3668424
Descripción
Fichas por Marissa Miller, actualizado hace más de 1 año
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Creado por Marissa Miller
hace casi 9 años
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| Pregunta | Respuesta |
| Number of Solutions to a Linear System | 0 solutions (inconsistent) 1 solution (consistent) infinitely many solutions (consistent) |
| Dot Product | \( \vec{u}\cdot\vec{v} = u_{1}v_{1} + u_{2}v_{2} + \ldots + u_{n}v_{n}\) |
| Socks and Shoes Theorem | \((AB)^{-1} = A^{-1}B^{-1} \) |
| Number of free variables = | number of unknowns - number of nonzero rows in echelon matrix |
| Solution to \(A\vec{x} = \vec{b} \) if \(A\) is invertible | \(\vec{x} = A^{-1}\vec{b} \) |
| Gaussian Elimination | 1) augmented matrix 2) reduce to echelon form 3) backwards sub |
| Reducing to Echelon form options | > multiply row by constant > interchange two rows > add multiple of another row |
| Inverse of a 2 by 2 Matrix \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] | \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] |
| Vector Space | > closed under addition and scalar multiplication > must contain \(\vec{0} \) |
| Nullspace | > set of all solutions to \(A\vec{x} = \vec{0} \) > if a matrix is invertible, the only solution to \(A\vec{x} = \vec{0} \) is the trivial one |
| Linear Combination | \(k_{1}\vec{v_{1}} + k_{2}\vec{v_{2}}+ \ldots k_{n}\vec{v_{n}}\) |
| Span | set of all linear combinations of vectors |
| Linearly Independent | If \(k_{1}\vec{v_{1}} + k_{2}\vec{v_{2}}+ \ldots k_{n}\vec{v_{n}} = \vec{0}\) is only true for \(k_{i} = 0\) |
| Basis | collection of linearly independent vectors that span a space |
| Dimension | number of vectors in a space |
| To determine of vectors are linearly independent... | ...echelon the vectors; any free variables means they are dependent |
| Cross Product | \(\vec{u}\times\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{vmatrix} \) |
| Equation for plane through two vectors | > cross product > use cross product to get a normal vector \( (a b c) \) > plane: \(ax +by +cz = 0\) |
| Column Rank | max number of linearly independent columns |
| Row Rank | max number of linearly independent rows |
| Rank = | Row rank = column rank |
| Column Space | for an m by n matrix, CS is a subspace of \( R^{m}\) spanning the columns |
| dim(CS(A)) = | rank(A) |
| Basis for CS | > echelon form of \(A^{T}\) > find number of independent columns > pick those columns |
| To see if \(\vec{b}\) is in CS(A)... | ...see if there is a solution to \(A\vec{x} = \vec{b}\) |
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