Topic 20: Square matrices, planesTopic 20 NotesJeremy Orloff20 Topic 20: Square matrices, planesOld, compressed version of topic 20 notes.Square SystemsWe look at 3 × 3 (and 2 × 2) but this applies to all dimensions.a1a2a3b1b2b3c1c2c3xyz=d1d2d3↔ A ·−→X =−→dThe goal is to solve for−→X when A and−→d are known.First we study the case where det(A) 6= 0, i.e. where A−1exists.Working carefully: A ·−→X =−→d⇒ A−1(A ·−→X) = A−1·−→d⇒ (A−1A)−→X = A−1·−→d⇒ I ·−→X = A−1·−→d⇒−→X = A−1·−→dConclusion: If det(A) 6= 0 then there is exactly one solution:−→X = A−1·−→d .Example: Using the example from last time:A =2 1 01 1 21 2 2, det(A) = −4, A−1= −14−2 −2 20 4 −41 −3 11.−→X = A−1111= −14−2 −2 20 4 −41 −3 1111=1/201/4(check this answer)2. Solve A ·−→X =000⇒−→X = A−1000=000(This one is easy to solve –but note the analysis above guarantees this is the onlysolution.)Now we look at the case det(A) = 0 (i.e. where A−1doesn’t exist).1. A ·−→X =−→0 has infinitely many solutions. (Homogeneous case)2. If−→d 6=−→0 then A ·−→X =−→d has either 0 or many solutions depending on−→d .Example: (1 × 1 case)7x = 5 one solution 7x = 0 one solution0x = 5 no solutions 0x = 0 infinitely many solution120 TOPIC 20: SQUARE MATRICES, PLANES 2Cramer’s rule: read the supplementary notes.(continued)20 TOPIC 20: SQUARE MATRICES, PLANES 3Here’s the reasoning in the 2 × 2 case.Example:1 23 6xy=d1d2⇒x + 2y = d13x + 6y = d2Each of these equations is the equation of a line. Geometrically solving the systemof equations means finding the intersection of these two lines. In this case the linesare parallel, if d2= 3d1the parallel lines are, in fact, the same and there are lotsof solutions. Otherwise the lines are parallel and don’t intersect, so there are nosolutions.In general, det(A) = 0 means the two rows are multiples of each other, i.e. the twolines are parallel.Two possibilities:1. The lines are different ⇒ no solutions:2. The lines are the same ⇒ infinitely many solutions:Lines are differ-entLines are thesameIn the homogeneous case the lines are automatically the same (parallel and throughthe origin –see above picture).Summary (valid for any size square system)det(A) 6= 0 det(A) = 0Homogeneous 1 solution:−→X = 0 many solutionsInhomogeneous 1 solution:−→X = A−1−→d depends on−→dNOTE: For the 1 solution cases –no matter how it’s found the solution is unique.Lines in the plane (this is a warmup for planes in space)Slope-intercept form: Given the slope m and the y-intercept b theequation of a line can be written y = mx + b.Point-normal form: Given a point (x0, y0) on the line and a vectorha, bi normal to the line the equation of the line can be writtena(x − x0) + b(y − y0) = 0.xyha, bi7G•(x0, y0)Planes in point-normal formP = (x0, y0, z0) = point in plane−→N = ha, b, ci = normal to plane⇒ a(x −x0) + b(y −y0) + c(z −z0) = 0 is the equation of the plane.Abstractly, let X = (x, y, z) and we can write:−−→PX ·−→N = 0 ⇔ (−→X −−→P ) ·−→N = 0 ⇔−→X ·−→N =−→P ·−→Nyzx•−→N77P(continued)20 TOPIC 20: SQUARE MATRICES, PLANES 4Example: Find the plane through the point (1,4,9) with normal h2, 3, 4i.answer: Point-normal form of the plane is 2(x − 1) + 3(y − 4) + 4(z −9) = 0.Example: Find the plane with normal N =bk containing the point(0,0,3)Eq. of plane: h0, 0, 1i · hx, y, z − 3i = 0 ⇔ z = 3.yzx•Example: Find the plane with x, y and z intercepts a, b and c.answer: Fast way: the plane has the points (a, 0, 0), (0, b, 0) and(0, 0, c) ⇒ equation is x/a + y/b + z/c = 1Slow way (but, works in general):The 3 points give us 2 vectors in the plane, h−a, b, 0i and h−a, 0, ci.⇒ N = h−a, b, 0i × h−a, 0, ci = hbc, ac, abi.Point-normal form: bc(x − a) + ac(y − 0) + ab(z − 0) = 0⇔ bc x + ac y + ab z = abc ⇔ x/a + y/b + z/c = 1.yzxabcVarious methods of solving systemsExample: Solve1 2 34 5 67 8 9xyz=−→0det = 0 ⇒ many solutions (all vectors perpendicular to all 3 rows).Cross product: h1, 2, 3i × h4, 5, 6i = h−3, 6, −3i ⇒ solutions = c−36−3Example: For what c does the following have a non-zero solution?2x + + cz = 0x − y + 2z = 0x − 2y + 2z = 0In matrix form:2 0 c1 −1 21 −2 2−→X =−→0 ⇒ want2 0 c1 −1 21 −2 2= 0⇒ 4 − c = 0 ⇒ c = 4In this case,−→x0= h2, 0, 4i ×h1, −1, 2i = h4, 0, −2i is a solution (as is a ·−→x0for anya).(continued)20 TOPIC 20: SQUARE MATRICES, PLANES 5Example: For what d1and d2does the following have a solution?1 00 0xy=d1d2⇒ x = d1and 0 = d2⇒if d26= 0 no solutionsif d2= 0d1ya solution for any yDistances:1. Distance point to plane:Ingredients: i) A point P , ii) A plane with normal−→N and point Q.The distance from P to the plane is d = |−−→PQ|cos θ =−−→PQ ·−→N|N|.Example: Let P = (1, 3, 2). Find the distance from P to the plane x + 2y = 3.OO−→NPQθQ = any point on the plane, we take Q = (3, 0, 0).N = normal to plane =bi + 2bj = h1, 2, 0i.Distance =ProjN−−→PQ=−−→PQ ·N|N|= |−−→PQ|cos θ.−−→PQ = h2, −3, −2i, ⇒ d = |−−→PQ ·N|N|| =4√5.2. Distance point to line:Ingredients: i) A point P , ii) A line with direction vector v and point Q.The distance from P to the line is d = |QP|sin θ =−−→QP ×v|v|.An alternate formula using projection is−−→QR = Projv−−→QP =−−→QP ·v|v|v|v|, d = |−−→RP| = |−−→QP −−−→QR|.Example: Let P = (1, 3, 2), find the distance from the point P to the line through(1, 0, 0) and (1, 2, 0).CC//PRQθvQ = any point on the line, we take Q = (1, 0, 0).v = direction vector of line = h1, 2, 0i − h1, 0, 0i = 2bj.R = point on line closest to P (unknown).−−→QP = 3bj + 2bk.Method 1: cross product formula: d = |−−→QP ×v|v|| = |(3bj + 2bk) ×bj| = 2.Method 2: projection formula:−−→QR =−−→QP ·v|v|v|v|= 3bj ⇒ d = |−−→QP −−−→QR| = |2bk| =
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