Calculus: Analytic Geometry and Calculus, with Vectors

(lu) #1
2.5 Determinants and applications 95

above and k is a scalar, we have


(3) kP= -3k^2 k -k 3k


k -k k/ P+Q=\ (^10) 0)
When we multiply a matrix by a scalar k, we multiply each element by k. When
we add two matrices, we add them elementwise. These rules are very different
from those applicable to determinants. The really crucial step in the develop-
ment of a useful algebra of matrices is the determination, in terms of two suitable
matrices 11 and B, of a third matrix C which we shall call the product '2B of -4
and B. Let J have n rows and p columns and let B have q rows and n columns,
so that the number of rows of .4 is the same as the number of columns of B.
The product AB is then a matrix C having p rows and q columns, the element
c,k in the 5th row and kth column of C being determined by the formula
(4) c3k = a,lblk + a,2bsk + ... +
a,,,are the elements of the 5th row of .4 and blk, bay, , b. k
are the elements of the kth column of B. To demonstrate that applications of
this ritual are not fearsome, we let J, B, C be the matrices P, Q, R defined above
and see how the result
(5)
C i -1 1/\0 1 -1/-\ 2
4 -6
-6 9)
3 -4)
is obtained.To obtain the element 7 in the first row and the first column of the
last matrix, we run one finger across the first row of the first matrix, and at the
same time, run another finger down the first column of the second matrix and
(with regret that we do not have three hands) write the sum
(6) 2.3 + 1.1 + 3.0 = 7
of the products of the elements that our fingers encounter. To obtain the ele-
ment 4 in the first row and second column of the last matrix, we apply the fingers
to the first row of the first matrix and the second column of the second matrix and
obtain
(7)
To obtain the element 3 in the third row and second column of the third matrix,
we follow the third row of the first matrix and the second column of the second
matrix to obtain
(8) 1.1 + (-1)(-1) + 1.1 = 3.
Nine such excursions suffice to work out the product of two matrices of order 3.
Only three such excursions suffice to give the general formula
(9)
an
aSY
83
32
ass sJ



  • 31 Ca3ixi + a31x2 +a33xJ

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