Linear Algebra Test 2
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Is it possible for a 5×5 matrix to be invertible when its columns do not span set of real numbers ℝ5? Why or why not? : No
If {a b, c d} (that is a 2x2 matrix) and ad = bc then A is not invertible : True, it is undefined
Explain why the columns of an nxn matrix A span Rn when A is invertible : Since A is invertible, for each b in Rn the equation Ax = b has a unique solution. Since the equation Ax = b has a solution for all b in Rn, the columns of A span Rn
when a row is replaced by itself plus k times another row how does it affect the determinant : it does not affect the determinant
is det(5A) the same as 5det(A)? : no
A row replacement operation does not affect the determinant of a matrix. : True. If a multiple of one row of a matrix A is added to another to produce a matrix B, then det B equals det A.
The determinant of A is the product of the pivots in any echelon form U of A, multiplied by (−1)^r, where r is the number of row interchanges made during row reduction from A to U. : False. Reduction to an echelon form may also include scaling a row by a nonzero constant, which can change the value of the determinant.
If det A is zero, then two rows or two columns are the same, or a row or a column is zero. : False
det (A^-1) = (-1)detA : False, Det A^-1 = (detA)^-1
The dimension of the column space of A is rank A. Choose the correct answer below. : True
The dimension of Nul A is the number of variables in the equation Axequals=0. Choose the correct answer below. : False, it is the number of free variables
A matrix A is not invertible if and only if 0 is an eigenvalue of A. Choose the correct answer below. : True
A number c is an eigenvalue of A if and only if the equation (A−cI)x=0 has a nontrivial solution. Choose the correct answer below. : True
A steady-state vector for a stochastic matrix is actually an eigenvector. Choose the correct answer below. : True
The eigenvalues of a matrix are on its main diagonal. Choose the correct answer below. : False, it has to be a triangular matrix
. If set of real numbers R Superscript nℝnhas a basis of eigenvectors of A, then A is diagonalizable. Choose the correct answer below. : The statement is true because A is diagonalizable if and only if there are enough eigenvectors to form a basis of set of real numbers R Superscript nℝn.
A matrix A is diagonalizable if A has n eigenvectors. : The statement is false. A diagonalizable matrix must have n linearly independent eigenvectors.
