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Differential Equations and Linear Algebra (4th Edition)
- Consider the matrices below. X=[1201],Y=[1032],Z=[3412],W=[3241] Find scalars a,b, and c such that W=aX+bY+cZ. Show that there do not exist scalars a and b such that Z=aX+bY. Show that if aX+bY+cZ=0, then a=b=c=0.arrow_forwardplease solve it as soon as possiblearrow_forward(2) Find a basis of the following subspaces and find their dimensions. (a) {. {(:) a + 2c b — с : а, b, с€ R a + b+c (Ъ) а — 36 + с — 0 b — 2с — 0 a : 2b — с — 0arrow_forward
- Use the fact that matrices A and B are row-equivalent. 2 10 0 5 1 1 0 3 7 22-2 15 34 11 2 4 10 30-4 A = B = 123 0 1 -1 0 2 0 1 -2 00 00 00 0 (a) Find the rank and nullity of A. rank 3 nullity X (b) Find a basis for the nullspace of A.arrow_forwardConsider the following ordered basis for P₂ (R): B = {x² + 2x + 1, x + 1, 1} (a) The quadratic polynomial that has [[1], [3], [2]] as its coordinate vector relative to B is: (b) The coordinate vector (relative to B) of x² + x + 1 is:arrow_forward13 The set of solutions to the system of linear equations x1 – 2x2 + x3 = 0 2x1 - 3x2 + x3 = 0 is a subspace of R3. Find a basis for this subspace.arrow_forward
- Find a basis for the subspace of R4 determined by the equation 2x-3y+4z-w=0arrow_forward(a) Check the set of functions A = {x, 3x²-1, 5x3 3x} for orthogonality on [−1, 1]. - (b) Determine if S is a subspace of R4. If it is, find a basis and the dimension. X1 x2 - S = 2x2 3x40 CR X3 ᏆᎪarrow_forwardPlease find the following: Dimension of null(A): Basis for null(A): Dimension of col(A): Basis for col (A):arrow_forward
- The set B = {− (1+3x²), − (2+2x+6x²), ��� (6+4x+21x²)} is a basis for P2. Find the coordinates of p(x) = − (17 + 12x + 57x²) relative to this basis: -B CB(P(x)) = [p(x)] B =arrow_forwardThe set B = {4x2-1, a-3+ 12x2, 12 – 3x - 40x2} is a basis for P2. Find the coordinates of p(x) = 52– 13x - 184x2 relative to this basis: %3D [p(x)]B =arrow_forward[3 -3 A =6 9 -9 -5 a. A basis for the row space of A is { vectors, such as , . }. You should be able to explain and justify your answer. Enter a coordinate vector, such as , or a comma separated list of coordinate b. The dimension of the row space of A is because (select all correct answers -- there may be more than one correct answer): OA. Two of the three columns in rref(A) are free variable columns. OB. rref(A) has a pivot in every row. OC. rref(A) is the identity matrix. OD. The basis we found for the row space of A has two vectors. OE. Two of the three rows in rref(A) do not have a pivot. OF. Two of the three rows in rref(A) have pivots. c. The row space of A is a subspace of because choose d. The geometry of the row space of A is choosearrow_forward
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