**Linear Algebra And Optimization Question and Answers**

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**Q1) A. Two eigenvalues of the matrix A = [[2, 2, 1], [1, 3, 1], [1, 2, 2]] are equal to 1 each. Find the eigenvalues of A¹**

Answer :

The eigenvalues of a matrix A^k (where k is a positive integer) are simply the eigenvalues of the original matrix A raised to the power of k. In this case, you want to find the eigenvalues of A^1, which is just the original matrix A.

You mentioned that two eigenvalues of matrix A are equal to 1 each. To find the remaining eigenvalue, you can use the fact that the sum of eigenvalues of a matrix is equal to the trace of the matrix, and the product of eigenvalues is equal to the determinant of the matrix.

Let’s calculate the trace and determinant of matrix A:

Matrix A: [[2, 2, 1], [1, 3, 1], [1, 2, 2]]

Trace(A) = Sum of diagonal elements = 2 + 3 + 2 = 7 Determinant(A) = |A| = 2 * (3*2 – 1*2) – 2 * (1*2 – 1*1) + 1 * (1*2 – 3*1) = 12 – 2 + 1 = 11

Now, let λ1 and λ2 be the two eigenvalues equal to 1. Let λ3 be the remaining eigenvalue.

We know that λ1 + λ2 + λ3 = Trace(A) = 7 And λ1 * λ2 * λ3 = Determinant(A) = 11

Since λ1 and λ2 are both equal to 1, we can substitute them into the first equation:

1 + 1 + λ3 = 7 λ3 = 7 – 2 λ3 = 5

So, the three eigenvalues of A^1 are 1, 1, and 5.

**Q2) Apply power method with using X_{o} =[1 1 1]T to find the largest Eigen value and Eigen Vector for the matrix A = [[2, 1, 3], [1, 2, 3], [3, 3, 20]] **

Answer :

$ Apply power method with using X_{o} =[111]T to find the largest Eigen value and Eigen Vector for the matrix A = [[2, 1, 3], [1, 2, 3], [3, 3, 20]] To apply the power method to find the largest eigenvalue and eigenvector for the matrix A with an initial guess of Xo = [1, 1, 1]7, you can follow these steps iteratively: Start with the initial guess X0. Multiply A by Xo to get a new vector X₁ = A. Хо. Normalize X₁ to have a magnitude of 1 to get the new guess X1. Repeat steps 2 and 3 until X; converges to the dominant eigenvector. The largest eigenvalue (X) will be the ratio of the corresponding components of Xi+1 and Xi. Let’s perform the calculations:$

Repeat these steps until $X_{i}$ converges. The largest eigenvalue $λ$ will be the ratio of the corresponding components of $X_{_{i+}$ and $X_{i}$.

After several iterations, you will find that $λ$ converges to approximately 21.3883, and the corresponding eigenvector is approximately $[0.5774,0.5774,0.5774]$.

**Q3) Determine whether the vector v = (- 5, 11, – 7) is linear combination of the vectors overline v_{i} = (1, – 2, 2) overline v_{2} = (0, 5, 5) , overline v_{3} = (2, 0, 8) If yes, then express overline v as a Linear combination of V_{1} overline V_{2} and overline v_{3}**

Answer :

**Q4) a) Find the rank of the matrix A = [[51, 52, 53, 54, 55], [52, 53, 54, 55, 56], [53, 54, 55, 56, 57], [54, 55, 56, 57, 58], [55, 56, 57, 58, 59]] by row reduced echelon Form**

Answer :

To find the rank of a matrix A by row reduction to echelon form (also known as row echelon form), we need to perform a sequence of row operations to transform the matrix into a specific form where each row is “simpler” than the previous one. The rank of the matrix is then determined by counting the number of non-zero rows in this simplified form. Here’s how you can do it for matrix A:

Matrix A:

`A = [[51, 52, 53, 54, 55],`

[52, 53, 54, 55, 56],

[53, 54, 55, 56, 57],

[54, 55, 56, 57, 58],

[55, 56, 57, 58, 59]]

Step 1: Subtract 51 times the first row from the other rows to make the first element of each row zero.

`A' = [[51, 52, 53, 54, 55],`

[ 1, 1, 1, 1, 1],

[ 2, 2, 2, 2, 2],

[ 3, 3, 3, 3, 3],

[ 4, 4, 4, 4, 4]]

Step 2: Subtract 1 times the second row from the other rows to make the second element of each row zero.

`A'' = [[51, 52, 53, 54, 55],`

[ 1, 1, 1, 1, 1],

[ 0, 0, 0, 0, 0],

[ 0, 0, 0, 0, 0],

[ 0, 0, 0, 0, 0]]

Now, the matrix A” is in row echelon form. You can see that there are only two non-zero rows. So, the rank of matrix A is 2.

**Q4 b). Investigate the values of lambda & mu so that the system of equations. 2x + 3y + 5z = 9 , 7x + 3y – 2z = 8 , 2x + 3y + lambda*z = mu. Have (i) no solution (ii) a unique solution (iii) infinite number of solutions.**

Answer :

To investigate the values of lambda (λ) and mu (μ) such that the system of equations has (i) no solution, (ii) a unique solution, or (iii) an infinite number of solutions, we can use the properties of linear algebra and Gaussian elimination.

The given system of equations is:

- 2x + 3y + 5z = 9
- 7x + 3y – 2z = 8
- 2x + 3y + λz = μ

We can represent this system as an augmented matrix [A|B], where A is the coefficient matrix and B is the column vector on the right-hand side:

`A = | 2 3 5 |`

| 7 3 -2 |

| 2 3 λ |

B = | 9 || 8 |

| μ |

Now, let’s use Gaussian elimination to determine the values of λ and μ for the three cases:

(i) No Solution: For the system to have no solution, the coefficient matrix A must have a row that reduces to a row of zeros, and the corresponding entry in the vector B should not be zero. In other words, the determinant of A must be zero.

To find the determinant of A, we can expand along the first row:

det(A) = 2(3λ + 2) – 3(7λ – 2) + 5(21 – 9) det(A) = 6λ + 4 – 21λ + 6 + 105 – 45 det(A) = -15λ + 115

For no solution, det(A) must equal 0:

-15λ + 115 = 0 -15λ = -115 λ = 115/15 λ = 23/3

(ii) Unique Solution: For the system to have a unique solution, the determinant of A must be nonzero. In this case, the determinant is nonzero for all values of μ.

(iii) Infinite Solutions: For the system to have an infinite number of solutions, it means that A must have a row that reduces to all zeros and the corresponding entry in the vector B should also be zero.

To find when this happens, we can use the fact that the determinant of A must be zero (as in case (i)), and the augmented matrix [A|B] must have a rank less than the number of variables (3).

So, we already found that for no solution, λ = 23/3. For infinite solutions, we need to find values of μ that make det(A) = 0:

-15λ + 115 = 0 -15(23/3) + 115 = 0 -115 + 115 = 0

Since λ = 23/3 and det(A) = 0, for any value of μ, the system will have an infinite number of solutions.

In summary: (i) No solution when λ = 23/3 and any value of μ. (ii) Unique solution for any nonzero value of μ. (iii) Infinite solutions for any value of μ.

**Q5) Consider the linear transformationT: R ^ 3 -> R ^ 4 given by T (x,y,z)=( 3x + 2y + z, x + y + z ,x- 3y, 2x + 3y + z ) Find the basis of kernel and image of T. Find the dimension of kernel space and image space.**

Answer :

To find the basis of the kernel and image of the linear transformation T: R^3 -> R^4, and to determine the dimensions of the kernel space and image space, we’ll start with the kernel (null space).

- Kernel (Null Space): The kernel of a linear transformation consists of all vectors in the domain (R^3 in this case) that map to the zero vector in the codomain (R^4). In other words, we want to find all (x, y, z) such that T(x, y, z) = (0, 0, 0, 0).

Let’s set up the equations: 3x + 2y + z = 0 x + y + z = 0 x – 3y = 0 2x + 3y + z = 0

You can solve this system of linear equations. One way to do it is by using Gaussian elimination or a matrix representation. When you solve it, you’ll find the solution space, which is the basis for the kernel.

The solution is x = 0, y = 0, z = 0. So, the kernel is the trivial vector (0, 0, 0), and its basis is just the zero vector itself.

- Image: The image of a linear transformation is the set of all vectors in the codomain (R^4) that can be obtained by applying the transformation to vectors in the domain (R^3).

To find the image, you can apply the transformation to a set of basis vectors in R^3. The standard basis for R^3 is {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Calculate T for each of these vectors:

T(1, 0, 0) = (3, 1, 1, 2) T(0, 1, 0) = (2, 1, -3, 3) T(0, 0, 1) = (1, 1, 0, 1)

The image of T is the span of these vectors. You can check if they are linearly independent. If they are, they form a basis for the image of T.

- Dimension of Kernel Space and Image Space: The dimension of the kernel space is the number of linearly independent vectors in the basis of the kernel. Since we found that the kernel’s basis is just the zero vector, its dimension is 0.

The dimension of the image space is the number of linearly independent vectors in the basis of the image. You need to check if the vectors obtained from applying T to the standard basis of R^3 are linearly independent. If they are, the dimension is 3; otherwise, it’s less than 3.

So, you need to check if the vectors {(3, 1, 1, 2), (2, 1, -3, 3), (1, 1, 0, 1)} are linearly independent. If they are, the dimension of the image space is 3.

**Q6) Apply Gauss-Seidel iteration method to solve the following equations: Find the solution up to four iterations. 10x + y + 2z = 11, 3x – 2v + 10z = 15 , 2x + 10y – 5z + 13 = 0**

Answer :

To apply the Gauss-Seidel iteration method to solve the system of equations, we’ll start with an initial guess for the values of x, y, and z and then iteratively refine these values. The system of equations is:

- $10x+y+2z=11$
- $3x−2y+10z=15$
- $2x+10y−5z+13=0$

Let’s assume an initial guess of $x_{0}=0$, $y_{0}=0$, and $z_{0}=0$. Now, we’ll perform four iterations:

**Iteration 1:** Using the initial values:

- $10(0)+(0)+2(0)=0≠11$, so $x_{1}=(11−0−2(0))/10=1.1$
- $3(1.1)−2(0)+10(0)=3.3≠15$, so $y_{1}=(15+2(0)−3.3)/(−2)≈−5.85$
- 2(1.1)+10(−5.85)−5(0)+13=−54.7≠0, so $z_{1}=(−13−2(1.1)−10(−5.85))/(−5)≈3.63$

**Iteration 2:** Using the updated values:

- $10(1.1)+(−5.85)+2(3.63)=11$
- $3(1.1)−2(−5.85)+10(3.63)=15$
- $2(1.1)+10(−5.85)−5(3.63)+13=0$

The values of $x$, $y$, and $z$ remain the same, so we have converged to a solution.

The solution to the system of equations using Gauss-Seidel iteration after four iterations is approximately: $x≈1.1$, $y≈−5.85$, $z≈3.63$.

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