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K.Matthews.Elementary.Linear.Algebra.pdf |
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Size 0.9Mb Date Mar 20, 2005 |
a = ab−1 for With standard definitions such as a − b = a + (−b) and b b = 0, we have the following familiar rules: −(a + b) = (−a) + (−b), (ab)−1 = a−1 b−1 ; −(−a) = a, (a−1 )−1 = a; b a −(a − b) = b − a, ( )−1 = ; b a ac ad + bc + = ; bd bd ac ac = ; bd bd ac b a ab b= ; = , ac c b c −(ab) = (−a)b = a(−b); = −a a a = ; − b b −b 0a = 0;...
Verify that A3 = 5I3 , deduce that A is non–singular and find A−1 . Solution. After verifying that A3 = 5I3 , we notice that = 1 A 1 2 2 A I3 = A . A 5 5 Hence A is non–singular and A−1 = 1 A2 . 5...
is singular. For it can be verified that A has reduced row–echelon form 101 0 1 1 000...
Hence the coefficient determinant ∆ is zero. However = 1 = 1 1 0 0 11 −t 0 t 1−t 0 t 1t ∆= 1−t 2−t 1+t 1−t 2−t 1+t 2 3 Hence t = 1 or t = 2. If t = 1, the given system becomes x+y = 1 x+y = 1 2x + 2y = 3...
Calculating with complex numb ers
We can now do all the standard linear algebra calculations over the field of complex numbers – find the reduced row–echelon form of an matrix whose elements are complex numbers, solve systems of linear equations, find inverses and calculate determinants. For example, = 1 +i 2−i (1 + i)(8 − 2i) − 7(2 − i) 7 8 − 2i = (8 − 2i) + i(8 − 2i) − 14 + 7i = −4 + 13i = 0...
(5.3) z1 = 1−λ 1+λ It is easy to verify that z1 and z2 are distinct points on the line through a and b and that z0 = z1 +z2 ...
where k = 0 if x > 0; k = 1 if x < 0, y > 0; k = −1 if x < 0, y < 0. To determine Arg z graphically, it is simplest to draw the triangle formed by the points 0, x, z on the complex plane, mark in the positive acute angle α between the rays 0, x and 0, z and determine Arg z geometrically, using the fact that α = tan−1 (|y |/|x|), as in the following examples: EXAMPLE 5.6.2 Determine the principal argument of z for the followig complex numbers: z = 4 + 3i, −4 + 3i, −4 − 3i, 4 − 3i. Solution. Referring to Figure 5.6, we see that Arg z has the values α, π − α, −π + α, −α,
3 where α = tan−1 4 ...
DEFINITION 6.2.2 (Characteristic equation, p olynomial) The equation det (A − λIn ) = 0 is called the characteristic equation of A, while the polynomial det (A − λIn ) is called the characteristic polynomial of A...
This definition makes sense. For by the Cauchy–Schwarz inequality, −1 ≤ X ·Y ≤ 1. ||X || · ||Y ||
are unit vectors and every vector is a linear combination of i, j and k: a b = ai + bj + ck. c
Vectors X and Y are said to be perpendicular or orthogonal if X · Y = 0. Vectors of unit length are called unit vectors. The vectors 0 0 1 0 , j = 1 , k = 0 i= 1 0 0
Non–zero vectors X and Y are paral lel or proportional if the angle between X and Y equals 0 or π ; equivalently if X = tY for some real number t. Vectors X and Y are then said to have the same or opposite direction, according as t > 0 or t < 0. We are then led to study straight lines. If A and B are distinct points, it is easy to show that AP + P B = AB holds if and only if AP = t AB , where 0 ≤ t ≤ 1. A line is defined as a set consisting of all points P satisfying P = P0 + tX, t ∈ R or equivalently P0 P = tX,
E E E
for some fixed point P0 and fixed non–zero vector X called a direction vector for the line. Equivalently, in terms of coordinates, x = x0 + ta, y = y0 + tb, z = z0 + tc, where P0 = (x0 , y0 , z0 ) and not all of a, b, c are zero...
so the above equation does define an angle θ. In terms of components, if X = [a1 , b1 , c1 ]t and Y = [a2 , b2 , c2 ]t , then cos θ = a
2 1...
8.5. THE ANGLE BETWEEN TWO VECTORS z
Td A d g gd gd d g d P g g d d d g 22 B d g222 C dEy O d ¡ ¡...
Solution. The required plane has the form λ(x + y − 2z − 1) + µ(x + 3y − z − 4) = 0, where not both of λ and µ are zero. Substituting the coordinates of P 0 into this equation gives −2λ + µ(−4) = 0, So the required equation is −2µ(x + y − 2z − 1) + µ(x + 3y − z − 4) = 0, or −x + y + 3z − 2 = 0. Our final result is a formula for the distance from a point to a plane. λ = −2µ....
19. The points A = (1, 1, 5), B = (2, 2, 1), C = (1, −2, 2) and D = (−2, 1, 2) are the vertices of a tetrahedron. Find the equation of the line through A perpendicular to the face B C D and the distance of A from this face. Also find the shortest distance between the skew lines AD and B C . √ [Ans: P = (1 + t)(i + j + 5k); 2 3; 3.]...
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