Calculus: Analytic Geometry and Calculus, with Vectors

(lu) #1

96 Vectors and geometry in three dimensions


or the special case

(10)

(13 _ I
1 -') 1 2 2
0 1 -1C3) \-1
If we let

(11)
=

(all a12


(^33) X =
x
21 a22 a),
3/'
a Y =
31 32 Y
a33
y
then (9) shows that the whole system of equations
a11x1 +a12'X2 + a13x3 = yl
(12) a21x1 + a22x2 + a23x3 = y2
asix1 + a32x2 + a33x3 = y3
is equivalent to the single matrix equation !IX = Y. It is standard practice
to think of X and Y as being vectors having scalar components xl, x2, x3 and yl,
y2, y3 and to think of the matrix .4 as being an operator (or transformer) which
transforms (or carries or converts) the vector X into the vector Y. In this and
other contexts, matrices have very many important applications. One who
wishes an easy exercise can prove that the formula
(13)
Caa31ll aa.12 aa2aall: all a21
(^2) le a22
a31 a33 a33 a13 a23
a311^100
a32)= 0 1 0)
a33^001
is valid whenever the rows of the first matrix are the scalar components of three
orthonormal vectors. The last matrix in (13) is called the identity matrix I
(of order 3) because Al = IX = B whenever .I is a square matrix (of order 3).
If 4 is a square matrix of order n and det J 0 0, then there is a unique (that is,
one and only one) matrix B such that 14B = I, where I is the identity matrix
of order n. This matrix B is such that -4B = BA = I. It is called the inverse
of J and is denoted by -4-1. If AX = Y and det r4 0 0, then X = d-1Y. If
det 4 = 0, the matrix .4 cannot have an inverse because
(14) det (AB) = (det fl) (det B)
when 14 and B are square matrices of the same order and, moreover, det I = 1.
One who is really ambitious may attack the famous eigenvalue problem for
matrices. The problem is to start with a given square matrix fl and learn about
the scalars A (the eigenvalues) and the nonzero vectors X (the eigenvectors) for
which AX = AX.
(^14) This problem involves square matrices of order 2; analogous results hold
for square matrices of greater order. Suppose
%l = a11y1 + a12y2 yl = bunt + b1Px2
st = aalyl + a2sya, y2 = h2lx1 + b:2x2.
Let
X
= (X2 /, y = `Y2), s =z2), -
(aal
a2talt B
(^14) \b21 b22),

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