A First Course in Linear Algebra, 2017a

3.2. Applications of the Determinant 137
Example 3.46: Use Cramer’s Rule for Non-Constant Matrix
Solve for z if
⎡
⎣
1 0 0
0 e t cost e t sint
0 −e t sint e t cost
⎤⎡
⎦⎣
x
y
z
⎤
⎡
⎦ = ⎣
⎤
1
t ⎦
t 2
Solution. We are asked to find the value of z in the solution. We will solve using Cramer’s rule. Thus
∣ 1 0 1 ∣∣∣∣∣
0 e t cost t
∣ 0 −e t sint t 2
z =
1 0 0
0 e t cost e t sint
∣ 0 −e t sint e t cost ∣
= t ((cost)t + sint)e −t ♠
3.2.3 Polynomial Interpolation
In studying a set of data that relates variables x and y, it may be the case that we can use a polynomial to
“fit” to the data. If such a polynomial can be established, it can be used to estimate values of x and y which
have not been provided.
Consider the following example.
Example 3.47: Polynomial Interpolation
Given data points (1,4),(2,9),(3,12), find an interpolating polynomial p(x) of degree at most 2 and
then estimate the value corresponding to x = 1 2 .
Solution. We want to find a polynomial given by
p(x)=r 0 + r 1 x 1 + r 2 x 2 2
such that p(1)=4, p(2)=9andp(3)=12. To find this polynomial, substitute the known values in for x
and solve for r 0 ,r 1 ,andr 2 .
Writing the augmented matrix, we have
⎡
p(1) = r 0 + r 1 + r 2 = 4
p(2) = r 0 + 2r 1 + 4r 2 = 9
p(3) = r 0 + 3r 1 + 9r 2 = 12
⎣
1 1 1 4
1 2 4 9
1 3 9 12
⎤
⎦