1. Find the rank of the matrix by Echelon form method. 2. Show that X1 = (1, -1, 1), X2 = (2, 1, 1), X3 = (1, 2, 2) are linearly dependent. 3. Test the consistency of the following... 1. Find the rank of the matrix by Echelon form method. 2. Show that X1 = (1, -1, 1), X2 = (2, 1, 1), X3 = (1, 2, 2) are linearly dependent. 3. Test the consistency of the following system of linear equations: 2x - y + 3z = 8, -x + 2y + z = 4, 3x + y - 4z = 0 and solve them. 4. Find the eigenvalues and corresponding eigenvectors for the matrix A = [[1, -6, -4], [0, 4, 2], [0, -6, -3]]. 5. Express (1,1,2) as a linear combination of vectors (0,2,1), (2,2,4) in V3(R). 6. Define the following: (i) Linear Transformation, (ii) Basis and dimension, (iii) Subspace, (iv) Kernel of transformation.
Understand the Problem
The questions presented involve solving mathematical problems related to linear algebra. This includes finding the rank of a matrix, determining linear dependency among vectors, testing the consistency of a system of linear equations, finding eigenvalues and eigenvectors, and expressing a vector as a linear combination of others. Additionally, definitions related to linear transformations and concepts from linear algebra are requested.
Answer
Answer for screen readers
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Rank of the matrix: 3
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The vectors ( X_1, X_2,
Steps to Solve
- Finding the Rank of the Matrix To find the rank of the matrix
$$ \begin{bmatrix} 2 & 4 & 3 & -2 \ -3 & -2 & -1 & 4 \ 6 & -1 & 7 & 2 \end{bmatrix} $$
we will use the Echelon form method.
- Transform the Matrix Perform row operations to transform the matrix into an upper triangular form.
- Start with the original matrix:
$$ \begin{bmatrix} 2 & 4 & 3 & -2 \ -3 & -2 & -1 & 4 \ 6 & -1 & 7 & 2 \end{bmatrix} $$
- Replace Row 2 with ( R_2 + \frac{3}{2}R_1 ) and Row 3 with ( R_3 - 3R_1 ):
$$ \begin{bmatrix} 2 & 4 & 3 & -2 \ 0 & 4 & 3.5 & 2 \ 0 & -13 & -2.5 & 8 \end{bmatrix} $$
- Next, simplify further by performing more row operations to eliminate below the first pivot.
- Continue to Row Echelon Form Perform ( R_3 = R_3 + \frac{13}{4}R_2 ):
$$ \begin{bmatrix} 2 & 4 & 3 & -2 \ 0 & 4 & 3.5 & 2 \ 0 & 0 & 7.125 & 9.5 \end{bmatrix} $$
From here, we can conclude that the non-zero rows give us the rank.
- Count the Non-zero Rows The matrix in row echelon form has 3 non-zero rows, so the rank is:
$$ \text{Rank}(A) = 3 $$
- Show Linear Dependency of Vectors To show that
$$ \mathbf{X_1} = (1, -1, 1), \quad \mathbf{X_2} = (2, 1, 1), \quad \mathbf{X_3} = (1, 2, 2) $$
are linearly dependent, we need to find scalars ( a, b, c ) such that:
$$ a\mathbf{X_1} + b\mathbf{X_2} + c\mathbf{X_3} = 0 $$
By setting this up in matrix form and solving for ( a, b, c ), we can determine dependencies.
- Testing Consistency of the System of Linear Equations Set up the augmented matrix for the equations:
$$ \begin{cases} 2x - y + 3z = 8 \ -x + 2y + z = 4 \ 3x + y - 4z = 0 \end{cases} $$
Use row reduction to find if a solution exists.
- Finding Eigenvalues and Eigenvectors For the matrix
$$ A = \begin{bmatrix} 1 & -6 & -4 \ 0 & 4 & 2 \ 0 & -6 & -3 \end{bmatrix} $$
Find the characteristic polynomial by calculating ( \det(A - \lambda I) ) and solve for ( \lambda ).
- Express as Linear Combination To express ( (1, 1, 2) ) as a linear combination of ( (0, 2, 1) ) and ( (2, 2, 4) ), you set:
$$ c_1(0, 2, 1) + c_2(2, 2, 4) = (1, 1, 2) $$
and solve for ( c_1, c_2 ).
- Define the Concepts
- Linear Transformation: A function between two vector spaces that preserves the operations of vector addition and scalar multiplication.
- Basis and Dimension: A basis of a vector space is a set of linearly independent vectors that spans the space; the dimension is the number of vectors in a basis.
- Subspace: A subset of a vector space that is itself a vector space under the same operations.
- Kernel of Transformation: The set of vectors that are mapped to the zero vector by a linear transformation.
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Rank of the matrix: 3
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The vectors ( X_1, X_2,
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