0
0−1
Step 2: Recall that for a diagonal matrix D,Dnis obtained by raising each
diagonal element to the power of n. Therefore, A2= (P DP −1)2=P D2P−1.
Step 3: Calculate D2:
D2=2 0
0−12
=4 0
0 1
Step 4: Now, compute A2:
A2=P D2P−1=1 1
1−14 0
0 1−1
2
1
2
1
2
1
2
Step 5: Simplify the expression to find A2and identify a suitable matrix B
such that B2=A.
A2=4 0
4 0−1
2
1
2
1
2
1
2=−2 2
−2 2
Therefore, a suitable matrix Bsuch that B2=Ais:
B=0 1
1 1
29
Question 33
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 4, λ2= 1,and λ3=−2. Find a
basis for each eigenspace corresponding to these eigenvalues.
Solution
Step 1: To find the eigenvectors corresponding to λ= 4, we need to solve the
system (A−4I)v=0, where Iis the identity matrix.
Step 2: Setting up the equation (A−4I)v=0, we have
0 2 −1
3 0 1
1 2 0
x
y
z
=
0
0
0
Step 3: Solving the system of equations, we get x= 0, y = 1, z = 3. There-
fore, an eigenvector corresponding to λ= 4 is v1=
0
1
3
.
Step 4: Following the same process, we can find eigenvectors corresponding
to λ= 1 and λ=−2. The eigenvectors corresponding to λ= 1 and λ=−2 are
v2=
1
1
1
and v3=
−1
1
−1
, respectively.
Therefore, bases for the eigenspaces are: - Eigenspace corresponding to
λ= 4: {
0
1
3
}- Eigenspace corresponding to λ= 1: {
1
1
1
}- Eigenspace corre-
sponding to λ=−2: {
−1
1
−1
}
Question 34
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−1. If the
corresponding eigenvectors are given by
v1=
1
0
−1
,v2=
1
1
0
,and v3=
1
2
−2
,
find the matrix A.
30
Solution
Step 1: Recall that a matrix can be constructed using its eigenvectors and
eigenvalues. Since Ais a 3 ×3 matrix, we can write it as
A=P DP −1,
where
P=v1v2v3and D=
λ10 0
0λ20
0 0 λ3
.
Step 2: Substituting the given values into Pand D, we have
P=
111
012
−1 0 −2
and D=
2 0 0
0 1 0
0 0 −1
.
Step 3: Calculate P−1. To find the inverse of P, we solve the equation
P−1P=I, where Iis the identity matrix. Therefore,
P−1=
−2 2 −1
1 0 −1
2−1 1
.
Step 4: Compute A=P DP −1. We have
A=
111
012
−1 0 −2
2 0 0
0 1 0
0 0 −1
−2 2 −1
1 0 −1
2−1 1
.
Step 5: Multiplying the matrices, we get
A=
111
012
−1 0 −2
2 0 0
0 1 0
0 0 −1
−2 2 −1
1 0 −1
2−1 1
=
2 1 −1
2 1 −2
6 2 −3
.
Therefore, the matrix Ais
2 1 −1
2 1 −2
6 2 −3
.
Question 35
Question
Let Abe a 3 ×3 matrix given by
A=
311
131
113
.
Find the eigenvalues of matrix Aand determine the corresponding eigenvec-
tors.
31
Solution
Step 1: Find the characteristic polynomial of matrix A.The character-
istic polynomial of a matrix Ais given by det(A−λI), where λis the eigenvalue
and Iis the identity matrix of the same size as A.
For our matrix A, we have:
A−λI =
3−λ1 1
1 3 −λ1
1 1 3 −λ
.
The determinant of A−λI is:
det(A−λI) = (3 −λ)((3 −λ)2−1) −(3 −λ−1) + 1 −(3 −λ)
det(A−λI) = (3 −λ)(9 −6λ+λ2−1) −2+1−3 + λ
det(A−λI) = (3 −λ)(λ2−6λ+ 8) −4 + λ
det(A−λI) = 3λ2−18λ+ 24 −λ3+ 6λ2−8λ−4 + λ
det(A−λI) = −λ3+ 9λ2−21λ+ 20.
So, the characteristic polynomial is −λ3+ 9λ2−21λ+ 20.
Step 2: Find the eigenvalues of matrix A.To find the eigenvalues, we
need to solve the equation −λ3+ 9λ2−21λ+ 20 = 0. By inspection, we can see
that one of the eigenvalues is λ= 1.
Let’s factor the polynomial to find the other eigenvalues:
−λ3+ 9λ2−21λ+ 20 = −(λ−1)(λ2−8λ+ 20).
Using the quadratic formula, we can find the remaining eigenvalues to be
λ= 4 and λ= 5.
Step 3: Find the eigenvectors corresponding to each eigenvalue.
Now, we will find the eigenvectors corresponding to each eigenvalue.
For λ= 1, we need to solve the system of equations (A−I)v= 0:
(A−I)v=
211
121
112
v=
0
0
0
.
Solving this system, we get v=
1
−1
0
.
Similarly, for λ= 4 and λ= 5, we can find the eigenvectors to be v=
1
1
−2
and v=
1
0
−1
, respectively.
Therefore, the eigenvalues of matrix Aare 1, 4, and 5, with corresponding
eigenvectors
1
−1
0
,
1
1
−2
, and
1
0
−1
.
32
