🧮 Unit 1 – Part A (2-Mark Q&A)

1. If 2, -1, -3 are the eigen values of a matrix A, then the eigen values of the matrix A²⁴ is

Step 1: Recall property of eigenvalues

If λ is an eigenvalue of A, then for any positive integer k, λ^k is an eigenvalue of A^k.

Step 2: Apply for k=24

The eigenvalues of A are λ₁ = 2, λ₂ = -1, λ₃ = -3.

So the eigenvalues of A²⁴ are 2²⁴, (-1)²⁴, and (-3)²⁴.

Step 3: Simplify powers

• 2²⁴
• (-1)²⁴ = 1 (since 24 is even)
• (-3)²⁴ = 3²⁴ (positive since 24 is even)

Final Answer:
The eigenvalues of A²⁴ are 2²⁴, 1, 3²⁴.

2. If 1, -1, 2 are the eigen values of a matrix A, then find the eigen values of the matrix A^-1.

Step 1: Recall property

If λ is an eigenvalue of A, then 1/λ is an eigenvalue of A^-1 (provided λ ≠ 0).

Step 2: Given eigenvalues of A
1, -1, 2

Step 3: Take reciprocals

• For λ = 1 ⇒ 1/1 = 1
• For λ = -1 ⇒ 1/(-1) = -1
• For λ = 2 ⇒ 1/2

Final Answer:
Eigenvalues of A^-1 are 1, -1, 1/2.

3. Write down the matrix for the following quadratic form 2x₁² - 2x₂² + 4x₃² + 2x₁x₂ - 6x₁x₃ + 6x₂x₃.

Step 1: General quadratic form

A quadratic form can be written as Q(x) = x^TAx, where A is a symmetric matrix.

Step 2: Match coefficients

• Coefficient of x₁² → a₁₁ = 2
• Coefficient of x₂² → a₂₂ = -2
• Coefficient of x₃² → a₃₃ = 4
• Cross terms (General form 2a_ijx_ix_j):
• 2x₁x₂ → 2a₁₂ = 2 ⇒ a₁₂ = 1
• -6x₁x₃ → 2a₁₃ = -6 ⇒ a₁₃ = -3
• 6x₂x₃ → 2a₂₃ = 6 ⇒ a₂₃ = 3

Step 3: Write the matrix

Final Answer:
The matrix of the quadratic form is:

4. If the Matrix A =

3	1	4
0	2	6
0	0	5

then eigen value of A^-1 is 3, 2, 5. (True/False)

Step 1: Recall properties

• Eigenvalues of a triangular matrix (upper or lower) are just its diagonal entries.
• If λ is an eigenvalue of A, then 1/λ is an eigenvalue of A^-1.

Step 2: Eigenvalues of A

The matrix A is upper triangular. Therefore, its eigenvalues are the diagonal entries: 3, 2, 5.

Step 3: Eigenvalues of A^-1

Take the reciprocals of A's eigenvalues: 1/3, 1/2, 1/5.

Step 4: Compare with statement

The statement says the eigenvalues of A^-1 are 3, 2, 5. This is incorrect.

Final Answer: The statement is False.

5. Is the Quadratic form x² - y² + 4z² + 4xy + 2yz + 6xz indefinite?

Step 1: Write the matrix of the quadratic form

• x² → a₁₁ = 1
• -y² → a₂₂ = -1
• 4z² → a₃₃ = 4
• 4xy → 2a₁₂ = 4 ⇒ a₁₂ = 2
• 2yz → 2a₂₃ = 2 ⇒ a₂₃ = 1
• 6xz → 2a₁₃ = 6 ⇒ a₁₃ = 3

The symmetric matrix is:

Step 2: Check definiteness (using Sylvester's criterion)

A quadratic form is indefinite if it has both positive and negative eigenvalues. We check the leading principal minors:

1. D₁ = |1| = 1 (Positive)
2. D₂ =

   1	2
   2	-1

   = (1)(-1) - (2)(2) = -1 - 4 = -5 (Negative)

Step 3: Conclusion

Since the principal minors are not all positive (D₂ < 0), the form is not positive definite. Since D₁ > 0 and D₂ < 0, the eigenvalues must have mixed signs.

Final Answer: Yes, the quadratic form is indefinite.

6. Find the sum of eigen values and product of eigen values of A =

2	-3
4	-2

Step 1: Recall formulas

• Sum of eigenvalues = trace(A) = Sum of main diagonal elements.
• Product of eigenvalues = det(A) = Determinant of the matrix.

Step 2: Apply formulas

• Sum (Trace) = 2 + (-2) = 0
• Product (Determinant) = (2)(-2) - (-3)(4) = -4 + 12 = 8

Final Answer:
Sum of eigenvalues = 0
Product of eigenvalues = 8

7. Two of the eigen values of A =

3	-1	1
-1	5	-1
1	-1	3

are 3 and 6. Find the eigen value of A^-1.

Step 1: Find the third eigenvalue of A

Property: Sum of eigenvalues = trace(A) (sum of diagonal elements).

Trace(A) = 3 + 5 + 3 = 11.

Let the eigenvalues be λ₁, λ₂, λ₃. We are given λ₁ = 3 and λ₂ = 6.

3 + 6 + λ₃ = 11 ⇒ 9 + λ₃ = 11 ⇒ λ₃ = 2.

The eigenvalues of A are 3, 6, 2.

Step 2: Find eigenvalues of A^-1

Property: If λ is an eigenvalue of A, then 1/λ is an eigenvalue of A^-1.

Take the reciprocals: 1/3, 1/6, 1/2.

Final Answer:
The eigenvalues of A^-1 are 1/3, 1/6, 1/2.

8. For A =

1	0
0	5

, write A² in terms of A and I, using Cayley Hamilton theorem.

Step 1: Find the characteristic equation of A

The characteristic equation is det(A - λI) = 0.

1-λ	0
0	5-λ

= (1-λ)(5-λ) = 0

This simplifies to λ² - 6λ + 5 = 0.

Step 2: Apply Cayley-Hamilton theorem

The theorem states that a matrix satisfies its own characteristic equation.

So, we replace λ with A and the constant 5 with 5I:
A² - 6A + 5I = 0

Step 3: Solve for A²

A² = 6A - 5I

Final Answer: A² = 6A - 5I

9. State the nature of Quadratic form 2xy + 2yz + 2zx.

Step 1: Write in matrix form

The diagonal entries (x², y², z²) are 0.

For cross terms (like 2a_ijx_ix_j), we divide the coefficient by 2.

• 2xy ⇒ a₁₂ = 1
• 2yz ⇒ a₂₃ = 1
• 2zx ⇒ a₁₃ = 1

The matrix is:

Step 2: Check nature

The nature is determined by the eigenvalues of A.

The eigenvalues for this specific matrix are λ₁ = 2, λ₂ = -1, and λ₃ = -1.

Since there are some positive eigenvalues (2) and some negative eigenvalues (-1), the form is indefinite.

Final Answer: The quadratic form is indefinite.

10. State Cayley Hamilton theorem

Cayley-Hamilton Theorem:
Every square matrix satisfies its own characteristic equation.

Explanation:

If A is a square matrix, and its characteristic equation (found from det(A - λI) = 0) is p(λ) = 0.

Then, according to the theorem, if you replace λ with the matrix A and any constant c with cI (where I is the identity matrix), the equation still holds true, resulting in the zero matrix.

p(A) = 0

Example:

If A =

2	0
0	3

, the characteristic equation is (2-λ)(3-λ) = λ² - 5λ + 6 = 0.

By the theorem, A² - 5A + 6I = 0.