MIT OpenCourseWare http ocw mit edu 18 02 Multivariable Calculus Fall 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 i det P1 P P1 P2 P1 P3 0 or equivalently P1 P N 0 where N is the normal vector N P1 P2 P1 P3 I explained this geometrically and showed how we get the same equation both 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 u3 1 1 2 x3 Entries in the matrix product dot product between rows multiply a 3x3 matrix by a column vector 3x1 matrix More generally matrix multiplication AB 0 1 2 3 4 3 0 2 of A and columns of X here we 14 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 rst 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 even make sense if sizes not compatible 1 0 0 I3 3 0 1 0 Identity matrix identity transformation IX X 0 0 1 Example R 0 1 1 0 plane rotation by 90 degrees counterclockwise R j Rj R2 I Inverse matrix Inverse of a matrix A necessarily square is a matrix M A 1 such that AM M A 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 1 U 1 2 Cofactor method to nd A 1 e cient for small matrices for large matrices computer software uses other algorithms A 1 det 1A 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 3 1 2 3 4 2 3 1 2 1 1 2 cofactors ip signs according to checkerboard diagram get 3 3 4 2 3 transpose exchange read horizontally write vertically get the adjoint rows columns 3 3 3 1 4 matrix M T adj A 1 2 1 2 3 3 3 1 1 4 4 divide by det A here 3 get A 1 1 3 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 de nes a plane 1 plane through origin with normal vector N 1 5 10 P x y z is in the plane N OP 0 1 5 10 x y z x 5y 10z 0 Coe cients of the equation are the 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 P0 P 0 x 2 5 y 1 10 z 1 0 or x 5y 10z 3 Again coe cients of equation components of normal vector Note the equation multiplied by a constant still de nes the same plane So to nd the equation of a plane we really need to look for the normal vector N we can e g nd 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 A perpendicular vector would be proportional to the coe cients 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 1 B 3 Exception if the 3rd plane is parallel to the line of intersection of the rst two What can happen asked on ashcards for possibilities If the line P1 P2 is contained in P3 there are in nitely 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 special cases correspond to systems with det A 0 Then we can t invert A to solve the system recall A 1 det 1A 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 in nitely many solutions forming a line or a plane Note det A 0 means det N 1 N 2 N 3 0 where N i 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 N i For example we can get a vector along the line of intersection by taking a cross product N 1 N 2 General systems AX B compared to AX 0 all the planes are shifted to parallel positions from their initial ones If det A 0 then unique solution is X A 1 B If det A 0 either there are in nitely many solutions or there are no solutions We don t have tools to decide whether it s in nitely many or none although elimination will let us nd out 18 02 Lecture 5 Fri Sept 14 2007 Lines We ve seen a line as intersection of 2 planes Other representation parametric equation as trajectory of a moving point E g line through Q0 1 2 2 Q1 1 3 1 moving point Q t starts at Q0 at t 0 moves at constant speed along line reaches Q1 at …
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