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Linear Algebra pre-requisites

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We say that  b1,,bn are linearly independent if c1b1+c2b2++cnbn=d1b1+d2b2++dnbn implies c1=d1,c2=d2,cn=dn or, equivalently,b1,,bn are linearly independent if c1b1+c2b2++cnbn=0 implies c1=c2==cn=0

Example 1: Lets look at the vectors b1=(1,0,0),b2=(0,1,0),b3=(0,0,1) . Observe that if ab1+bb2+cb3=(1,2,3), then a=1,b=2, and c=3 and there is no other values for a,b, and c to give the vector (1,2,3). Next let's for the same values of  b1,b2, and b3, the only values of a,b, and c such that ab1+bb2+cb3=(0,0,0) is a=0,b=0, and c=0. Therefore vectors b1=(1,0,0),b2=(0,1,0),b3=(0,0,1) are linearly independent.

Example 2: b1=1,b2=t,b3=t2.
1+2t+3t21+2t+4t2.

If a+bt+ct2=1+2t+3t2 then a=1,b=2,c=3.

Application: Partial Fractions
4(x2+1)(x3)=Ax+Bx2+1+Cx3

If you don't like denominators, get rid of them:
4=(Ax+B)(x3)+C(x2+1)4=Ax2+Bx3Ax3B+Cx2+C4=(A+C)x2+(B3A)x3B+C
I.e., 0x2+0x+4=(A+C)x2+(B3A)x3B+C

Thus 0=A+C,0=B3A,4=3B+C.
C=A,B=3A,4=3(3A)+A4=10A

Hence A=25,B=3(25)=65,C=25.

Thus,
4(x2+1)(x3)=25x65x2+1+25x3=2x65(x2+1)+25(x3)

Alternatively, can plug in x=3 to quickly find C and then solve for A and B. We can also use matrices to solve linear equations.


This page titled Linear Algebra pre-requisites is shared under a not declared license and was authored, remixed, and/or curated by Isabel K. Darcy.

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