Eigenvalues MCQ Quiz in বাংলা - Objective Question with Answer for Eigenvalues - বিনামূল্যে ডাউনলোড করুন [PDF]

The correct answer is "option 4".

CONCEPT:

According to Cayley Hamilton theorem:

If matrix A has Eigenvalue ‘x’ then An has Eigenvalue as xn.

DATA:

According to the given values,

a₁₁ = a₁₂ = a₂₁ = 1, a₂₂ = -1

The 2 x 2 matrix will be: $A=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]$

CALCULATION:

Considering characteristic equation is |A - λ I| = 0

where λ is eigen value and I is identity matrix

∴ Characteristics equation is given by

$\left|\begin{array}{rr}1-\lambda & 1\\ 1& -1-\lambda \end{array}\right|=0$

Solving the matrix,

−(1 − λ).(1 + λ) − 1 = 0

−(1 − λ2) – 1 = 0

− 2 + λ2 = 0

λ2 = 2

λ = ±(2)1/2

Therefore the Eigen values are:

((2)^1/2)¹⁹= 2⁹ x (2)^1/2 = 512√2

((-2)1/2)19= (−1)⁹  x 29 x (2)1/2 = −512√2

Hence the correct answer is "option 4".

Concept:

Eigen values are the roots of the characteristic equation.

It is calculated by |A - λI| = 0 where λ = root or eigen value.

Calculation:

Given:

$A=\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]$

The characteristic equation is, |A – λI| = 0

$\Rightarrow \left|\begin{array}{cc}1-\lambda & 1\\ -1& 1-\lambda \end{array}\right|=0$

∴ (1 - λ)² - (-1) = 0

∴ (1 - λ)2 = -1

$\therefore \left(1-\lambda \right)=\sqrt{-1}\Rightarrow \pm i$

∴ λ = 1 + i and 1 - i

Given that, $A=\left(\begin{array}{ccc}2& 1& 1\\ 2& 3& 4\\ -1& -1& -2\end{array}\right)$

Eigen values of A = λ₁, λ₂, λ₃ = 1, -1 and 3

Properties of Eigen values:

(1) If λ is an eigen value of a matrix A, then λⁿ will be an eigen value of a matrix Aⁿ.

(2) If λ is an eigen value of a matrix A, then kλ will be an eigen value of a matrix kA where k is a scalar

(3) Sum of eigen values is equal to trace of that matrix.

Eigen values of A³ = 1, -1, 27

Eigen values of 3A² = 3, 3, 27

Eigen values of (A³ – 3A²) = -2, -4, 0

Trace of (A³ – 3A²) = -2 - 4 + 0 = -6

Concept:

Properties of eigenvalues:

1.) Sum of eigenvalues = Sum of the diagonal elements of the matrix

2.) Product of eigenvalues = Determinant of the matrix

Calculation:

The eigenvalues are in the ratio of 3:1, then the eigenvalues are 3λ and λ.

x + 2 = 4λ 

$\lambda =\frac{x+2}{4}$............(i)

2x - 1 = 3λ² .........(ii)

Putting the value of (i) in (ii), we get:

$2x-1=\frac{3(x+2{)}^2}{16}$

3(x+2)² = 32x - 16

3x² - 20x + 28 = 0

x (3x - 14) -2 (3x - 14) = 0

(3x - 14)(x - 2) = 0

$x=2,\frac{14}{3}$

Sum of the eigen values of matrix is = Sum of diagonal values present in the matrix

∴ 1 + 0 + P = 3 + λ₂ + λ₃

⇒ P + 1 = 3 + λ₂ + λ₃

Given:

The square matrix of order 3 x 3 is $\left[\begin{array}{ccc}1& 2& 3\\ 0& -4& 2\\ 0& 0& 7\end{array}\right]$

Concept:

In order to determine a matrix's eigenvalues, take the following actions:
1 - Verify that the specified matrix A is a square matrix. Determine the same order's identity matrix I as well.
2 - Calculate the matrix $A-\lambda I$, where $\lambda$ is a scalar value.
3 - Locate the determinant of the matrix $A-\lambda I$ and set it equal to zero.
4 - Determine all feasible values of A, which are the necessary eigenvalues of matrix A, from the equation that has been created.

Solution:

$A=\left[\begin{array}{ccc}1& 2& 3\\ 0& -4& 2\\ 0& 0& 7\end{array}\right]$

$I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\lambda I=\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right]$

$A-\lambda I=\left[\begin{array}{ccc}1& 2& 3\\ 0& -4& 2\\ 0& 0& 7\end{array}\right]-\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right]$

$A-\lambda I=\left[\begin{array}{ccc}1-\lambda & 2& 3\\ 0& -4-\lambda & 2\\ 0& 0& 7-\lambda \end{array}\right]$

$|A-\lambda I|=(1-\lambda )[(-4-\lambda )(7-\lambda )-2\times 0]$

$|A-\lambda I|=(1-\lambda )[-28+4\lambda -7\lambda +{\lambda }^2]$

$|A-\lambda I|=(1-\lambda )[{\lambda }^2-7\lambda +4\lambda -28]$

$|A-\lambda I|=(1-\lambda )[\lambda (\lambda -7)+4(\lambda -7)]$

$|A-\lambda I|=(1-\lambda )(\lambda -7)(\lambda +4)$

$|A-\lambda I|=0$

$(1-\lambda )(\lambda -7)(\lambda +4)=0$

$\lambda =1,-4,7$

Hence, option 1 is correct.

For eigenvalues of the matrix:

$\left|\begin{array}{ccc}1-\lambda & 2& 0\\ 2& 1-\lambda & 0\\ 0& 0& -1-\lambda \end{array}\right|=0$

From the above, we can write:

(-1 - λ) [(1 - λ)² - 4] = 0

The above can be written as:

(-1 - λ) [(1 - λ) + 2][(1 - λ) - 2] = 0

(λ + 1)( -1 - λ)(3 - λ) = 0

λ = -1, -1, 3

∴ All the Eigen values are in the range:

-1 ≤ λ ≤ 3

-2 ≤ λ -1 ≤ 2

|λ - 1| ≤ 2

Concept:

If λ is an eigenvalue of a matrix A and k  is a scalar then:

1) λm is the eigenvalue of matrix Am (m belongs to N).

2) kλ  is an eigenvalue of matrix kA.

3) λ + k is an eigenvalue of the matrix A + kI.

4) λ - k  is an eigenvalue of matrix A - kI.

Calculation:

For a given matrix, if the eigenvalues are λ₁ and λ₂, then the eigenvalues of Aⁿ will be ${\lambda }_1^{n}and{\lambda }_2^n$.

1 and -2 will become 1 and -8;

Although the eigenvectors remain the same.

Eigen values and Eigen vector of a square matrix

"λ" is called eigen value and "x" is called eigen vector of a square matrix "A", if

Ax = λx

Characteristics of eigen values:

• Tr (A) = Summation of eigen values
• |A| = Product of eigen values
• If A = Upper triangular matrix or lower triangular matrix or diagonal matrix, then its eigen values will be diagonal elements.
• Eigen values of the hermitian matrix and real symmetric matrix are always real.
• Eigen values of skew-symmetric and skew hermitian matrix are either zero or purely imaginary.
• Eigen values of the orthogonal matrix and unitary matrix have unit modulus.

Concept:

If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Properties of Eigenvalues:

• The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A
• The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A
• If λ is an eigenvalue of a matrix A, then λn will be an eigenvalue of a matrix An.
• If λ is an eigenvalue of a matrix A, then kλ will be an eigenvalue of a matrix kA where k is a scalar.

Calculation:

$A=\left[\begin{array}{ccc}1& 0& 0\\ 2& 3& 1\\ 0& 2& 4\end{array}\right]$

Eigenvalues are given by

|A - λI| = 0

$\Rightarrow \left|\begin{array}{ccc}1-\lambda & 0& 0\\ 2& 3-\lambda & 1\\ 0& 2& 4-\lambda \end{array}\right|=0$

⇒ (1-λ){(3-λ)(4-λ) - 2} = 0

⇒ (1-λ){(12- 3λ - 4λ + λ²) - 2} = 0

⇒ 12 - 3λ - 4λ + λ2  - 2 - 12λ + 3λ² + 4λ² - λ³  + 2λ = 0 

⇒ λ³- 8 λ² + 17 λ - 10 = 0 ..... (A)

⇒ (λ -1) (λ² - 7 λ + 10) = 0

⇒ (λ -1) (λ 2 - 5 λ - 2λ + 10) = 0 

⇒ (λ -1) (λ (λ - 5) - 2 (λ -5)) = 0

⇒ (λ -1) (λ -2) (λ -5) = 0

⇒ λ = 1, 2, 5

or

putting values in the equation A from the options one by one

Eigen values are = 1, 2 and 5Shortcut TrickThe sum of the n eigenvalues of A is the same as the trace of A (i.e. the sum of the diagonal elements of A). 

The product of the n eigenvalues of A is the same as the determinant of A

So, go with the options

as in option 3, eigen values are 1,2 and 5

Sum of eigen values = Sum of diagonal elements

1+ 2 + 5 = 1 + 3 + 4

7 = 7

Det (A) = 1 × {(3× 4)- (1× 2 )}- 0 - 0 = 10

Product of eigen values = determinant of A

1 × 2 × 5 = 10

10 = 10

So, option 3 is correct answer.