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Problems #11 from pg. 32 and additional problem (A) are graded.
#11, pg.32: Determine if →b is a linear combination of →a1,→a2, and →a3 where
→a1=[1−20],→a2=[012],→a3=[5−68],→b=[2−16].
Solution: To write →b as a linear combination of →a1,→a2, and →a3 means we must find the weights x1,x2,x3 (if they exist) that satisfy
x1→a1+x2→a2+x3→a3=→b.
When the vector algebra is simplified (carry out the vector sums) we get the equation
[x1+5x3−2x1+x2−6x32x2+8x3]=[2−16],
which is a system of linear equations. We will solve this system using the augmented matrix
[1052−21−6−10286]r∗2=r2+2r1∼[105201430286]r∗3=r3−2r2∼[105201430000]
If we interpret this augmented marix as a system, we get
{x1+5x3=2x2+4x3=30=0
or equivalently
{x1=2−5x3x2=3−4x30=0
We are free to choose any value of x3 so we will choose x3=0 and doing so yields the solution
{x1=2x2=3x3=0.
This implies that we should be able to write the vector →b as
2→a1+3→a2+0→a3=[2−40]+[036]=[2−16]=→b,
as was to be shown.
NOTE: Every different choice of x3 yields a different linear combination that yields →b. Here we call x3 a free variable.
Additional Problem (A): Solve the matrix equation A→x=→b where A=[1010−1100−1] and →b=[b1b2b3].
NOTE: Many people wrote the matrix in this problem as [1010−1000−1]. This may because I made a typo in class...I will not penalize correct answers that start with this matrix instead of the actual problem.
Solution: We know the solution of this equation to be equivalent to the solution of the system whose augmented matrix is
[101b10−11b200−1b3]r∗1=r1+r3r∗2=r2+r3∼[100b1+b30−10b2+b300−1b3]r∗2=−r2r∗3=−r3∼[100b1+b3010−b2−b3001−b3]
hence we see that the solution is given by
→x=[b1+b3−b2−b3−b3.]
