MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable CalculusFall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � 18.02 Lecture 3. – Tue, Sept 11, 2007 Remark: A × B = −B × A, A × A = 0. Application of cross product: equation of plane through P1, P2, P3: P = (x, y, z) is in the plane iff det(−−P1P , P−−−1P→2 , −−−→ P1P N = 0, where N is the normal vectorP1P3 ) = 0, or equivalently, −−→· N = −−−P1P→2 ×−−−→ I explained this geometrically, and showed how we get the same equation bothP1P3 . ways. Matrices. Often quantities are related by linear transformations; e.g. changing coordinate systems, from P = (x1, x2, x3) to something more adapted to the problem, with new coordinates (u1, u2, u3). For example ⎧ ⎨ u1 = 2x1 + 3x2 + 3x3 u2 = 2x1 + 4x2 + 5x3⎩ u3 = x1 + x2 + 2x3⎡ ⎤⎡ ⎤ ⎡ ⎤ 2 3 3 x1 u1 Rewrite using matrix product: ⎣ 2 4 5 ⎦⎣ x2 ⎦ = ⎣ u2 ⎦, i.e. AX = U . 1 1 2 x3 u3 Entries in the matrix product = dot product between rows of A and columns of X. (here we multiply a 3x3 matrix by a column vector = 3x1 matrix). More generally, m atrix multiplication AB: ⎡ ⎤ ⎡ ⎤ . 0 ⎡ ⎤ 1 2 3 4 ⎢ . 3 ⎥ . 14 ⎣ . . . . ⎦⎢⎣ . 0 ⎥⎦ = ⎣ . . ⎦ . . . . . . . 2 (Also explained one can set up A to the left, B to the top, then each entry of AB = dot product between row to its left and column above it). Note: for this to make sense, width of A must equal height of B. What AB means: BX = apply transformation B to vector X, so (AB)X = A(BX) = apply first B then A. (so matrix multiplication is like composing transformations, but from right to left!) (Remark: matrix product is not commutative, AB is in general not the same as BA – one of the two need not eve n make sense if sizes not compatible). ⎡ ⎤ 1 0 0 Identity matrix: identity transformation IX = X. I3×3 = ⎣ 0 1 0 ⎦0 0 1 Example: R =0 −1 = plane rotation by 90 degrees counterclockwise.1 0 Rˆı =ˆj, Rˆj = −ˆı , R2 = −I. Inverse matrix. Inverse of a matrix A (necessarily square) is a matrix M = A−1 such that AM = MA = In. A−1 corresponds to the reciprocal linear relation. E.g., solution to linear system AX = U: can solve for X as function of U by X = A−1U. 1� 2 Cofactor method to find A−1 (efficient for small matrices; for large matrices com puter software uses other algorithms): A−1 = det(1 A) adj(A) (adj(A) = “adjoint matrix”). ⎡ ⎤ 2 3 3 Illustration on example: starting from A = ⎣ 2 4 5 ⎦ 1 1 2 1) matrix of minors (= determinants formed by deleting one row and one column from A):⎡ ⎤ 3 � �−1 −2 4 5 � ⎣ 3 1 −1 ⎦ (e.g. top-left is �� 1 2 �� = 3). 3 4 2 ⎡ ⎤ + − + 3 +1 −2 2) cofactors = flip signs according to checkerboard diagram − + − : get ⎣ −3 1 +1 ⎦ + − + 3 −4 2 3) transpose = exchange rows / columns (read horizontally, write vertically) get the adjoint⎡ ⎤ 3 −3 3 matrix MT = adj(A) = ⎣ 1 1 −4 ⎦ −2 1 2 ⎡ ⎤ 13 −3 3 4) divide by det(A) (here = 3): get A−1 =3 ⎣ 1 1 −4 ⎦. −2 1 2 18.02 Lecture 4. – Thu, Sept 13, 2007 Handouts: PS1 solutions; PS2. Equations of planes. Recall an equation of the form ax + by + cz = d defines a plane. 1) plane through origin with normal vector N = �1, 5, 10�: P = (x, y, z) is in the plane ⇔ N −−→= 0 ⇔ �1, 5, 10� · �x, y, z� = x + 5y + 10z = 0. Coefficients of the equation are theOP· components of the normal vector. 2) plane through P0 = (2, 1, −1) with same normal vector N = �1, 5, 10�: parallel to the previous one! P is in the plane ⇔ N · P0P = 0 ⇔ (x − 2) + 5(y − 1) + 10(z + 1) = 0, or x + 5y + 10z = −3.−−→Again coefficients of equation = components of normal vector. (Note: the equation multiplied by a constant still defines the same plane). So, to find the equation of a plane, we really need to look for the normal vector N ; we can e.g. find it by cross-product of 2 vectors that are in the plane. Flashcard question: the vector v = �1, 2, −1� and the plane x + y + 3z = 5 are 1) parallel, 2) perpendicular, 3) neither? perpendicular vector would be proportional to the coefficients, i.e. to �1, 1, 3�; let’s test if it’s in the plane: equivalent to being ⊥ N . We have v N = 1 + 2 − 3 = 0 so v is parallel to the plane.)· Interpretation of 3x3 systems. A 3x3 system asks for the intersection of 3 planes. Two planes intersect in a line, and usually the third plane intersects it in a single point (picture shown). The unique solution to AX = B is given by X = A−1B. (A3 Exception: if the 3rd plane is parallel to the line of intersection of the first two? What can happen? (asked on flashcards for possibilities). If the line P1 ∩ P2 is contained in P3 there are infinitely many solutions (the line); if it is parallel to P3 there are no solutions. (could also get a plane of solutions if all three equations are the same) These sp e cial cases correspond to systems with det(A) = 0. Then we can’t invert A to solve the system: recall A−1 = det(1 A) adj(A). Theorem: A is invertible ⇔ det A = 0.�Homogeneous systems: AX = 0. Then all 3 planes pass through the origin, so there is the obvious (”trivial”) solution X = 0. If det A =� 0 then this solution is unique: X = A−1 0 = 0. Otherwise, if det A = 0 there are infinitely many solutions (forming a line or a plane). Note: det A = 0 means det(N1, N 2, N3) = 0, where Ni are the normals to the planes Pi. This means the parallelepiped formed by the N i has no area, i.e. they are coplanar (showed picture of 3 planes intersecting in a line, and their coplanar normals). The line of solutions is then perpendicular to the plane containing Ni. For example we can get a vector along the line of interse ction by taking a cross-product N1 × N 2. General systems: AX = B: compared to …
View Full Document
Unlocking...