Mathematics
Linear Algebra
449 Questions
Linear algebra involves the study of matrices, vectors, and linear transformations. Common topics include finding the rank of a matrix, calculating eigenvalues and eigenvectors, and performing LU decomposition. These advanced mathematical concepts are frequently tested in engineering, statistics, and civil service examinations.
Matrix rank calculationLU decompositionEigenvalues and eigenvectorsMatrix invertibilityDeterminant properties
Linear Algebra Questions
What is the trace of the matrix (E = \begin{bmatrix} 5 & 2 \ 3 & -1 \end{bmatrix})?
A
Correct answer
Explanation
The trace of a matrix is the sum of its diagonal elements. In this case, the trace of (E) is (5 + (-1) = 4).
If (F) is a square matrix and (F^T = -F), then which of the following statements is true?
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All eigenvalues of \(F\) are zero.
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All eigenvalues of \(F\) are non-zero.
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The determinant of \(F\) is zero.
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The trace of \(F\) is zero.
C
Correct answer
Explanation
If (F^T = -F), then the determinant of (F) must be zero, as the determinant of a matrix is equal to the product of its eigenvalues.
Which of the following matrices is an orthogonal matrix?
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\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}
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\begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{bmatrix}
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\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}
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\begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}
A
Correct answer
Explanation
An orthogonal matrix is a square matrix whose inverse is equal to its transpose. In this case, only (\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}) is orthogonal.
What is the null space of the matrix (G = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix})?
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The set of all vectors \(x\) such that \(Gx = 0\).
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The set of all vectors \(x\) such that \(xG = 0\).
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The set of all vectors \(x\) such that \(G^Tx = 0\).
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The set of all vectors \(x\) such that \(x^TG = 0\).
A
Correct answer
Explanation
The null space of a matrix is the set of all vectors that, when multiplied by the matrix, result in the zero vector.
Which of the following matrices is a diagonalizable matrix?
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\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}
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\begin{bmatrix} 1 & 0 \ 0 & 2 \end{bmatrix}
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\begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix}
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\begin{bmatrix} 1 & 2 \ 3 & 6 \end{bmatrix}
B
Correct answer
Explanation
A diagonalizable matrix is a square matrix that can be expressed as a product of two matrices, one of which is a diagonal matrix. In this case, only (\begin{bmatrix} 1 & 0 \ 0 & 2 \end{bmatrix}) is diagonalizable.
What is the characteristic polynomial of the matrix (H = \begin{bmatrix} 2 & 3 \ -1 & 2 \end{bmatrix})?
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\(x^2 - 4x + 7\)
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\(x^2 - 3x + 2\)
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\(x^2 + 4x + 7\)
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\(x^2 + 3x + 2\)
A
Correct answer
Explanation
The characteristic polynomial of a matrix is a polynomial whose roots are the eigenvalues of the matrix. In this case, the characteristic polynomial of (H) is (x^2 - 4x + 7).
Which of the following matrices is a singular matrix?
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\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}
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\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}
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\begin{bmatrix} 1 & 2 \ 2 & 4 \end{bmatrix}
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\begin{bmatrix} 1 & 2 \ 3 & 6 \end{bmatrix}
B
Correct answer
Explanation
A singular matrix is a square matrix whose determinant is zero. In this case, only (\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}) is singular.
What is the cofactor of the element (a_{23}) of the matrix (J = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix})?
C
Correct answer
Explanation
The cofactor of an element (a_{ij}) of a matrix is the determinant of the submatrix obtained by deleting the (i)-th row and (j)-th column of the matrix, multiplied by ((-1)^{i+j}). In this case, the cofactor of (a_{23}) is ((-1)^{2+3}) times the determinant of the submatrix (\begin{bmatrix} 1 & 2 \ 7 & 8 \end{bmatrix}), which is (-9).
What is the general form of the Singular Value Decomposition (SVD) of a matrix?
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A = UΣV^T
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A = UΣV
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A = UΣ^T V^T
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A = UΣ^T V
A
Correct answer
Explanation
The SVD of a matrix A is given by A = UΣV^T, where U and V are orthogonal matrices and Σ is a diagonal matrix containing the singular values of A.
What are the singular values of a matrix?
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The eigenvalues of the matrix
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The square roots of the eigenvalues of the matrix
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The diagonal entries of the matrix
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The non-zero entries of the matrix
B
Correct answer
Explanation
The singular values of a matrix are the square roots of the eigenvalues of the matrix A^T A.
What is the relationship between the singular values and the rank of a matrix?
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The rank of a matrix is equal to the number of non-zero singular values
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The rank of a matrix is equal to the number of singular values
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The rank of a matrix is equal to the sum of the singular values
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The rank of a matrix is equal to the product of the singular values
A
Correct answer
Explanation
The rank of a matrix is equal to the number of linearly independent rows or columns of the matrix, which is also equal to the number of non-zero singular values.
What is the geometric interpretation of the singular value decomposition?
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It rotates and scales the matrix to make it diagonal
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It projects the matrix onto a subspace
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It decomposes the matrix into a sum of rank-one matrices
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It finds the eigenvectors and eigenvalues of the matrix
C
Correct answer
Explanation
The SVD decomposes the matrix into a sum of rank-one matrices, where each rank-one matrix is the product of a singular value and two orthogonal vectors.
What is the computational complexity of the singular value decomposition?
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O(n^3)
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O(n^2)
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O(n log n)
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O(n)
A
Correct answer
Explanation
The computational complexity of the SVD is typically O(n^3), where n is the size of the matrix.
What is the relationship between the SVD and the eigenvalue decomposition?
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The SVD is a generalization of the eigenvalue decomposition
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The eigenvalue decomposition is a generalization of the SVD
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The SVD and the eigenvalue decomposition are unrelated
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The SVD and the eigenvalue decomposition are equivalent
A
Correct answer
Explanation
The SVD is a generalization of the eigenvalue decomposition in the sense that the eigenvalue decomposition can be obtained from the SVD by setting some of the singular values to zero.